Gilbert Strang; Edwin “Jed” Herman

Checkpoint

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.

4.2

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).

4.3

$z = 3 - {(x - 1)}^{2} .$ This function describes a parabola opening downward in the plane $y = 3 .$

4.4

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

4.5

${(x - 1)}^{2} + {(y + 2)}^{2} + {(z - 3)}^{2} = 16$ describes a sphere of radius $4$ centered at the point $(1, −2, 3) .$

4.6

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

4.7

If $y = k (x - 2) + 1,$ 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.

4.8

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

4.9

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

4.10

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 $x y -plane.$

4.11

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

4.12

$\frac{\partial f}{\partial x} = 8 x + 2 y + 3, \frac{\partial f}{\partial y} = 2 x - 2 y - 2$

4.13

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

4.14

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

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

4.15

$\frac{\partial f}{\partial x} = 4 x - 8 x y + 5 z^{2} - 6, \frac{\partial f}{\partial y} = −4 x^{2} + 4 y, \frac{\partial f}{\partial z} = 10 x z + 3$

4.16

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

4.17

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

4.19

$z = 7 x + 8 y - 3$

4.20

$L (x, y) = 6 - 2 x + 3 y,$ so $L (4.1, 0.9) = 6 - 2 (4.1) + 3 (0.9) = 0.5$ $f (4.1, 0.9) = e^{5 - 2 (4.1) + 3 (0.9)} = e^{−0.5} \approx 0.6065 .$

4.21

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

$\begin{matrix} lim_{(x, y) \to (x_{0}, y_{0})} \frac{E (x, y)}{\sqrt{{(x - x_{0})}^{2} + {(y - y_{0})}^{2}}} & = lim_{(x, y) \to (−1, 2)} \frac{−4 {(y - 2)}^{2}}{\sqrt{{(x + 1)}^{2} + {(y - 2)}^{2}}} \\ \leq lim_{(x, y) \to (−1, 2)} \frac{−4 ({(x + 1)}^{2} + {(y - 2)}^{2})}{\sqrt{{(x + 1)}^{2} + {(y - 2)}^{2}}} \\ = lim_{(x, y) \to (2, −3)} - 4 \sqrt{{(x + 1)}^{2} + {(y - 2)}^{2}} \\ = 0. \end{matrix}$

4.22

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

4.23

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

4.24

$\frac{\partial z}{\partial u} = 0, \frac{\partial z}{\partial v} = \frac{−21}{{(3 sin 3 v + cos 3 v)}^{2}}$

4.25

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

4.26

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

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.

4.27

$\frac{d y}{d x} = {\frac{2 x + y + 7}{2 y - x + 3} |}_{(3, −2)} = \frac{2 (3) + (−2) + 7}{2 (−2) - (3) + 3} = - \frac{11}{4}$
Equation of the tangent line: $y = - \frac{11}{4} x + \frac{25}{4}$

4.28

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

4.29

$\nabla f (x, y) = \frac{2 x^{2} + 2 x y + 6 y^{2}}{{(2 x + y)}^{2}} i - \frac{x^{2} + 12 x y + 3 y^{2}}{{(2 x + y)}^{2}} j$

4.30

The gradient of $g$ at $(−2, 3)$ is $\nabla g (−2, 3) = i + 14 j .$ The unit vector that points in the same direction as $\nabla g (−2, 3)$ is $\frac{\nabla g (−2, 3)}{‖ \nabla g (−2, 3) ‖} = \frac{1}{\sqrt{197}} i + \frac{14}{\sqrt{197}} j = \frac{\sqrt{197}}{197} i + \frac{14 \sqrt{197}}{197} j,$ which gives an angle of $θ = arcsin ((14 \sqrt{197}) / 197) \approx 1.499 rad .$ The maximum value of the directional derivative is $‖ \nabla g (−2, 3) ‖ = \sqrt{197} .$

4.31

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

4.32

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

4.33

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

4.34

$(2, −5)$

4.35

$(\frac{4}{3}, \frac{1}{3})$ is a saddle point, $(- \frac{3}{2}, - \frac{3}{8})$ is a local maximum.

4.36

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

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

4.37

$f$ has a maximum value of $976$ at the point $(8, 2) .$

4.38

A maximum production level of $13890$ occurs with $5625$ labor hours and $$ 5500$ of total capital input.

