Content check — Calculus Vol. 3, Chapter 4: Differentiation of Functions of Several Variables

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 4: Differentiation of Functions of Several Variables

248 problems · 26 figures

Checkpoint 4.1 §4.1 source ↗

Find the domain and range of the function

f (x, y) =

\sqrt{36 - 9 x^{2} - 9 y^{2}} .

Answer

The domain is the shaded circle defined by the inequality $9 x^{2} + 9 y^{2} \leq 36,$ which has a circle of radius $2$ as its boundary. The range is $[0, 6] .$

A circle of radius two with center at the origin. The equation x2 + y2 ≤ 4 is given.

Checkpoint 4.2 §4.1 source ↗

Find and graph the level curve of the function

g (x, y) =

x^{2} +

y^{2} -

6 x +

2 y

corresponding to

c = 15 .

Answer

The equation of the level curve can be written as ${(x - 3)}^{2} +$ ${(y + 1)}^{2} =$ $25,$ which is a circle with radius $5$ centered at $(3, −1) .$

An circle of radius 5 with center (3, –1).

Checkpoint 4.3 §4.1 source ↗

Determine the equation of the vertical trace of the function

g (x, y) =

− x^{2} -

y^{2} +

2 x +

4 y -

1

corresponding to

y = 3,

and describe its graph.

Answer

z = 3 - {(x - 1)}^{2} .

This function describes a parabola opening downward in the plane

y = 3 .

Checkpoint 4.4 §4.1 source ↗

Find the domain of the function

h (x, y, t) =

(3 t - 6) \sqrt{y - 4 x^{2} + 4} .

Answer

domain (h) =

{(x, y, t) \in ℝ^{3} | y \geq 4 x^{2} - 4}

Checkpoint 4.5 §4.1 source ↗

Find an equation of the level surface of the function

g (x, y, z) = x^{2} + y^{2} + z^{2} - 2 x + 4 y - 6 z

corresponding to $c = 2,$ and describe the surface, if possible.

Answer

{(x - 1)}^{2} +

{(y + 2)}^{2} +

{(z - 3)}^{2} =

16

describes a sphere of radius

4

centered at the point

(1, −2, 3) .

4-1 §4.1 source ↗

For the following exercises, evaluate each function at the indicated values.

$W (x, y) =$ $4 x^{2} +$ $y^{2} .$ Find $W (2, −1),$ $W (−3, 6) .$

Answer

17, 72

4-3 §4.1 source ↗

For the following exercises, evaluate each function at the indicated values.

The volume of a right circular cylinder is calculated by a function of two variables, $V (x, y) =$ $π x^{2} y,$ where $x$ is the radius of the right circular cylinder and $y$ represents the height of the cylinder. Evaluate $V (2, 5)$ and explain what this means.

Answer

20 π .

This is the volume when the radius is

2

and the height is

5 .

4-5 §4.1 source ↗

For the following exercises, find the domain of the function.

$V (x, y) =$ $4 x^{2} +$ $y^{2}$

Answer

All points in the

x y -plane

4-7 §4.1 source ↗

For the following exercises, find the domain of the function.

$f (x, y) =$ $4 ln (y^{2} - x)$

Answer

x < y^{2}

4-9 §4.1 source ↗

For the following exercises, find the domain of the function.

$z (x, y) =$ $y^{2} -$ $x^{2}$

Answer

All real ordered pairs in the

x y -plane

of the form

(a, b)

4-11 §4.1 source ↗

Find the range of the functions.

$g (x, y) =$ $\sqrt{16 - 4 x^{2} - y^{2}}$

Answer

{z | 0 \leq z \leq 4}

4-13 §4.1 source ↗

Find the range of the functions.

$z = y^{2} - x^{2}$

Answer

The set

ℝ

4-15 §4.1 source ↗

For the following exercises, find the level curves of each function at the indicated value of $c$ to visualize the given function.

$z (x, y) =$ $y^{2} -$ $x^{2},$ $c = 4$

Answer

y^{2} - x^{2} = 4,

a hyperbola

4-17 §4.1 source ↗

For the following exercises, find the level curves of each function at the indicated value of $c$ to visualize the given function.

$g (x, y) =$ $4 -$ $x -$ $y;$ $c =$ $0,$ $4$

Answer

4 = x + y,

a line;

x + y = 0,

line through the origin

4-19 §4.1 source ↗

For the following exercises, find the level curves of each function at the indicated value of $c$ to visualize the given function.

$h (x, y) =$ $2 x -$ $y;$ $c =$ $0,$ $−2,$ $2$

Answer

\begin{matrix} 2 x - y = 0, & 2 x - y = −2 \end{matrix}, 2 x - y = 2;

three lines

4-21 §4.1 source ↗

For the following exercises, find the level curves of each function at the indicated value of $c$ to visualize the given function.

$g (x, y) =$ $\frac{x}{x + y};$ $c =$ $−1,$ $0,$ $2$

Answer

\frac{x}{x + y} =

−1,

\frac{x}{x + y} =

0,

\frac{x}{x + y} =

2

4-23 §4.1 source ↗

For the following exercises, find the level curves of each function at the indicated value of $c$ to visualize the given function.

$g (x, y) =$ $e_{}^{x y};$ $c =$ $\frac{1}{2},$ $3$

Answer

\begin{matrix} e^{x y} = \frac{1}{2}, & e^{x y} = 3 \end{matrix}

4-25 §4.1 source ↗

For the following exercises, find the level curves of each function at the indicated value of $c$ to visualize the given function.

$f (x, y) =$ $x y -$ $x;$ $c =$ $−2,$ $0,$ $2$

Answer

\begin{matrix} x y - x = −2, & x y - x = 0, \end{matrix} x y - x = 2

4-27 §4.1 source ↗

For the following exercises, find the level curves of each function at the indicated value of $c$ to visualize the given function.

$g (x, y) =$ $ln (\frac{y}{x^{2}});$ $c =$ $−2,$ $0,$ $2$

Answer

\begin{matrix} e^{−2} x^{2} = y, & y = x^{2}, y = e^{2} x^{2} \end{matrix}

4-29 §4.1 source ↗

For the following exercises, find the level curves of each function at the indicated value of $c$ to visualize the given function.

$f (x, y) =$ $\frac{y + 2}{x^{2}},$ $c =$ any constant

Answer

The level curves are parabolas of the form

y = c x^{2} - 2 .

4-31 §4.1 source ↗

For the following exercises, find the vertical traces of the functions at the indicated values of $x$ and y, and plot the traces.

$f (x, y) =$ $3 x +$ $y^{3},$ $x =$ $1$

Answer

$z = 3 + y^{3},$ a curve in the $z y -plane$ with rulings parallel to the $x -axis$

A planar version of the function y3 + 3 with results in the z axis and nothing mattering from the x axis.

4-33 §4.1 source ↗

Find the domain of the following functions.

$z =$ $\sqrt{100 - 4 x^{2} - 25 y^{2}}$

Answer

\frac{x^{2}}{25} + \frac{y^{2}}{4} \leq 1

4-35 §4.1 source ↗

Find the domain of the following functions.

$f (x, y, z) =$ $\frac{1}{\sqrt{36 - 4 x^{2} - 9 y^{2} - z^{2}}}$

Answer

\frac{x^{2}}{9} +

\frac{y^{2}}{4} +

\frac{z^{2}}{36} <

1

4-37 §4.1 source ↗

Find the domain of the following functions.

$f (x, y, z) =$ $\sqrt[3]{16 - x^{2} - y^{2} - z^{2}}$

Answer

All points in

x y z -space

4-39 §4.1 source ↗

For the following exercises, plot a graph of the function.

$z =$ $f (x, y) =$ $\sqrt{x^{2} + y^{2}}$

Answer

An upward facing, gently increasing paraboloid.

4-41 §4.1 source ↗

For the following exercises, plot a graph of the function.

Use technology to graph $z = x^{2} y .$

Answer

A twisted plane with corners at (1, –1, –1), (–1, –1, –1), (–1, 1, 0.5), and (1, 1, 0.5).

4-43 §4.1 source ↗

Sketch the following by finding the level curves. Verify the graph using technology.

$f (x, y) =$ $2 -$ $\sqrt{x^{2} + y^{2}}$

Answer

A downward facing, gently decreasing paraboloid.

4-45 §4.1 source ↗

Sketch the following by finding the level curves. Verify the graph using technology.

$z = cos \sqrt{x^{2} + y^{2}}$

Answer

A hemisphere in the center with edges then swooping up at the four corners.

4-47 §4.1 source ↗

Sketch the following by finding the level curves. Verify the graph using technology.

Describe the contour lines for several values of $c$ for $z =$ $x^{2} +$ $y^{2} -$ $2 x -$ $2 y .$

Answer

The contour lines are circles.

4-49 §4.1 source ↗

Find the level surface for the functions of three variables and describe it.

$w (x, y, z) =$ $x^{2} +$ $y^{2} +$ $z^{2},$ $c =$ $9$

Answer

x^{2} + y^{2} + z^{2} = 9,

a sphere of radius

3

4-51 §4.1 source ↗

Find the level surface for the functions of three variables and describe it.

$w (x, y, z) =$ $x^{2} +$ $y^{2} -$ $z^{2},$ $c =$ $4$

Answer

x^{2} + y^{2} - z^{2} = 4,

a hyperboloid of one sheet

4-53 §4.1 source ↗

For the following exercises, find an equation of the level curve of $f$ that contains the point $P .$

$f (x, y) =$ $1 -$ $4 x^{2} -$ $y^{2},$ $P (0, 1)$

Answer

4 x^{2} + y^{2} = 1,

4-55 §4.1 source ↗

For the following exercises, find an equation of the level curve of $f$ that contains the point $P .$

$g (x, y) =$ $e^{x y} (x^{2} + y^{2}),$ $P (1, 0)$

Answer

1 =

e^{x y} (x^{2} + y^{2})

4-57 §4.1 source ↗

For the following exercises, find an equation of the level curve of $f$ that contains the point $P .$

A thin plate made of iron is located in the $x y -plane.$ The temperature $T$ in degrees Celsius at a point $P (x, y)$ is inversely proportional to the square of its distance from the origin. Express $T$ as a function of $x and y .$

Answer

T (x, y) =

\frac{k}{x^{2} + y^{2}}

4-59 §4.1 source ↗

For the following exercises, find an equation of the level curve of $f$ that contains the point $P .$

Refer to the preceding problem. Find the level curves for $T =$ $40 ° C and T =$ $100 ° C,$ and describe what the level curves represent.

Answer

x^{2} + y^{2} = \frac{k}{40},

x^{2} + y^{2} = \frac{k}{100} .

The level curves represent circles of radii

\sqrt{10 k} / 20

and

\sqrt{k} / 10

Checkpoint 4.6 §4.2 source ↗

Evaluate the following limit:

lim_{(x, y) \to (5, −2)} \sqrt[3]{\frac{x^{2} - y}{y^{2} + x - 1}} .

Answer

lim_{(x, y) \to (5, −2)} \sqrt[3]{\frac{x^{2} - y}{y^{2} + x - 1}} =

\frac{3}{2}

Checkpoint 4.7 §4.2 source ↗

Show that

lim_{(x, y) \to (2, 1)} \frac{(x - 2) (y - 1)}{{(x - 2)}^{2} + {(y - 1)}^{2}}

does not exist.

Answer

y = k (x - 2) + 1,

then

lim_{(x, y) \to (2, 1)} \frac{(x - 2) (y - 1)}{{(x - 2)}^{2} + {(y - 1)}^{2}} =

\frac{k}{1 + k^{2}} .

Since the answer depends on

k,

the limit fails to exist.

Checkpoint 4.8 §4.2 source ↗

Evaluate the following limit:

lim_{(x, y) \to (5, −2)} \sqrt{29 - x^{2} - y^{2}} .

Answer

lim_{(x, y) \to (5, −2)} \sqrt{29 - x^{2} - y^{2}} =

0

Checkpoint 4.9 §4.2 source ↗

Show that the function

f (x, y) =

\sqrt{26 - 2 x^{2} - y^{2}}

is continuous at point

(2, −3) .

Answer

1. The domain of $f$ contains the ordered pair $(2, −3)$ because $f (a, b) =$ $f (2, −3) =$ $\sqrt{16 - 2 {(2)}^{2} - {(−3)}^{2}} =$ $3$

2. $lim_{(x, y) \to (a, b)} f (x, y) =$ $3$

3. $lim_{(x, y) \to (a, b)} f (x, y) =$ $f (a, b) =$ $3$

Checkpoint 4.10 §4.2 source ↗

Show that the functions

f (x, y) =

2 x^{2} y^{3} +

3

and

g (x, y) =

{(2 x^{2} y^{3} + 3)}^{4}

are continuous everywhere.

Answer

The polynomials

g (x) = 2 x^{2}

and

h (y) = y^{3}

are continuous at every real number; therefore, by the product of continuous functions theorem,

f (x, y) =

2 x^{2} y^{3}

is continuous at every point

(x, y)

in the

x y -plane.

Furthermore, any constant function is continuous everywhere, so

g (x, y) = 3

is continuous at every point

(x, y)

in the

x y -plane.

Therefore,

f (x, y) =

2 x^{2} y^{3} +

3

is continuous at every point

(x, y)

in the

x y -plane.

Last,

h (x) = x^{4}

is continuous at every real number

x,

so by the continuity of composite functions theorem

g (x, y) =

{(2 x^{2} y^{3} + 3)}^{4}

is continuous at every point

(x, y)

in the

x y -plane.

Checkpoint 4.11 §4.2 source ↗

Find

lim_{(x, y, z) \to (4, −1, 3)} \sqrt{13 - x^{2} - 2 y^{2} + z^{2}} .

Answer

lim_{(x, y, z) \to (4, −1, 3)} \sqrt{13 - x^{2} - 2 y^{2} + z^{2}} =

2

4-61 §4.2 source ↗

For the following exercises, find the limit of the function.

$lim_{(x, y) \to (1, 2)} \frac{5 x^{2} y}{x^{2} + y^{2}}$

Answer

2.0

4-63 §4.2 source ↗

For the following exercises, evaluate the limits at the indicated values of $x and y .$ If the limit does not exist, state this and explain why the limit does not exist.

$lim_{(x, y) \to (0, 0)} \frac{4 x^{2} + 10 y^{2} + 4}{4 x^{2} - 10 y^{2} + 6}$

Answer

\frac{2}{3}

4-65 §4.2 source ↗

For the following exercises, evaluate the limits at the indicated values of $x and y .$ If the limit does not exist, state this and explain why the limit does not exist.

$lim_{(x, y) \to (0, 1)} \frac{y^{2} sin x}{x}$

Answer

1

4-67 §4.2 source ↗

For the following exercises, evaluate the limits at the indicated values of $x and y .$ If the limit does not exist, state this and explain why the limit does not exist.

$lim_{(x, y) \to (π / 4, 1)} \frac{y tan x}{y + 1}$

Answer

\frac{1}{2}

4-69 §4.2 source ↗

For the following exercises, evaluate the limits at the indicated values of $x and y .$ If the limit does not exist, state this and explain why the limit does not exist.

$lim_{(x, y) \to (2, 5)} (\frac{1}{x} - \frac{5}{y})$

Answer

- \frac{1}{2}

4-71 §4.2 source ↗

For the following exercises, evaluate the limits at the indicated values of $x and y .$ If the limit does not exist, state this and explain why the limit does not exist.

$lim_{(x, y) \to (4, 4)} e^{− x^{2} - y^{2}}$

Answer

e^{−32}

4-73 §4.2 source ↗

For the following exercises, evaluate the limits at the indicated values of $x and y .$ If the limit does not exist, state this and explain why the limit does not exist.

$lim_{(x, y) \to (1, 2)} (x^{2} y^{3} - x^{3} y^{2} + 3 x + 2 y)$

Answer

11.0

4-75 §4.2 source ↗

For the following exercises, evaluate the limits at the indicated values of $x and y .$ If the limit does not exist, state this and explain why the limit does not exist.

$lim_{(x, y) \to (0, 0)} \frac{x y + 1}{x^{2} + y^{2} + 1}$

Answer

1.0

4-77 §4.2 source ↗

For the following exercises, evaluate the limits at the indicated values of $x and y .$ If the limit does not exist, state this and explain why the limit does not exist.

$lim_{(x, y) \to (0, 0)} ln (x^{2} + y^{2})$

Answer

The limit does not exist because when

x

and

y

both approach zero, the function approaches

ln 0,

which is undefined (approaches negative infinity).

4-79 §4.2 source ↗

For the following exercises, complete the statement.

A point $(x_{0}, y_{0})$ in a plane region $R$ is called a boundary point of $R$ if ___________.

Answer

every open disk centered at

(x_{0}, y_{0})

contains points inside

R

and outside

R

4-81 §4.2 source ↗

For the following exercises, use algebraic techniques to evaluate the limit.

$lim_{(x, y) \to (0, 0)} \frac{x^{4} - 4 y^{4}}{x^{2} + 2 y^{2}}$

Answer

0.0

4-83 §4.2 source ↗

For the following exercises, use algebraic techniques to evaluate the limit.

$lim_{(x, y) \to (0, 0)} \frac{x^{2} - x y}{\sqrt{x} - \sqrt{y}}$

Answer

0.00

4-85 §4.2 source ↗

For the following exercises, evaluate the limits of the functions of three variables.

$lim_{(x, y, z) \to (0, 0, 0)} \frac{x^{2} - y^{2} - z^{2}}{x^{2} + y^{2} - z^{2}}$

Answer

The limit does not exist.

4-87 §4.2 source ↗

For the following exercises, evaluate the limit of the function by determining the value the function approaches along the indicated paths. If the limit does not exist, explain why not.

Evaluate $lim_{(x, y) \to (0, 0)} \frac{x y + y^{3}}{x^{2} + y^{2}}$ using the results of previous problem.

Answer

The limit does not exist. The function approaches two different values along different paths.

4-89 §4.2 source ↗

For the following exercises, evaluate the limit of the function by determining the value the function approaches along the indicated paths. If the limit does not exist, explain why not.

Evaluate $lim_{(x, y) \to (0, 0)} \frac{x^{2} y}{x^{4} + y^{2}}$ using the results of previous problem.

Answer

The limit does not exist because the function approaches two different values along the paths.

4-91 §4.2 source ↗

Discuss the continuity of the following functions. Find the largest region in the $x y -plane$ in which the following functions are continuous.

$f (x, y) =$ $ln (x + y)$

Answer

The function

f

is continuous in the region

y > − x .

4-93 §4.2 source ↗

Discuss the continuity of the following functions. Find the largest region in the $x y -plane$ in which the following functions are continuous.

$f (x, y) = \frac{1}{x y}$

Answer

The function

f

is continuous on the set

R =

x, y | x \neq 0, y \neq 0

4-95 §4.2 source ↗

For the following two exercises, determine the region in which the function is continuous. Explain your answer.

$f (x, y) =$ $\frac{sin (x^{2} + y^{2})}{x^{2} + y^{2}}$

Answer

The function is discontinuous at

(0, 0)

since

(0, 0)

is not in the domain.

4-97 §4.2 source ↗

For the following two exercises, determine the region in which the function is continuous. Explain your answer.

Determine whether $g (x, y) =$ $\frac{x^{2} - y^{2}}{x^{2} + y^{2}}$ is continuous at $(0, 0) .$

Answer

The function is discontinuous at

(0, 0) .

The limit at

(0, 0)

fails to exist and

g (0, 0)

does not exist.

4-99 §4.2 source ↗

For the following two exercises, determine the region in which the function is continuous. Explain your answer.

Determine the region of the $x y -plane$ in which the function $g (x, y) =$ $arctan (\frac{x y^{2}}{x + y})$ is continuous. Use technology to support your conclusion.

Answer

Since the function

arctan x

is continuous over

(− \infty, \infty),

g (x, y) =

arctan (\frac{x y^{2}}{x + y})

is continuous where

z = \frac{x y^{2}}{x + y}

is continuous. The inner function

z

is continuous on all points of the

x y -plane

except where

y = − x .

Thus,

g (x, y) =

arctan (\frac{x y^{2}}{x + y})

is continuous on all points of the coordinate plane except at points at which

y = − x .

4-101 §4.2 source ↗

For the following two exercises, determine the region in which the function is continuous. Explain your answer.

At what points in space is $g (x, y, z) =$ $x^{2} +$ $y^{2} -$ $2 z^{2}$ continuous?

Answer

All points

P (x, y, z)

in space

4-103 §4.2 source ↗

For the following two exercises, determine the region in which the function is continuous. Explain your answer.

Show that $lim_{(x, y) \to (0, 0)} \frac{1}{x^{2} + y^{2}}$ does not exist at $(0, 0)$ by plotting the graph of the function.

Answer

The graph increases without bound as $x and y$ both approach zero.

On xyz space, there is a shape drawn that decreases to 0 as x and y increase or decrease but that increases greatly closer to the origin. It increases to such an extent that the graph is cut off above z = 10, which coincides with a circle of radius 0.6 around (0, 0, 10).

4-105 §4.2 source ↗

For the following two exercises, determine the region in which the function is continuous. Explain your answer.

[T]

a. Use a CAS to draw a contour map of $z = \sqrt{9 - x^{2} - y^{2}} .$

b. What is the name of the geometric shape of the level curves?

c. Give the general equation of the level curves.

d. What is the maximum value of $z ?$

e. What is the domain of the function?

f. What is the range of the function?

Answer

A series of five concentric circles, with radii 3, 2.75, 2.5, 2.2, and 1.75. The areas between the circles are colored, with the darkest color between the circles of radii 3 and 2.75.

b. The level curves are circles centered at $(0, 0)$ with radius $9 - c .$ c. $x^{2} + y^{2} = 9 - c$ d. $z = 3$ e. ${(x, y) \in ℝ^{2} | x^{2} + y^{2} \leq 9}$ f. ${z | 0 \leq z \leq 3}$

4-107 §4.2 source ↗

For the following two exercises, determine the region in which the function is continuous. Explain your answer.

Use polar coordinates to find $lim_{(x, y) \to (0, 0)} \frac{sin \sqrt{x^{2} + y^{2}}}{\sqrt{x^{2} + y^{2}}} .$ You can also find the limit using L’Hôpital’s rule.

Answer

1.0

4-109 §4.2 source ↗

For the following two exercises, determine the region in which the function is continuous. Explain your answer.

Discuss the continuity of $f (g (x, y))$ where $f (t) = 1 / t$ and $g (x, y) =$ $2 x -$ $5 y .$

Answer

f (g (x, y))

is continuous at all points

(x, y)

that are not on the line

2 x - 5 y = 0 .

4-111 §4.2 source ↗

For the following two exercises, determine the region in which the function is continuous. Explain your answer.

Given $f (x, y) =$ $x^{2} -$ $4 y,$ find $lim_{h \to 0} \frac{f (1 + h, y) - f (1, y)}{h} .$

Answer

2.0

Checkpoint 4.12 §4.3 source ↗

Use the definition of the partial derivative as a limit to calculate $\partial f / \partial x$ and $\partial f / \partial y$ for the function

f (x, y) = 4 x^{2} + 2 x y - y^{2} + 3 x - 2 y + 5 .

Answer

\frac{\partial f}{\partial x} =

8 x +

2 y +

3,

\frac{\partial f}{\partial y} =

2 x -

2 y -

2

Checkpoint 4.13 §4.3 source ↗

Calculate

\partial f / \partial x

and

\partial f / \partial y

for the function

f (x, y) =

tan (x^{3} - 3 x^{2} y^{2} + 2 y^{4})

by holding the opposite variable constant, then differentiating.

Answer

\begin{matrix} \frac{\partial f}{\partial x} = (3 x^{2} - 6 x y^{2}) {sec}^{2} (x^{3} - 3 x^{2} y^{2} + 2 y^{4}) \\ \frac{\partial f}{\partial y} = (−6 x^{2} y + 8 y^{3}) {sec}^{2} (x^{3} - 3 x^{2} y^{2} + 2 y^{4}) \end{matrix}

Checkpoint 4.14 §4.3 source ↗

Use a contour map to estimate $\partial f / \partial y$ at point $(0, \sqrt{2})$ for the function

f (x, y) = x^{2} - y^{2} .

Compare this with the exact answer.

Answer

Using the curves corresponding to $c =$ $−2 and c =$ $−3,$ we obtain

${\frac{\partial f}{\partial y} |}_{(x, y) = (0, \sqrt{2})} \approx$ $\frac{f (0, \sqrt{3}) - f (0, \sqrt{2})}{\sqrt{3} - \sqrt{2}} =$ $\frac{−3 - (−2)}{\sqrt{3} - \sqrt{2}} \cdot \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} =$ $− \sqrt{3} -$ $\sqrt{2} \approx$ $−3.146 .$

The exact answer is

${\frac{\partial f}{\partial y} |}_{(x, y) = (0, \sqrt{2})} =$ $({−2 y |}_{(x, y) = (0, \sqrt{2})} = −2 \sqrt{2} \approx −2.828 .$

Checkpoint 4.15 §4.3 source ↗

Use the limit definition of partial derivatives to calculate $\partial f / \partial x$ for the function

f (x, y, z) = 2 x^{2} - 4 x^{2} y + 2 y^{2} + 5 x z^{2} - 6 x + 3 z - 8 .

Then find $\partial f / \partial y$ and $\partial f / \partial z$ by setting the other two variables constant and differentiating accordingly.

Answer

\frac{\partial f}{\partial x} =

4 x -

8 x y +

5 z^{2} -

6,

\frac{\partial f}{\partial y} =

−4 x^{2} +

4 y,

\frac{\partial f}{\partial z} =

10 x z +

3

Checkpoint 4.16 §4.3 source ↗

Calculate

\partial f / \partial x,

\partial f / \partial y,

and

\partial f / \partial z

for the function

f (x, y, z) =

sec (x^{2} y) -

tan (x^{3} y z^{2}) .

Answer

\begin{matrix} \frac{\partial f}{\partial x} = 2 x y sec (x^{2} y) tan (x^{2} y) - 3 x^{2} y z^{2} {sec}^{2} (x^{3} y z^{2}) \\ \frac{\partial f}{\partial y} = x^{2} sec (x^{2} y) tan (x^{2} y) - x^{3} z^{2} {sec}^{2} (x^{3} y z^{2}) \\ \frac{\partial f}{\partial z} = −2 x^{3} y z {sec}^{2} (x^{3} y z^{2}) \end{matrix}

Checkpoint 4.17 §4.3 source ↗

Calculate all four second partial derivatives for the function

f (x, y) = sin (3 x - 2 y) + cos (x + 4 y) .

Answer

\begin{matrix} \frac{\partial^{2} f}{\partial x^{2}} = −9 sin (3 x - 2 y) - cos (x + 4 y) \\ \frac{\partial^{2} f}{\partial x \partial y} = 6 sin (3 x - 2 y) - 4 cos (x + 4 y) \\ \frac{\partial^{2} f}{\partial y \partial x} = 6 sin (3 x - 2 y) - 4 cos (x + 4 y) \\ \frac{\partial^{2} f}{\partial y^{2}} = −4 sin (3 x - 2 y) - 16 cos (x + 4 y) \end{matrix}

4-113 §4.3 source ↗

For the following exercises, calculate the partial derivative using the limit definitions only.

$\frac{\partial z}{\partial y}$ for $z = x^{2} - 3 x y + y^{2}$

Answer

\frac{\partial z}{\partial y} = −3 x + 2 y

4-115 §4.3 source ↗

For the following exercises, calculate the sign of the partial derivative using the graph of the surface.

A partial paraboloid with vertex at the origin and pointing up.

$f_{x} (−1, 1)$

Answer

The sign is negative.

4-117 §4.3 source ↗

For the following exercises, calculate the sign of the partial derivative using the graph of the surface.

$f_{x} (0, 0)$

Answer

The partial derivative is zero at the origin.

4-119 §4.3 source ↗

For the following exercises, calculate the partial derivatives.

$\frac{\partial z}{\partial y}$ for $z =$ $sin (3 x) cos (3 y)$

Answer

\frac{\partial z}{\partial y} =

−3 sin (3 x) sin (3 y)

4-121 §4.3 source ↗

For the following exercises, calculate the partial derivatives.

$\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ for $z = ln (x^{6} + y^{4})$

Answer

\frac{\partial z}{\partial x} =

\frac{6 x^{5}}{x^{6} + y^{4}};

\frac{\partial z}{\partial y} =

\frac{4 y^{3}}{x^{6} + y^{4}}

4-123 §4.3 source ↗

For the following exercises, calculate the partial derivatives.

Let $z = e^{x y} .$ Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y} .$

Answer

\frac{\partial z}{\partial x} =

y e^{x y};

\frac{\partial z}{\partial y} =

x e^{x y}

4-125 §4.3 source ↗

For the following exercises, calculate the partial derivatives.

Let $z =$ $tan (2 x - y) .$ Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y} .$

Answer

\frac{\partial z}{\partial x} =

2 {sec}^{2} (2 x - y),

\frac{\partial z}{\partial y} =

− {sec}^{2} (2 x - y)

4-127 §4.3 source ↗

For the following exercises, calculate the partial derivatives.

Let $f (x, y) =$ $arctan (\frac{y}{x}) .$ Evaluate $f_{x} (2, −2)$ and $f_{y} (2, −2) .$

Answer

f_{x} (2, −2) =

\frac{1}{4} =

f_{y} (2, −2)

4-129 §4.3 source ↗

For the following exercises, calculate the partial derivatives.

Find $\frac{\partial z}{\partial x}$ at $(0, 1)$ for $z =$ $e^{− x} cos (y) .$ Evaluate the partial derivatives at point $P (0, 1) .$

Answer

\frac{\partial z}{\partial x} =

− cos (1)

4-131 §4.3 source ↗

For the following exercises, calculate the partial derivatives.

Given $f (x, y, z) =$ $2 sin (x + y),$ find $f_{x} (0, \frac{π}{2}, −4),$ $f_{y} (0, \frac{π}{2}, −4),$ and $f_{z} (0, \frac{π}{2}, −4) .$

Answer

\begin{matrix} f_{x} = 0, & f_{y} = 0, & f_{z} = 0 \end{matrix}

4-133 §4.3 source ↗

For the following exercises, calculate the partial derivatives.

Express the volume of a right circular cylinder as a function of two variables:

a. its radius $r$ and its height $h .$

b. Show that the rate of change of the volume of the cylinder with respect to its radius is the product of its circumference multiplied by its height.

c. Show that the rate of change of the volume of the cylinder with respect to its height is equal to the area of the circular base.

Answer

V (r, h) = π r^{2} h

\frac{\partial V}{\partial r} = 2 π r h

\frac{\partial V}{\partial h} = π r^{2}

4-135 §4.3 source ↗

Find the indicated higher-order partial derivatives.

$f_{x y}$ for $z = ln (x - y)$

Answer

f_{x y} = \frac{1}{{(x - y)}^{2}}

4-137 §4.3 source ↗

Find the indicated higher-order partial derivatives.

Let $z =$ $x^{2} +$ $3 x y +$ $2 y^{2} .$ Find $\frac{\partial^{2} z}{\partial x^{2}}$ and $\frac{\partial^{2} z}{\partial y^{2}} .$

Answer

\frac{\partial^{2} z}{\partial x^{2}} =

2,

\frac{\partial^{2} z}{\partial y^{2}} =

4

4-139 §4.3 source ↗

Find the indicated higher-order partial derivatives.

Given $f (x, y, z) =$ $x y z,$ find $f_{x y y}, f_{y x y},$ and $f_{y y x} .$

Answer

f_{x y y} =

f_{y x y} =

f_{y y x} =

0

4-141 §4.3 source ↗

Find the indicated higher-order partial derivatives.

Show that $z =$ $\frac{1}{2} (e^{y} - e^{− y}) sin x$ is a solution of the differential equation $\frac{\partial^{2} z}{\partial x^{2}} +$ $\frac{\partial^{2} z}{\partial y^{2}} =$ $0 .$

Answer

\begin{matrix} \frac{d^{2} z}{d x^{2}} & = & - \frac{1}{2} (e^{y} - e^{− y}) sin x \\ \frac{d^{2} z}{d y^{2}} & = & \frac{1}{2} (e^{y} - e^{− y}) sin x \\ \frac{d^{2} z}{d x^{2}} + \frac{d^{2} z}{d y^{2}} & = & 0 \end{matrix}

4-143 §4.3 source ↗

Find the indicated higher-order partial derivatives.

Let $f (x, y, z) =$ $x^{2} y^{3} z -$ $3 x y^{2} z^{3} +$ $5 x^{2} z -$ $y^{3} z .$ Find $f_{x y z} .$

Answer

f_{x y z} =

6 y^{2} x -

18 y z^{2}

4-145 §4.3 source ↗

Find the indicated higher-order partial derivatives.

Given $f (x, y) =$ $x^{2} +$ $x -$ $3 x y +$ $y^{3} -$ $5,$ find all points at which $f_{x} = f_{y} = 0$ simultaneously.

Answer

(\frac{1}{4}, \frac{1}{2}),

(1, 1)

4-147 §4.3 source ↗

Find the indicated higher-order partial derivatives.

Given $f (x, y) =$ $y^{3} -$ $3 y x^{2} -$ $3 y^{2} -$ $3 x^{2} +$ $1,$ find all points on $f$ at which $f_{x} = f_{y} = 0$ simultaneously.

Answer

(0, 0),

(0, 2),

(\sqrt{3}, −1),

(− \sqrt{3}, −1)

4-149 §4.3 source ↗

Find the indicated higher-order partial derivatives.

Show that $z = e^{x} sin y$ satisfies the equation $\frac{\partial^{2} z}{\partial x^{2}} +$ $\frac{\partial^{2} z}{\partial y^{2}} =$ $0 .$

Answer

\frac{\partial^{2} z}{\partial x^{2}} +

\frac{\partial^{2} z}{\partial y^{2}} =

e^{x} sin (y) -

e^{x} sin y =

0

4-151 §4.3 source ↗

Find the indicated higher-order partial derivatives.

Show that $z =$ $e^{− t} cos (\frac{x}{c})$ satisfies the heat equation $\frac{\partial z}{\partial t} =$ $c^{2} \frac{\partial^{2} z}{\partial x^{2}}$

Answer

c^{2} \frac{\partial^{2} z}{\partial x^{2}} =

e^{− t} cos (\frac{x}{c})

4-153 §4.3 source ↗

Find the indicated higher-order partial derivatives.

Find $lim_{Δ y \to 0} \frac{f (x, y + Δ y) - f (x, y)}{Δ y}$ for $f (x, y) =$ $−7 x -$ $2 x y +$ $7 y .$

Answer

\frac{\partial f}{\partial y} = −2 x + 7

4-155 §4.3 source ↗

Find the indicated higher-order partial derivatives.

Find $lim_{Δ x \to 0} \frac{Δ f}{Δ x} =$ $lim_{Δ x \to 0} \frac{f (x + Δ x, y) - f (x, y)}{Δ x}$ for $f (x, y) =$ $sin (x y) .$

Answer

\frac{\partial f}{\partial x} = y cos x y

4-159 §4.3 source ↗

Find the indicated higher-order partial derivatives.

The law of cosines can be thought of as a function of three variables. Let $x, y,$ and $θ$ be two sides of any triangle where the angle $θ$ is the included angle between the two sides. Then, $F (x, y, θ) =$ $x^{2} +$ $y^{2} -$ $2 x y cos θ$ gives the square of the third side of the triangle. Find $\frac{\partial F}{\partial θ}$ and $\frac{\partial F}{\partial x}$ when $x = 2, y = 3,$ and $θ = \frac{π}{6} .$

Answer

\frac{\partial F}{\partial θ} =

6,

\frac{\partial F}{\partial x} =

4 -

3 \sqrt{3}

4-161 §4.3 source ↗

Find the indicated higher-order partial derivatives.

A Cobb-Douglas production function is $f (x, y) =$ $200 x^{0.7} y^{0.3},$ where $x and y$ represent the amount of labor and capital available. Let $x = 500$ and $y = 1000 .$ Find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ at these values, which represent the marginal productivity of labor and capital, respectively.

Answer

\frac{\partial f}{\partial x}

(500, 1000) =

172.36,

\frac{\partial f}{\partial y}

(500, 1000) =

36.93

Checkpoint 4.19 §4.4 source ↗

Find an equation of the tangent plane to the surface defined by the function

f (x, y) =

x^{3} -

x^{2} y +

y^{2} -

2 x +

3 y -

2

at point

(−1, 3) .

Answer

z = 7 x + 8 y - 3

Checkpoint 4.20 §4.4 source ↗

Given the function

f (x, y) =

e^{5 - 2 x + 3 y},

approximate

f (4.1, 0.9)

using point

(4, 1)

for

(x_{0}, y_{0}) .

What is the approximate value of

f (4.1, 0.9)

to four decimal places?

Answer

L (x, y) =

6 -

2 x +

3 y,

L (4.1, 0.9) =

6 -

2 (4.1) +

3 (0.9) =

0.5

f (4.1, 0.9) =

e^{5 - 2 (4.1) + 3 (0.9)} =

e^{−0.5} \approx

0.6065 .

Checkpoint 4.21 §4.4 source ↗

Show that the function

f (x, y) =

3 x -

4 y^{2}

is differentiable at point

(−1, 2) .

Answer

$f (−1, 2) =$ $−19,$ $f_{x} (−1, 2) =$ $3,$ $f_{y} (−1, 2) =$ $−16,$ $E (x, y) =$ $−4 {(y - 2)}^{2} .$

$\begin{matrix} lim_{(x, y) \to (x_{0}, y_{0})} \frac{E (x, y)}{\sqrt{{(x - x_{0})}^{2} + {(y - y_{0})}^{2}}} & = lim_{(x, y) \to (−1, 2)} \frac{−4 {(y - 2)}^{2}}{\sqrt{{(x + 1)}^{2} + {(y - 2)}^{2}}} \\ \leq lim_{(x, y) \to (−1, 2)} \frac{−4 ({(x + 1)}^{2} + {(y - 2)}^{2})}{\sqrt{{(x + 1)}^{2} + {(y - 2)}^{2}}} \\ = lim_{(x, y) \to (–1, 2)} - 4 \sqrt{{(x + 1)}^{2} + {(y - 2)}^{2}} \\ = 0. \end{matrix}$

Checkpoint 4.22 §4.4 source ↗

Find the differential

d z

of the function

f (x, y) =

4 y^{2} +

x^{2} y -

2 x y

and use it to approximate

Δ z

at point

(1, −1) .

Use

Δ x = 0.03

and

Δ y = −0.02 .

What is the exact value of

Δ z ?

Answer

\begin{matrix} d z & = & 0.18 \\ Δ z & = & f (1.03, −1.02) - f (1, −1) = 0.180682 \end{matrix}

4-163 §4.4 source ↗

For the following exercises, find a unit normal vector to the surface at the indicated point.

$z =$ $f (x, y) =$ $x^{3},$ $(2, −1, 8)$

Answer

(\frac{\sqrt{145}}{145}) (12 i - k)

4-165 §4.4 source ↗

For the following exercises, as a useful review for techniques used in this section, find a normal vector and a tangent vector at point $P .$

$x^{2} +$ $x y +$ $y^{2} =$ $3,$ $P (−1, −1)$

Answer

Normal vector:

i + j,

tangent vector:

i - j

4-167 §4.4 source ↗

For the following exercises, as a useful review for techniques used in this section, find a normal vector and a tangent vector at point $P .$

$x y^{2} -$ $2 x^{2} +$ $y +$ $5 x =$ $6,$ $P (4, 2)$

Answer

Normal vector:

7 i - 17 j,

tangent vector:

17 i + 7 j

4-169 §4.4 source ↗

For the following exercises, as a useful review for techniques used in this section, find a normal vector and a tangent vector at point $P .$

$z e^{x^{2} - y^{2}} -$ $3 =$ $0,$ $P (2, 2, 3)$

Answer

\begin{matrix} Normal vector & - 12 i + 12 j - k \\ Tangent vector & 0 i + j + 12 k or 0 i + j - 12 k \end{matrix}

4-171 §4.4 source ↗

For the following exercises, find the equation for the tangent plane to the surface at the indicated point. (Hint: Solve for $z$ in terms of $x$ and $y .)$

$z =$ $−9 x^{2} -$ $3 y^{2},$ $P (2, 1, −39)$

Answer

−36 x -

6 y -

z =

−39

4-173 §4.4 source ↗

For the following exercises, find the equation for the tangent plane to the surface at the indicated point. (Hint: Solve for $z$ in terms of $x$ and $y .)$

$z =$ $ln (10 x^{2} + 2 y^{2} + 1),$ $P (0, 0, 0)$

Answer

z = 0

4-175 §4.4 source ↗

For the following exercises, find the equation for the tangent plane to the surface at the indicated point. (Hint: Solve for $z$ in terms of $x$ and $y .)$

$x y +$ $y z +$ $z x =$ $11,$ $P (1, 2, 3)$

Answer

5 x +

4 y +

3 z -

22 =

0

4-177 §4.4 source ↗

For the following exercises, find the equation for the tangent plane to the surface at the indicated point. (Hint: Solve for $z$ in terms of $x$ and $y .)$

$x^{3} +$ $y^{3} =$ $3 x y z,$ $P (1, 2, \frac{3}{2})$

Answer

4 x - 5 y + 4 z = 0

4-179 §4.4 source ↗

For the following exercises, find the equation for the tangent plane to the surface at the indicated point. (Hint: Solve for $z$ in terms of $x$ and $y .)$

$z =$ $sin x +$ $sin y +$ $sin (x + y),$ $P (0, 0, 0)$

Answer

2 x + 2 y - z = 0

4-181 §4.4 source ↗

For the following exercises, find the equation for the tangent plane to the surface at the indicated point. (Hint: Solve for $z$ in terms of $x$ and $y .)$

$z =$ $x^{2} -$ $2 x y +$ $y^{2},$ $P (1, 2, 1)$

Answer

−2 (x - 1) +

2 (y - 2) -

(z - 1) =

0

4-183 §4.4 source ↗

For the following exercises, find parametric equations for the normal line to the surface at the indicated point. (Recall that to find the equation of a line in space, you need a point on the line, $P_{0} (x_{0,} y_{0}, z_{0}),$ and a vector $n = 〈 a, b, c 〉$ that is parallel to the line. Then the equation of the line is $x -$ $x_{0} =$ $a t,$ $y -$ $y_{0} =$ $b t,$ $z -$ $z_{0} =$ $c t .)$

$z =$ $5 x^{2} -$ $2 y^{2},$ $P (2, 1, 18)$

Answer

x =

20 t +

2,

y =

−4 t +

1,

z =

− t +

18

4-185 §4.4 source ↗

$z =$ $ln (3 x^{2} + 7 y^{2} + 1),$ $P (0, 0, 0)$

Answer

x = 0, y = 0, z = t

4-187 §4.4 source ↗

$z = x^{2} - 2 x y + y^{2}$ at point $P (1, 2, 1)$

Answer

x -

1 =

2 t;

y -

2 =

−2 t;

z -

1 =

t

4-189 §4.4 source ↗

For the following exercises, use the figure shown here.

A surface in the xyz plane is marked as z = f(x, y). This surface has a tangent plane at (x0, y0, z0), with the corresponding point (x0, y0) marked on the xy plane. Also marked on the xy plane is the point (x0 + Δx, y0 + Δy). From this point, a line is drawn to the surface and three points are marked. The first point is C, which is (x0 + Δx, y0 + Δy, z0), then there is B, which is on the tangent plane, and then there is A, which is on the surface. The distance between B and C is marked df(x0, y0).

The length of line segment $B C$ is equal to what mathematical expression?

Answer

The differential of the function

z (x, y) =

d z =

f_{x} d x +

f_{y} d y

4-191 §4.4 source ↗

For the following exercises, complete each task.

Show that $f (x, y) = e^{x y} x$ is differentiable at point $(1, 0) .$

Answer

Using the definition of differentiability, we have

e^{x y} x \approx x + y .

4-193 §4.4 source ↗

For the following exercises, complete each task.

Show that $f (x, y) =$ $x^{2} +$ $3 y$ is differentiable at every point. In other words, show that $Δ z =$ $f (x + Δ x, y + Δ y) -$ $f (x, y) =$ $f_{x} Δ x +$ $f_{y} Δ y +$ $ε_{1} Δ x +$ $ε_{2} Δ y,$ where both $ε_{1}$ and $ε_{2}$ approach zero as $(Δ x, Δ y)$ approaches $(0, 0) .$

Answer

Δ z =

2 x Δ x +

3 Δ y +

{(Δ x)}^{2} .

{(Δ x)}^{2} \to 0

for small

Δ x

and

z

satisfies the definition of differentiability.

4-195 §4.4 source ↗

For the following exercises, complete each task.

Let $z =$ $f (x, y) =$ $x e^{y} .$ Compute $Δ z$ from $P (1, 2)$ to $Q (1.05, 2.1)$ and then find the approximate change in $z$ from point $P$ to point $Q .$ Recall $Δ z =$ $f (x + Δ x, y + Δ y) -$ $f (x, y),$ and $d z$ and $Δ z$ are approximately equal.

Answer

Δ z \approx 1.185422

and

d z \approx 1.108 .

They are relatively close.

4-197 §4.4 source ↗

For the following exercises, complete each task.

See the preceding problem. Use differentials to estimate the volume of aluminum in an enclosed aluminum can with diameter $8.0 cm$ and height $12 cm$ if the aluminum is $0.04$ cm thick.

Answer

16

cm³

4-199 §4.4 source ↗

For the following exercises, complete each task.

Let $z =$ $f (x, y) =$ $x^{2} +$ $3 x y -$ $y^{2} .$ Find the exact change in the function and the approximate change in the function as $x$ changes from $2.00 to 2.05$ and $y$ changes from $3.00 to 2.96 .$

Answer

Δ z =

exact change

= 0.6449,

approximate change is

d z = 0.65 .

The two values are close.

4-201 §4.4 source ↗

For the following exercises, complete each task.

The radius $r$ and height $h$ of a right circular cylinder are measured with possible errors of $4 % and 5 %,$ respectively. Approximate the maximum possible percentage error in measuring the volume (Recall that the percentage error is the ratio of the amount of error over the original amount. So, in this case, the percentage error in $V$ is given by $\frac{d V}{V} .)$

Answer

13 % or 0.13

4-203 §4.4 source ↗

For the following exercises, complete each task.

The electrical resistance $R$ produced by wiring resistors $R_{1}$ and $R_{2}$ in parallel can be calculated from the formula $\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} .$ If $R_{1}$ and $R_{2}$ are measured to be $7 Ω$ and $6 Ω,$ respectively, and if these measurements are accurate to within $0.05 Ω,$ estimate the maximum possible error in computing $R .$ (The symbol $Ω$ represents an ohm, the unit of electrical resistance.)

Answer

0.025

4-205 §4.4 source ↗

For the following exercises, complete each task.

The period $T$ of a simple pendulum with small oscillations is calculated from the formula $T = 2 π \sqrt{\frac{L}{g}},$ where $L$ is the length of the pendulum and $g$ is the acceleration resulting from gravity. Suppose that $L$ and $g$ have errors of, at most, $0.5 %$ and $0.1 %,$ respectively. Use differentials to approximate the maximum percentage error in the calculated value of $T .$

Answer

0.3 %

4-207 §4.4 source ↗

For the following exercises, find the linear approximation of each function at the indicated point.

$f (x, y) =$ $x \sqrt{y},$ $P (1, 4)$

Answer

2 x + \frac{1}{4} y - 1

4-209 §4.4 source ↗

For the following exercises, find the linear approximation of each function at the indicated point.

$f (x, y) =$ $arctan (x + 2 y),$ $P (1, 0)$

Answer

\frac{1}{2} x +

y +

\frac{1}{4} π -

\frac{1}{2}

4-211 §4.4 source ↗

For the following exercises, find the linear approximation of each function at the indicated point.

$f (x, y, z) =$ $\sqrt{x^{2} + y^{2} + z^{2}},$ $P (3, 2, 6)$

Answer

\frac{3}{7} x + \frac{2}{7} y + \frac{6}{7} z

4-213 §4.4 source ↗

For the following exercises, find the linear approximation of each function at the indicated point.

[T] Find the equation for the tangent plane to the surface at the indicated point, and graph the surface and the tangent plane: $z =$ $ln (10 x^{2} + 2 y^{2} + 1),$ $P (0, 0, 0) .$

Answer

$z = 0$

A curved surface is shown with tangent plane at (0, 0, 0). The curved surface looks like the middle part of the bottom of a boat, and the tangent plane is z = 0.

Checkpoint 4.23 §4.5 source ↗

Calculate $d z / d t$ given the following functions. Express the final answer in terms of $t .$

z = f (x, y) = x^{2} - 3 x y + 2 y^{2}, x = x (t) = 3 sin 2 t, y = y (t) = 4 cos 2 t

Answer

\begin{matrix} \frac{d z}{d t} & = \frac{\partial f}{\partial x} \frac{d x}{d t} + \frac{\partial f}{\partial y} \frac{d y}{d t} \\ = (2 x - 3 y) (6 cos 2 t) + (−3 x + 4 y) (−8 sin 2 t) \\ = −92 sin 2 t cos 2 t - 72 ({cos}^{2} 2 t - {sin}^{2} 2 t) \\ = −46 sin 4 t - 72 cos 4 t . \end{matrix}

Checkpoint 4.24 §4.5 source ↗

Calculate $\partial z / \partial u$ and $\partial z / \partial v$ given the following functions:

z = f (x, y) = \frac{2 x - y}{x + 3 y}, x (u, v) = e^{2 u} cos 3 v, y (u, v) = e^{2 u} sin 3 v .

Answer

\frac{\partial z}{\partial u} =

0,

\frac{\partial z}{\partial v} =

\frac{−21}{{(3 sin 3 v + cos 3 v)}^{2}}

Checkpoint 4.25 §4.5 source ↗

Calculate $\partial w / \partial u$ and $\partial w / \partial v$ given the following functions:

\begin{matrix} w & = & f (x, y, z) = \frac{x + 2 y - 4 z}{2 x - y + 3 z} \\ x & = & x (u, v) = e^{2 u} cos 3 v \\ y & = & y (u, v) = e^{2 u} sin 3 v \\ z & = & z (u, v) = e^{2 u} . \end{matrix}

Answer

\begin{matrix} \frac{\partial w}{\partial u} = 0 \\ \frac{\partial w}{\partial v} = \frac{15 - 33 sin 3 v + 6 cos 3 v}{{(3 + 2 cos 3 v - sin 3 v)}^{2}} \end{matrix}

Checkpoint 4.26 §4.5 source ↗

Create a tree diagram for the case when

w = f (x, y), x = x (t, u, v), y = y (t, u, v)

and write out the formulas for the three partial derivatives of $w .$

Answer

$\begin{matrix} \frac{\partial w}{\partial t} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial t} \\ \frac{\partial w}{\partial u} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial u} \\ \frac{\partial w}{\partial v} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial v} \end{matrix}$

A diagram that starts with w = f(x, y). Along the first branch, it is written ∂w/∂x, then x = x(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then ∂w/∂x ∂x/∂t; the second subbranch says u and then ∂w/∂x ∂x/∂u; and the third subbranch says v and then ∂w/∂x ∂x/∂v. Along the second branch, it is written ∂w/∂y, then y = y(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then ∂w/∂y ∂y/∂t; the second subbranch says u and then ∂w/∂y ∂y/∂u; and the third subbranch says v and then ∂w/∂y ∂y/∂v.

Checkpoint 4.27 §4.5 source ↗

Find

d y / d x

y

is defined implicitly as a function of

x

by the equation

x^{2} +

x y -

y^{2} +

7 x -

3 y -

26 =

0 .

What is the equation of the tangent line to the graph of this curve at point

(3, −2) ?

Answer

\frac{d y}{d x} =

{\frac{2 x + y + 7}{2 y - x + 3} |}_{(3, −2)} =

\frac{2 (3) + (−2) + 7}{2 (−2) - (3) + 3} =

-

\frac{11}{4}

Equation of the tangent line:

y = - \frac{11}{4} x + \frac{25}{4}

4-215 §4.5 source ↗

For the following exercises, use the information provided to solve the problem.

Let $w (x, y, z) =$ $x y cos z,$ where $x = t, y = t^{2},$ and $z = arcsin t .$ Find $\frac{d w}{d t} .$

Answer

\frac{d w}{d t} =

\frac{3 t^{2} - 4 t^{2}}{\sqrt{1 - t^{2}}}

4-217 §4.5 source ↗

For the following exercises, use the information provided to solve the problem.

If $w =$ $5 x^{2} +$ $2 y^{2},$ $x =$ $−3 s +$ $t,$ and $y = s - 4 t,$ find $\frac{\partial w}{\partial s}$ and $\frac{\partial w}{\partial t} .$

Answer

\frac{\partial w}{\partial s} =

94 s - 46 t,

\frac{\partial w}{\partial t} =

–46 s +

74 t

4-219 §4.5 source ↗

For the following exercises, use the information provided to solve the problem.

If $f (x, y) =$ $x y,$ $x =$ $r cos θ,$ and $y = r sin θ,$ find $\frac{\partial f}{\partial r}$ and express the answer in terms of $r$ and $θ .$

Answer

\frac{\partial f}{\partial r} =

r sin (2 θ)

4-221 §4.5 source ↗

For the following exercises, find $\frac{d f}{d t}$ using the chain rule and direct substitution.

$f (x, y) =$ $x^{2} +$ $y^{2},$ $x = t, y = t^{2}$

Answer

\frac{d f}{d t} = 2 t + 4 t^{3}

4-223 §4.5 source ↗

For the following exercises, find $\frac{d f}{d t}$ using the chain rule and direct substitution.

$f (x, y) =$ $x y,$ $x =$ $1 -$ $\sqrt{t},$ $y =$ $1 +$ $\sqrt{t}$

Answer

\frac{d f}{d t} = −1

4-225 §4.5 source ↗

For the following exercises, find $\frac{d f}{d t}$ using the chain rule and direct substitution.

$f (x, y) =$ $ln (x + y),$ $x = e^{t}, y = e^{t}$

Answer

\frac{d f}{d t} = 1

4-227 §4.5 source ↗

For the following exercises, find $\frac{d f}{d t}$ using the chain rule and direct substitution.

Let $w (x, y, z) =$ $x^{2} +$ $y^{2} +$ $z^{2},$ $x =$ $cos t,$ $y =$ $sin t,$ and $z = e^{t} .$ Express $w$ as a function of $t$ and find $\frac{d w}{d t}$ directly. Then, find $\frac{d w}{d t}$ using the chain rule.

Answer

\frac{d w}{d t} = 2 e^{2 t}

in both cases

4-229 §4.5 source ↗

For the following exercises, find $\frac{d f}{d t}$ using the chain rule and direct substitution.

Let $u = e^{x} sin y,$ where $x = -ln 2 t$ and $y = π t .$ Find $\frac{d u}{d t}$ when $x = ln 2$ and $y = \frac{π}{4} .$

Answer

\frac{d u}{d t} = \sqrt{2} (π -4)

4-231 §4.5 source ↗

For the following exercises, find $\frac{d y}{d x}$ using partial derivatives.

$x^{3} + y^{2} x - 3 = 0$

Answer

\frac{d y}{d x} =

-

\frac{3 x^{2} + y^{2}}{2 x y}

4-233 §4.5 source ↗

For the following exercises, find $\frac{d y}{d x}$ using partial derivatives.

$x^{2} - 2 x y + y^{4} = 4$

Answer

\frac{d y}{d x} =

\frac{y - x}{− x + 2 y^{3}}

4-235 §4.5 source ↗

For the following exercises, find $\frac{d y}{d x}$ using partial derivatives.

$x^{2 / 3} +$ $y^{2 / 3} =$ $a^{2 / 3}$

Answer

\frac{d y}{d x} = − \sqrt[3]{\frac{y}{x}}

4-237 §4.5 source ↗

For the following exercises, find $\frac{d y}{d x}$ using partial derivatives.

$e^{x y} + y e^{y} = 1$

Answer

\frac{d y}{d x} =

-

\frac{y e^{x y}}{x e^{x y} + e^{y} (1 + y)}

4-239 §4.5 source ↗

For the following exercises, find $\frac{d y}{d x}$ using partial derivatives.

Find $\frac{d z}{d t}$ using the chain rule where $z =$ $3 x^{2} y^{3},$ $x =$ $t^{4},$ and $y = t^{2} .$

Answer

\frac{d z}{d t} = 42 t^{13}

4-241 §4.5 source ↗

For the following exercises, find $\frac{d y}{d x}$ using partial derivatives.

Let $z =$ $e^{1 - x y},$ $x =$ $t^{1 / 3},$ and $y = t^{3} .$ Find $\frac{d z}{d t} .$

Answer

\frac{d z}{d t} =

-

\frac{10}{3} t^{7 / 3} \times e^{1 - t^{10 / 3}}

4-243 §4.5 source ↗

For the following exercises, find $\frac{d y}{d x}$ using partial derivatives.

Let $z =$ $\frac{x}{y},$ $x =$ $2 cos u,$ and $y = 3 sin v .$ Find $\frac{\partial z}{\partial u}$ and $\frac{\partial z}{\partial v} .$

Answer

\frac{\partial z}{\partial u} =

\frac{−2 sin u}{3 sin v}

and

\frac{\partial z}{\partial v} =

\frac{−2 cos u cos v}{3 {sin}^{2} v}

4-245 §4.5 source ↗

For the following exercises, find $\frac{d y}{d x}$ using partial derivatives.

If $z = x y e^{x / y},$ $x = r cos θ,$ and $y = r sin θ,$ find $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial θ}$ when $r = 2$ and $θ = \frac{π}{6} .$

Answer

\frac{\partial z}{\partial r} = \sqrt{3} e^{\sqrt{3}},

\frac{\partial z}{\partial θ} =

(2 - 4 \sqrt{3}) e^{\sqrt{3}}

4-247 §4.5 source ↗

For the following exercises, find $\frac{d y}{d x}$ using partial derivatives.

If $w =$ $sin (x y z),$ $x =$ $1 -$ $3 t,$ $y =$ $e^{1 - t},$ and $z = 4 t,$ find $\frac{\partial w}{\partial t} .$

Answer

\frac{\partial w}{\partial t} =

- 12 t e^{1 - t} \cos (1 - 3 t) e^{1 - t} (- 4 t) + 4 t (1 - 3 t) e^{1 - t} \cos (1 - 3 t) e^{1 - t} (- 4 t) - 4 (1 - 3 t) e^{1 - t} \cos (1 - 3 t) e^{1 - t} (- 4 t)

4-249 §4.5 source ↗

For the following exercises, use this information: A function $f (x, y)$ is said to be homogeneous of degree $n$ if $f (t x, t y) =$ $t^{n} f (x, y) .$ For all homogeneous functions of degree $n,$ the following equation is true: $x \frac{\partial f}{\partial x} +$ $y \frac{\partial f}{\partial y} =$ $n f (x, y) .$ Show that the given function is homogeneous and verify that $x \frac{\partial f}{\partial x} +$ $y \frac{\partial f}{\partial y} =$ $n f (x, y) .$

$f (x, y) =$ $\sqrt{x^{2} + y^{2}}$

Answer

f (t x, t y) =

\sqrt{t^{2} x^{2} + t^{2} y^{2}} =

t^{1} f (x, y),

\frac{\partial f}{\partial y} =

x \frac{1}{2} {(x^{2} + y^{2})}^{− 1 / 2} \times 2 x +

y \frac{1}{2} {(x^{2} + y^{2})}^{− 1 / 2} \times 2 y =

1 f (x, y)

4-251 §4.5 source ↗

The volume of a right circular cylinder is given by $V (x, y) =$ $π x^{2} y,$ where $x$ is the radius of the cylinder and y is the cylinder height. Suppose $x$ and $y$ are functions of $t$ given by $x = \frac{1}{2} t$ and $y = \frac{1}{3} t$ so that $x and y$ are both increasing with time. How fast is the volume increasing when $x = 2$ and $y = \frac{4}{3} ?$

Answer

V' = 4 π

4-253 §4.5 source ↗

The radius of a right circular cone is increasing at $3$ cm/min whereas the height of the cone is decreasing at $2$ cm/min. Find the rate of change of the volume of the cone when the radius is $13$ cm and the height is $18$ cm.

Answer

\frac{d V}{d t} =

\frac{1066 π}{3} {cm}^{3} / min

4-255 §4.5 source ↗

A closed box is in the shape of a rectangular solid with dimensions $x, y, and z .$ (Dimensions are in inches.) Suppose each dimension is changing at the rate of $0.5$ in./min. Find the rate of change of the total surface area of the box when $x =$ $2 in ., y =$ $3 in., and z =$ $1 in .$

Answer

\frac{d A}{d t} =

12 in .^{2} / min

4-257 §4.5 source ↗

The temperature $T$ at a point $(x, y)$ is $T (x, y)$ and is measured using the Celsius scale. A fly crawls so that its position after $t$ seconds is given by $x = \sqrt{1 + t}$ and $y = 2 + \frac{1}{3} t,$ where $x and y$ are measured in centimeters. The temperature function satisfies $T_{x} (2, 3) = 4$ and $T_{y} (2, 3) = 3 .$ How fast is the temperature increasing on the fly’s path after $3$ sec?

Answer

2 ° C/sec

4-259 §4.5 source ↗

Let $u = u (x, y, z),$ where $x =$ $x (w, t),$ $y =$ $y (w, t),$ $z =$ $z (w, t),$ $w =$ $w (r, s),$ $and t =$ $t (r, s) .$ Use a tree diagram and the chain rule to find an expression for $\frac{\partial u}{\partial r} .$

Answer

\begin{matrix} \frac{\partial u}{\partial r} & = \frac{\partial u}{\partial x} (\frac{\partial x}{\partial w} \frac{\partial w}{\partial r} + \frac{\partial x}{\partial t} \frac{\partial t}{\partial r}) + \frac{\partial u}{\partial y} (\frac{\partial y}{\partial w} \frac{\partial w}{\partial r} + \frac{\partial y}{\partial t} \frac{\partial t}{\partial r}) \\ + \frac{\partial u}{\partial z} (\frac{\partial z}{\partial w} \frac{\partial w}{\partial r} + \frac{\partial z}{\partial t} \frac{\partial t}{\partial r}) \end{matrix}

Checkpoint 4.28 §4.6 source ↗

Find the directional derivative

D_{u} f (x, y)

f (x, y) =

3 x^{2} y -

4 x y^{3} +

3 y^{2} -

4 x

in the direction of

u =

(cos \frac{π}{3}) i +

(sin \frac{π}{3}) j

using Equation 4.37. What is

D_{u} f (3, 4) ?

Answer

\begin{matrix} D_{u} f (x, y) = \frac{(6 x y - 4 y^{3} - 4) (1)}{2} + \frac{(3 x^{2} - 12 x y^{2} + 6 y) \sqrt{3}}{2} \\ D_{u} f (3, 4) = \frac{72 - 256 - 4}{2} + \frac{(27 - 576 + 24) \sqrt{3}}{2} = −94 - \frac{525 \sqrt{3}}{2} \end{matrix}

Checkpoint 4.29 §4.6 source ↗

Find the gradient

\nabla f (x, y)

f (x, y) =

(x^{2} - 3 y^{2}) / (2 x + y) .

Answer

\nabla f (x, y) =

\frac{2 x^{2} + 2 x y + 6 y^{2}}{{(2 x + y)}^{2}} i -

\frac{x^{2} + 12 x y + 3 y^{2}}{{(2 x + y)}^{2}} j

Checkpoint 4.30 §4.6 source ↗

Find the direction for which the directional derivative of

g (x, y) =

4 x -

x y +

2 y^{2}

(−2, 3)

is a maximum. What is the maximum value?

Answer

The gradient of

g

(−2, 3)

\nabla g (−2, 3) =

i +

14 j .

The unit vector that points in the same direction as

\nabla g (−2, 3)

\frac{\nabla g (−2, 3)}{‖ \nabla g (−2, 3) ‖} =

\frac{1}{\sqrt{197}} i +

\frac{14}{\sqrt{197}} j =

\frac{\sqrt{197}}{197} i +

\frac{14 \sqrt{197}}{197} j,

which gives an angle of

θ =

arcsin ((14 \sqrt{197}) / 197) \approx

1.499 rad .

The maximum value of the directional derivative is

‖ \nabla g (−2, 3) ‖ =

\sqrt{197} .

Checkpoint 4.31 §4.6 source ↗

For the function

f (x, y) =

x^{2} -

2 x y +

5 y^{2} +

3 x -

2 y +

4,

find the tangent to the level curve at point

(1, 1) .

Draw the graph of the level curve corresponding to

f (x, y) = 9

and draw

\nabla f (1, 1)

and a tangent vector.

Answer

$\nabla f (x, y) =$ $(2 x - 2 y + 3) i +$ $(−2 x + 10 y - 2) j$
$\nabla f (1, 1) =$ $3 i +$ $6 j$
Tangent vector: $6 i - 3 j$ or $−6 i + 3 j$

A rotated ellipse with equation f(x, y) = 8. At the point (1, 1) on the ellipse, there are drawn two arrows, one tangent vector and one normal vector. The normal vector is marked ∇f(1, 1) and is perpendicular to the tangent vector.

Checkpoint 4.32 §4.6 source ↗

Find the gradient

\nabla f (x, y, z)

f (x, y, z) =

\frac{x^{2} - 3 y^{2} + z^{2}}{2 x + y - 4 z} .

Answer

\begin{matrix} \nabla f (x, y, z) & = \frac{2 x^{2} + 2 x y + 6 y^{2} - 8 x z - 2 z^{2}}{{(2 x + y - 4 z)}^{2}} i - \frac{x^{2} + 12 x y + 3 y^{2} - 24 y z + z^{2}}{{(2 x + y - 4 z)}^{2}} j \\ + \frac{4 x^{2} - 12 y^{2} - 4 z^{2} + 4 x z + 2 y z}{{(2 x + y - 4 z)}^{2}} k . \end{matrix}

Checkpoint 4.33 §4.6 source ↗

Calculate

D_{u} f (x, y, z)

and

D_{u} f (0, −2, 5)

in the direction of

v =

−3 i +

12 j -

4 k

for the function

f (x, y, z) =

3 x^{2} +

x y -

2 y^{2} +

4 y z -

z^{2} +

2 x z .

Answer

\begin{matrix} D_{u} f (x, y, z) & = & - \frac{3}{13} (6 x + y + 2 z) + \frac{12}{13} (x - 4 y + 4 z) - \frac{4}{13} (2 x + 4 y - 2 z) \\ D_{u} f (0, −2, 5) & = & \frac{384}{13} \end{matrix}

4-261 §4.6 source ↗

For the following exercises, find the directional derivative using the limit definition only.

$f (x, y) =$ $y^{2} cos (2 x)$ at point $P (\frac{π}{3}, 2)$ in the direction of $u =$ $(cos \frac{π}{4}) i +$ $(sin \frac{π}{4}) j$

Answer

- 2 \sqrt{6} - \sqrt{2}

4-263 §4.6 source ↗

For the following exercises, find the directional derivative of the function at point $P$ in the direction of $u$ or $v$ as appropriate.

$f (x, y) = x y,$ $P (0, −2),$ $v = \frac{1}{2} i + \frac{\sqrt{3}}{2} j$

Answer

−1

4-265 §4.6 source ↗

For the following exercises, find the directional derivative of the function at point $P$ in the direction of $u$ or $v$ as appropriate.

$h (x, y, z) =$ $x y z,$ $P (2, 1, 1),$ $v =$ $2 i +$ $j -$ $k$

Answer

\frac{2}{\sqrt{6}}

4-267 §4.6 source ↗

For the following exercises, find the directional derivative of the function at point $P$ in the direction of $u$ or $v$ as appropriate.

$f (x, y) = x^{2} - y^{2}, \begin{matrix} u = 〈 \frac{\sqrt{3}}{2}, \frac{1}{2} 〉, & P (1, 0) \end{matrix}$

Answer

\sqrt{3}

4-269 §4.6 source ↗

For the following exercises, find the directional derivative of the function at point $P$ in the direction of $u$ or $v$ as appropriate.

$\begin{matrix} f (x, y) = e^{x} cos y, & u = 〈 0, 1 〉, & P = (0, \frac{π}{2}) \end{matrix}$

Answer

−1.0

4-271 §4.6 source ↗

For the following exercises, find the directional derivative of the function at point $P$ in the direction of $u$ or $v$ as appropriate.

$f (x, y) = ln (x^{2} + y^{2}), \begin{matrix} u = 〈 \frac{3}{5}, \frac{4}{5} 〉, & P (1, 2) \end{matrix}$

Answer

\frac{22}{25}

4-273 §4.6 source ↗

For the following exercises, find the directional derivative of the function at point $P$ in the direction of $u$ or $v$ as appropriate.

$f (x, y, z) = y^{2} + x z, \begin{matrix} P (1, 2, 2), & v = 〈 2, −1, 2 〉 \end{matrix}$

Answer

\frac{2}{3}

4-275 §4.6 source ↗

For the following exercises, find the directional derivative of the function in the direction of the unit vector $u =$ $cos θ i +$ $sin θ j .$

$f (x, y) =$ $\frac{y}{x + 2 y},$ $θ =$ $-$ $\frac{π}{4}$

Answer

\frac{− \sqrt{2} (x + y)}{2 {(x + 2 y)}^{2}}

4-277 §4.6 source ↗

For the following exercises, find the directional derivative of the function in the direction of the unit vector $u =$ $cos θ i +$ $sin θ j .$

$w (x, y) =$ $y e^{x},$ $θ =$ $\frac{π}{3}$

Answer

\frac{e^{x} (y + \sqrt{3})}{2}

4-279 §4.6 source ↗

For the following exercises, find the directional derivative of the function in the direction of the unit vector $u =$ $cos θ i +$ $sin θ j .$

$\begin{matrix} f (x, y) = ln (x + 2 y), & θ = \frac{π}{3} \end{matrix}$

Answer

\frac{1 + 2 \sqrt{3}}{2 (x + 2 y)}

4-281 §4.6 source ↗

For the following exercises, find the gradient.

Find the gradient of $f (x, y, z) =$ $x y +$ $y z +$ $x z$ at point $P (1, 2, 3) .$

Answer

〈 5, 4, 3 〉

4-283 §4.6 source ↗

For the following exercises, find the gradient.

$f (x, y, z) = 4 x^{5} y^{2} z^{3}, \begin{matrix} P (2, −1, 1), & u = \frac{1}{3} i + \frac{2}{3} j - \frac{2}{3} k \end{matrix}$

Answer

−320

4-285 §4.6 source ↗

For the following exercises, find the directional derivative of the function at point $P$ in the direction of $Q .$

$f (x, y, z) = \frac{y}{x + z}, \begin{matrix} P (2, 1, −1), & Q (−1, 2, 0) \end{matrix}$

Answer

\frac{3}{\sqrt{11}}

4-287 §4.6 source ↗

For the following exercises, find the derivative of the function at $P$ in the direction of $u .$

$f (x, y) = ln (5 x + 4 y), \begin{matrix} P (3, 9), & u = 6 i + 8 j \end{matrix}$

Answer

\frac{31}{255}

4-289 §4.6 source ↗

For the following exercises, find the derivative of the function at $P$ in the direction of $u .$

[T] Use technology to sketch the level curve of $f (x, y) =$ $x^{2} +$ $4 y^{2}$ that passes through $P (−2, 0)$ and draw the gradient vector at $P .$

Answer

The top of half of an ellipse centered at the origin with major axis horizontal and of length 4 and minor axis 2. The point (–2, 0) is marked, and there is an arrow pointing out from it to the left marked –4i.

4-291 §4.6 source ↗

For the following exercises, find the gradient vector at the indicated point.

$f (x, y) =$ $x e^{y} -$ $ln (x),$ $P (3, 0)$

Answer

\frac{2}{3} i + 3 j

4-293 §4.6 source ↗

For the following exercises, find the gradient vector at the indicated point.

$f (x, y, z) =$ $x \sqrt{y^{2} + z^{2}},$ $P (−2, −1, −1)$

Answer

\sqrt{2} i + \sqrt{2} j + \sqrt{2} k

4-295 §4.6 source ↗

For the following exercises, find the derivative of the function.

$f (x, y) = e^{x y}$ at point $(6, 7)$ in the direction the function increases most rapidly

Answer

1.6 (10^{19})

4-297 §4.6 source ↗

For the following exercises, find the derivative of the function.

$f (x, y, z) =$ $ln (x y + y z + z x)$ at point $(−9, −18, −27)$ in the direction the function increases most rapidly

Answer

\frac{5 \sqrt{2}}{99}

4-299 §4.6 source ↗

For the following exercises, find the maximum rate of change of $f$ at the given point and the direction in which it occurs.

$f (x, y) =$ $x e^{− y},$ $(1, 0)$

Answer

\sqrt{2}, 〈 1, -1 〉

4-301 §4.6 source ↗

For the following exercises, find the maximum rate of change of $f$ at the given point and the direction in which it occurs.

$f (x, y) =$ $cos (3 x + 2 y),$ $(\frac{π}{6}, - \frac{π}{8})$

Answer

\sqrt{\frac{13}{2}}, 〈 −3, −2 〉

4-303 §4.6 source ↗

For the following exercises, find equations of

$f (x, y, z) =$ $x y +$ $y z +$ $x z =$ $3$ at point $(1, 1, 1)$

Answer

x + y + z = 3,

x - 1 = y - 1 = z - 1

4-305 §4.6 source ↗

For the following exercises, find equations of

$f (x, y, z) =$ $x e^{y} cos z -$ $z =$ $1$ at point $(1, 0, 0)$

Answer

x + y - z = 1,

x - 1 = y = − z

4-307 §4.6 source ↗

For the following exercises, solve the problem.

The electrical potential (voltage) in a certain region of space is given by the function $V (x, y, z) =$ $5 x^{2} -$ $3 x y +$ $x y z .$

a. Find the rate of change of the voltage at point $(3, 4, 5)$ in the direction of the vector $〈 1, 1, −1 〉 .$

b. In which direction does the voltage change most rapidly at point $(3, 4, 5) ?$

c. What is the maximum rate of change of the voltage at point $(3, 4, 5) ?$

Answer

\frac{32}{\sqrt{3}},

〈 38, 6, 12 〉,

2 \sqrt{406}

4-309 §4.6 source ↗

For the following exercises, solve the problem.

In two dimensions, the motion of an ideal fluid is governed by a velocity potential $φ .$ The velocity components of the fluid $u$ in the x-direction and $v$ in the y-direction, are given by $〈 u, v 〉 = \nabla φ .$ Find the velocity components associated with the velocity potential $φ (x, y) =$ $sin π x sin 2 π y .$

Answer

〈 u,

v 〉 =

〈 π cos (π x) sin (2 π y),

2 π sin (π x) cos (2 π y) 〉

Checkpoint 4.34 §4.7 source ↗

Find the critical point of the function

f (x, y) =

x^{3} +

2 x y -

2 x -

4 y .

Answer

(2, −5)

Checkpoint 4.35 §4.7 source ↗

Use the second derivative to find the local extrema of the function

f (x, y) = x^{3} + 2 x y - 6 x - 4 y^{2} .

Answer

(\frac{4}{3}, \frac{1}{3})

is a saddle point,

(- \frac{3}{2}, - \frac{3}{8})

is a local maximum.

Checkpoint 4.36 §4.7 source ↗

Use the problem-solving strategy for finding absolute extrema of a function to find the absolute extrema of the function

f (x, y) = 4 x^{2} - 2 x y + 6 y^{2} - 8 x + 2 y + 3

on the domain defined by $0 \leq x \leq 2$ and $−1 \leq y \leq 3 .$

Answer

The absolute minimum occurs at $(1, 0) :$ $f (1, 0) = −1 .$

The absolute maximum occurs at $(0, 3) :$ $f (0, 3) = 63 .$

4-311 §4.7 source ↗

For the following exercises, find all critical points.

$f (x, y) =$ ${(3 x - 2)}^{2} +$ ${(y - 4)}^{2}$

Answer

(\frac{2}{3}, 4)

4-313 §4.7 source ↗

For the following exercises, find all critical points.

$f (x, y) =$ $15 x^{3} -$ $3 x y +$ $15 y^{3}$

Answer

(0, 0)

(\frac{1}{15}, \frac{1}{15})

4-315 §4.7 source ↗

For the following exercises, find the critical points of the function by using algebraic techniques (completing the square) or by examining the form of the equation. Verify your results using the partial derivatives test.

$f (x, y) =$ $− x^{2} -$ $5 y^{2} +$ $8 x -$ $10 y -$ $13$

Answer

Maximum at

(4, −1, 8)

4-317 §4.7 source ↗

$f (x, y) =$ $\sqrt{x^{2} + y^{2}} +$ $1$

Answer

Relative minimum at

(0, 0, 1)

4-319 §4.7 source ↗

For the following exercises, use the second derivative test to classify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these.

$f (x, y) = x^{2} y^{2}$

Answer

The second derivative test fails. Since

x^{2} y^{2} > 0

for all x and y different from zero, and

x^{2} y^{2} = 0

when either x or y equals zero (or both), then the absolute minimum occurs at

(0, 0) .

4-321 §4.7 source ↗

For the following exercises, use the second derivative test to classify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these.

$f (x, y) =$ $2 x y +$ $3 x +$ $4 y$

Answer

f (−2, - \frac{3}{2}) =

−6

is a saddle point.

4-323 §4.7 source ↗

For the following exercises, use the second derivative test to classify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these.

$f (x, y) =$ $x^{2} +$ $4 x y +$ $y^{2}$

Answer

f (0, 0) = 0;

(0, 0, 0)

is a saddle point.

4-325 §4.7 source ↗

For the following exercises, use the second derivative test to classify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these.

$f (x, y) =$ $9 -$ $x^{4} y^{4}$

Answer

The only critical point(s) are where both

f_{x} = 0

and

f_{y} = 0

and that is

f (0, 0) =

9

. However, since

f_{x x} f_{x y} - {f_{x y}}^{2} =

0

at the critical point, the Second Derivative Test fails. By graphing, we can see that this point is a local maximum.

4-327 §4.7 source ↗

For the following exercises, use the second derivative test to classify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these.

$f (x, y) =$ $3 x^{2} -$ $2 x y +$ $y^{2} -$ $8 y$

Answer

Relative minimum located at

(2, 6) .

4-329 §4.7 source ↗

For the following exercises, use the second derivative test to classify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these.

$f (x, y) =$ $y^{2} +$ $x y +$ $3 y +$ $2 x +$ $3$

Answer

(1, −2)

is a saddle point.

4-331 §4.7 source ↗

For the following exercises, use the second derivative test to classify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these.

$f (x, y) =$ $x^{2} +$ $2 y^{2} -$ $x^{2} y$

Answer

(2, 1)

and

(−2, 1)

are saddle points;

(0, 0)

is a relative minimum.

4-333 §4.7 source ↗

For the following exercises, use the second derivative test to classify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these.

$f (x, y) =$ $e^{− (x^{2} + y^{2} + 2 x)}$

Answer

(−1, 0)

is a relative maximum.

4-335 §4.7 source ↗

For the following exercises, use the second derivative test to classify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these.

$f (x, y) =$ $x^{2} +$ $10 x y +$ $y^{2}$

Answer

(0, 0)

is a saddle point.

4-337 §4.7 source ↗

For the following exercises, use the second derivative test to classify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these.

$f (x, y) =$ $120 x +$ $120 y -$ $x y -$ $x^{2} -$ $y^{2}$

Answer

The relative maximum is at

(40, 40) .

4-339 §4.7 source ↗

For the following exercises, use the second derivative test to classify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these.

$f (x, y) =$ $x^{2} +$ $x -$ $3 x y +$ $y^{3} -$ $5$

Answer

(\frac{1}{4}, \frac{1}{2})

is a saddle point and

(1, 1)

is the relative minimum.

4-341 §4.7 source ↗

For the following exercises, determine the extreme values and the saddle points. Use a CAS to graph the function.

[T] $f (x, y) =$ $y e^{x} -$ $e^{y}$

Answer

A saddle point is located at $(0, 0) .$

A complicated graph that starts near (1, 1, 1), and decreases significantly along the x and y axes, so much so along the y axis that it is cut off. The rest of the graph stays near 0.

4-343 §4.7 source ↗

For the following exercises, determine the extreme values and the saddle points. Use a CAS to graph the function.

[T] $f (x, y) =$ $sin (x) sin (y),$ $x \in (0, 2 π),$ $y \in (0, 2 π)$

Answer

There is a saddle point at $(π, π),$ local maxima at $(\frac{π}{2}, \frac{π}{2}) and (\frac{3 π}{2}, \frac{3 π}{2}),$ and local minima at $(\frac{π}{2}, \frac{3 π}{2}) and (\frac{3 π}{2}, \frac{π}{2}) .$

A series of hills and holes alternating through a space with amplitude 1.

4-345 §4.7 source ↗

Find the absolute extrema of the given function on the indicated closed and bounded set $R .$

Find the absolute maximum and minimum values of $f (x, y) =$ $x^{2} +$ $y^{2} -$ $2 y +$ $1$ on the region $R =$ ${(x, y) | x^{2} + y^{2} \leq 4} .$

Answer

(0, 1, 0)

is the absolute minimum and

(0, −2, 9)

is the absolute maximum.

4-347 §4.7 source ↗

Find the absolute extrema of the given function on the indicated closed and bounded set $R .$

$f (x, y) =$ $\frac{−2 y}{x^{2} + y^{2} + 1}$ on $R =$ ${(x, y) : x^{2} + y^{2} \leq 4}$

Answer

There is an absolute minimum at

(0, 1, −1)

and an absolute maximum at

(0, −1, 1) .

4-349 §4.7 source ↗

Find the absolute extrema of the given function on the indicated closed and bounded set $R .$

Find the points on the surface $x^{2} - y z = 5$ that are closest to the origin.

Answer

(\sqrt{5}, 0, 0),

(− \sqrt{5}, 0, 0)

4-351 §4.7 source ↗

Find the absolute extrema of the given function on the indicated closed and bounded set $R .$

The sum of the length and the girth (perimeter of a cross-section) of a package carried by a delivery service cannot exceed $108$ in. Find the dimensions of the rectangular package of largest volume that can be sent.

Answer

18 by 36 by 18 in .

4-353 §4.7 source ↗

Find the absolute extrema of the given function on the indicated closed and bounded set $R .$

Find the point on the surface $f (x, y) =$ $x^{2} +$ $y^{2} +$ $10$ nearest the plane $x + 2 y - z = 0 .$ Identify the point on the plane.

Answer

(\frac{1}{2}, 1, \frac{45}{4})

4-355 §4.7 source ↗

Find the absolute extrema of the given function on the indicated closed and bounded set $R .$

A company manufactures two types of athletic shoes: jogging shoes and cross-trainers. The total revenue from $x$ jogging shoes and $y$ cross-trainers is given by $R (x, y) =$ $−5 x^{2} -$ $8 y^{2} -$ $2 x y +$ $42 x +$ $102 y,$ where $x$ and $y$ are in thousands of units. Find the values of x and y to maximize the total revenue.

Answer

x = 3

and

y = 6

4-357 §4.7 source ↗

Find the absolute extrema of the given function on the indicated closed and bounded set $R .$

Find the maximum volume of a cylindrical soda can such that the sum of its height and circumference is $120$ cm.

Answer

V =

\frac{64,000}{π} \approx

20,

372

cm³

Checkpoint 4.37 §4.8 source ↗

Use the method of Lagrange multipliers to find the maximum value of

f (x, y) =

9 x^{2} +

36 x y -

4 y^{2} -

18 x -

8 y

subject to the constraint

3 x + 4 y = 32 .

Answer

f

has a maximum value of

976

at the point

(8, 2) .

Checkpoint 4.38 §4.8 source ↗

A company has determined that its production level is given by the Cobb-Douglas function

f (x, y) =

2.5 x^{0.45} y^{0.55}

where x represents the total number of labor hours in

1

year and y represents the total capital input for the company. Suppose

1

unit of labor costs

$ 40

and

1

unit of capital costs

$ 50 .

Use the method of Lagrange multipliers to find the maximum value of

f (x, y) =

2.5 x^{0.45} y^{0.55}

subject to a budgetary constraint of

$ 500,000

per year.

Answer

A maximum production level of

13890

occurs with

5625

labor hours and

$ 5500

of total capital input.

Checkpoint 4.39 §4.8 source ↗

Use the method of Lagrange multipliers to find the minimum value of the function

f (x, y, z) = x + y + z

subject to the constraint $x^{2} + y^{2} + z^{2} = 1 .$

Answer

\begin{matrix} f (\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}) & = & \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = \sqrt{3} \\ f (- \frac{\sqrt{3}}{3}, - \frac{\sqrt{3}}{3}, - \frac{\sqrt{3}}{3}) & = & - \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} = − \sqrt{3} . \end{matrix}

Checkpoint 4.40 §4.8 source ↗

Use the method of Lagrange multipliers to find the minimum value of the function

f (x, y, z) = x^{2} + y^{2} + z^{2}

subject to the constraints $2 x + y + 2 z = 9$ and $5 x +$ $5 y +$ $7 z =$ $29 .$

Answer

f (2, 1, 2) = 9

is a minimum.

4-359 §4.8 source ↗

For the following exercises, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

$f (x, y, z) =$ $x y z,$ $x^{2} +$ $2 y^{2} +$ $3 z^{2} =$ $6$

Answer

maximum:

2 \frac{\sqrt{3}}{3},

minimum:

\frac{−2 \sqrt{3}}{3}

4-361 §4.8 source ↗

For the following exercises, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

$f (x, y) =$ $4 x^{3} +$ $y^{2};$ $2 x^{2} +$ $y^{2} =$ $1$

Answer

maximum:

(\frac{\sqrt{2}}{2}, 0, \sqrt{2}),

minimum:

(\frac{− \sqrt{2}}{2}, 0, − \sqrt{2})

4-363 §4.8 source ↗

For the following exercises, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

$f (x, y, z) =$ $y z +$ $x y,$ $x y =$ $1,$ $y^{2} +$ $z^{2} =$ $1$

Answer

maximum:

\frac{3}{2},

minimum =

\frac{1}{2}

4-365 §4.8 source ↗

For the following exercises, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

$f (x, y) =$ $4 x y,$ $\frac{x^{2}}{9} +$ $\frac{y^{2}}{16} =$ $1$

Answer

maxima:

f (\frac{3 \sqrt{2}}{2}, 2 \sqrt{2}) =

24,

f (- \frac{3 \sqrt{2}}{2}, −2 \sqrt{2}) =

24;

minima:

f (- \frac{3 \sqrt{2}}{2}, 2 \sqrt{2}) =

−24,

f (\frac{3 \sqrt{2}}{2}, −2 \sqrt{2}) =

−24

4-367 §4.8 source ↗

For the following exercises, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

$f (x, y, z) =$ $x +$ $3 y -$ $z,$ $x^{2} +$ $y^{2} +$ $z^{2} =$ $4$

Answer

maximum:

2 \sqrt{11}

f (\frac{2}{\sqrt{11}}, \frac{6}{\sqrt{11}}, \frac{−2}{\sqrt{11}});

minimum:

−2 \sqrt{11}

f (\frac{−2}{\sqrt{11}}, \frac{−6}{\sqrt{11}}, \frac{2}{\sqrt{11}})

4-369 §4.8 source ↗

For the following exercises, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

Minimize $f (x, y) =$ $x^{2} +$ $y^{2}$ on the hyperbola $x y = 1 .$

Answer

2.0

4-371 §4.8 source ↗

For the following exercises, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

Maximize $f (x, y, z) =$ $2 x +$ $3 y +$ $5 z$ on the sphere $x^{2} +$ $y^{2} +$ $z^{2} =$ $19 .$

Answer

19 \sqrt{2}

4-373 §4.8 source ↗

For the following exercises, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

The curve $x^{3} - y^{3} = 1$ is asymptotic to the line $y = x .$ Find the point(s) on the curve $x^{3} - y^{3} = 1$ farthest from the line $y = x .$

Answer

(\frac{1}{\sqrt[3]{2}}, \frac{−1}{\sqrt[3]{2}})

4-375 §4.8 source ↗

For the following exercises, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

Minimize $f (x, y) =$ $x^{2} +$ $y^{2},$ $x +$ $2 y -$ $5 =$ $0 .$

Answer

f (1, 2) = 5

4-377 §4.8 source ↗

For the following exercises, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

Minimize $f (x, y, z) =$ $x^{2} +$ $y^{2} +$ $z^{2},$ $x +$ $y +$ $z =$ $1 .$

Answer

f (\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) =

\frac{1}{3}

4-379 §4.8 source ↗

For the following exercises, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

Minimize $f (x, y, z) =$ $x^{2} +$ $y^{2} +$ $z^{2}$ when $x + y + z = 9$ and $x + 2 y + 3 z = 20 .$

Answer

minimum:

f (2, 3, 4) = 29

4-381 §4.8 source ↗

For the next group of exercises, use the method of Lagrange multipliers to solve the following applied problems.

A rectangular box without a top (a topless box) is to be made from $12$ ft² of cardboard. Find the maximum volume of such a box.

Answer

The maximum volume is

4

ft³. The dimensions are

1 \times 2 \times 2

ft.

4-383 §4.8 source ↗

For the next group of exercises, use the method of Lagrange multipliers to solve the following applied problems.

Find the point on the surface $x^{2} -$ $2 x y +$ $y^{2} -$ $x +$ $y =$ $0$ closest to the point $(1, 2, −3) .$

Answer

The point on the surface

x^{2} - 2 x y +

y^{2} - x +

y =

0

closest to the point

(1, 2, - 3)

\frac{3}{2}, \frac{3}{2}, −3

4-385 §4.8 source ↗

For the next group of exercises, use the method of Lagrange multipliers to solve the following applied problems.

Find the minimum distance from point $(0, 1)$ to the parabola $x^{2} = 4 y .$

Answer

1.0

4-387 §4.8 source ↗

For the next group of exercises, use the method of Lagrange multipliers to solve the following applied problems.

Find the minimum distance from the plane $x + y + z = 1$ to point $(2, 1, 1) .$

Answer

\sqrt{3}

4-389 §4.8 source ↗

For the next group of exercises, use the method of Lagrange multipliers to solve the following applied problems.

Find the point on the line $y = 2 x + 3$ that is closest to point $(4, 2) .$

Answer

(\frac{2}{5}, \frac{19}{5})

4-391 §4.8 source ↗

For the next group of exercises, use the method of Lagrange multipliers to solve the following applied problems.

Find the maximum value of $f (x, y) =$ $sin x sin y,$ where $x and y$ denote the acute angles of a right triangle. Draw the contours of the function using a CAS.

Answer

$\frac{1}{2}$

An alternating series of hills and holes of amplitude 1 across xyz space.

4-393 §4.8 source ↗

For the next group of exercises, use the method of Lagrange multipliers to solve the following applied problems.

[T] By investing x units of labor and y units of capital, a watch manufacturer can produce $P (x, y) =$ $50 x^{0.4} y^{0.6}$ watches. Find the maximum number of watches that can be produced on a budget of $$ 20,000$ if labor costs $100/unit and capital costs $200/unit. Use a CAS to sketch a contour plot of the function.

Answer

Roughly 3365 watches at the critical point $(80, 60)$

A series of curves in the first quadrant, with the first starting near (2, 120), decreasing sharply to near (20, 20), and then decreasing slowly to (120, 5). The next curve starts near (10, 120), decreases sharply to near (40, 40), and then decreases slowly to (120, 20). The next curve starts near (20, 120), decreases sharply to near (60, 60), and then decreases slowly to (120, 40). The next curve starts near (40, 120), decreases to near (80, 80), and then decreases a little slowly to (120, 60). The last curve starts near (60, 120) and decreases rather evenly through (100, 100) to (120, 90).

4-395 §4.R source ↗

For the following exercises, determine whether the statement is true or false. Justify your answer with a proof or a counterexample.

If the function $f (x, y)$ and all its second derivatives are continuous everywhere, then $f_{x y} = f_{y x} .$

Answer

True, by Clairaut’s theorem

4-397 §4.R source ↗

For the following exercises, determine whether the statement is true or false. Justify your answer with a proof or a counterexample.

$(\frac{3}{4}, \frac{9}{16})$ is a critical point of $g (x, y) =$ $4 x^{3} -$ $2 x^{2} y +$ $y^{2} -$ $2 .$

Answer

False

4-399 §4.R source ↗

For the following exercises, sketch the function in one graph and, in a second, sketch several level curves.

$f (x, y) =$ $x +$ $4 y^{2} .$

Answer

Answers may vary

4-401 §4.R source ↗

For the following exercises, evaluate the following limits, if they exist. If they do not exist, prove it.

$lim_{(x, y) \to (0, 0)} \frac{4 x y}{x - 2 y^{2}}$

Answer

Does not exist

4-403 §4.R source ↗

For the following exercises, find the largest region of continuity for the function.

$g (x, y) =$ $ln (4 - x^{2} - y^{2})$

Answer

Continuous at all points on the

x, y -plane,

except where

x^{2} + y^{2} > 4 .

4-405 §4.R source ↗

For the following exercises, find all first derivatives, full or partial, as appropriate.

$u (x, y) =$ $x^{4} -$ $3 x y +$ $1,$ $x =$ $2 t,$ $y =$ $t^{3}$

Answer

\frac{\partial u}{\partial x} =

4 x^{3} -

3 y,

\frac{\partial u}{\partial y} =

−3 x,

\frac{d x}{d t} =

2,

\frac{d y}{d t} =

3 t^{2},

\frac{d u}{d t} =

40 t^{3}

4-407 §4.R source ↗

For the following exercises, find all second partial derivatives.

$h (x, y, z) =$ $\frac{x^{3} e^{2 y}}{z}$

Answer

h_{x x} (x, y, z) =

\frac{6 x e^{2 y}}{z},

h_{x y} (x, y, z) =

\frac{6 x^{2} e^{2 y}}{z},

h_{x z} (x, y, z) =

-

\frac{3 x^{2} e^{2 y}}{z^{2}},

h_{y x} (x, y, z) =

\frac{6 x^{2} e^{2 y}}{z},

h_{y y} (x, y, z) =

\frac{4 x^{3} e^{2 y}}{z},

h_{y z} (x, y, z) =

-

\frac{2 x^{3} e^{2 y}}{z^{2}},

h_{z x} (x, y, z) =

-

\frac{3 x^{2} e^{2 y}}{z^{2}},

h_{z y} (x, y, z) =

-

\frac{2 x^{3} e^{2 y}}{z^{2}},

h_{z z} (x, y, z) =

\frac{2 x^{3} e^{2 y}}{z^{3}}

4-409 §4.R source ↗

For the following exercises, find an equation of the tangent plane to the specified surface at the given point.

$z = e^{x} + \frac{2}{y}$ at point $(0, 1, 3)$

Answer

z = x - 2 y + 5

4-411 §4.R source ↗

For the following exercises, find an equation of the tangent plane to the specified surface at the given point.

Find the differential $d z$ of $h (x, y) =$ $4 x^{2} +$ $2 x y -$ $3 y$ and approximate $Δ z$ at the point $(1, −2) .$ Let $Δ x = 0.1$ and $Δ y = 0.01 .$

Answer

d z = 4 d x - d y,

d z (0.1, 0.01) =

0.39,

Δ z = 0.432

4-413 §4.R source ↗

For the following exercises, find an equation of the tangent plane to the specified surface at the given point.

Find the maximal directional derivative magnitude and direction for the function $f (x, y) =$ $x^{3} +$ $2 x y -$ $cos (π y)$ at point $(3, 0) .$

Answer

3 \sqrt{85}, 〈 27, 6 〉

4-415 §4.R source ↗

For the following exercises, find the gradient.

$f (x, y) =$ $\frac{\sqrt{x} + y^{2}}{x y}$

Answer

\nabla f (x, y) =

-

\frac{\sqrt{x} + 2 y^{2}}{2 x^{2} y} i +

(\frac{1}{x} - \frac{1}{\sqrt{x} y^{2}}) j

4-417 §4.R source ↗

For the following exercises, use Lagrange multipliers to find the maximum and minimum values for the functions with the given constraints.

$f (x, y) =$ $x^{2} y,$ $x^{2} +$ $y^{2} =$ $4$

Answer

maximum:

\frac{16}{3 \sqrt{3}},

minimum:

- \frac{16}{3 \sqrt{3}}

4-419 §4.R source ↗

For the following exercises, use Lagrange multipliers to find the maximum and minimum values for the functions with the given constraints.

A machinist is constructing a right circular cone out of a block of aluminum. The machine gives an error of $5 %$ in height and $2 %$ in radius. Find the maximum error in the volume of the cone if the machinist creates a cone of height $6$ cm and radius $2$ cm.

Answer

2.3228

cm³

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Problems and answers are adapted from Calculus Volume 3 by Gilbert Strang and Edwin “Jed” Herman, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/calculus-volume-3/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.