4.39

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

4.40

$f (2, 1, 2) = 9$ is a minimum.

Section 4.1 Exercises

1.

$17, 72$

3.

$20 π .$ This is the volume when the radius is $2$ and the height is $5 .$

5.

All points in the $x y -plane$

7.

$x < y^{2}$

9.

All real ordered pairs in the $x y -plane$ of the form $(a, b)$

11.

${z | 0 \leq z \leq 4}$

13.

The set $ℝ$

15.

$y^{2} - x^{2} = 4,$ a hyperbola

17.

$4 = x + y,$ a line; $x + y = 0,$ line through the origin

19.

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

21.

$\frac{x}{x + y} = −1, \frac{x}{x + y} = 0, \frac{x}{x + y} = 2$

23.

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

25.

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

27.

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

29.

The level curves are parabolas of the form $y = c x^{2} - 2 .$

31.

$z = 3 + y^{3},$ 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.

33.

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

35.

$\frac{x^{2}}{9} + \frac{y^{2}}{4} + \frac{z^{2}}{36} < 1$

37.

All points in $x y z -space$

39.

An upward facing, gently increasing paraboloid.

41.

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

43.

A downward facing, gently decreasing paraboloid.

45.

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

47.

The contour lines are circles.

49.

$x^{2} + y^{2} + z^{2} = 9,$ a sphere of radius $3$

51.

$x^{2} + y^{2} - z^{2} = 4,$ a hyperboloid of one sheet

53.

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

55.

$1 = e^{x y} (x^{2} + y^{2})$

57.

$T (x, y) = \frac{k}{x^{2} + y^{2}}$

59.

$x^{2} + y^{2} = \frac{k}{40},$ $x^{2} + y^{2} = \frac{k}{100} .$ The level curves represent circles of radii $\sqrt{10 k} / 20$ and $\sqrt{k} / 10$

Section 4.2 Exercises

61.

2.0

63.

$\frac{2}{3}$

65.

$1$

67.

$\frac{1}{2}$

69.

$- \frac{1}{2}$

71.

$e^{−32}$

73.

$11.0$

75.

$1.0$

77.

The limit does not exist because when $x$ and $y$ both approach zero, the function approaches $ln 0,$ which is undefined (approaches negative infinity).

79.

every open disk centered at $(x_{0}, y_{0})$ contains points inside $R$ and outside $R$

81.

$0.0$

83.

$0.00$

85.

The limit does not exist.

87.

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

89.

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

91.

The function $f$ is continuous in the region $y > − x .$

93.

The function $f$ is continuous at all points in the $x y -plane$ except at $(0, 0) .$

95.

The function is continuous at $(0, 0)$ since the limit of the function at $(0, 0)$ is $0,$ the same value of $f (0, 0) .$

97.

The function is discontinuous at $(0, 0) .$ The limit at $(0, 0)$ fails to exist and $g (0, 0)$ does not exist.

99.

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 $y = − x .$

101.

All points $P (x, y, z)$ in space

103.

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).

105.

a.

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. ${z | 0 \leq z \leq 3}$

107.

$1.0$

109.

$f (g (x, y))$ is continuous at all points $(x, y)$ that are not on the line $2 x - 5 y = 0 .$

111.

$2.0$

Section 4.3 Exercises

113.

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

115.

The sign is negative.

117.

The partial derivative is zero at the origin.

119.

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

121.

$\frac{\partial z}{\partial x} = \frac{6 x^{5}}{x^{6} + y^{4}}; \frac{\partial z}{\partial y} = \frac{4 y^{3}}{x^{6} + y^{4}}$

123.

$\frac{\partial z}{\partial x} = y e^{x y}; \frac{\partial z}{\partial y} = x e^{x y}$

125.

$\frac{\partial z}{\partial x} = 2 {sec}^{2} (2 x - y), \frac{\partial z}{\partial y} = − {sec}^{2} (2 x - y)$

127.

$f_{x} (2, −2) = \frac{1}{4} = f_{y} (2, −2)$

129.

$\frac{\partial z}{\partial x} = − cos (1)$

131.

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

133.

a. $V (r, h) = π r^{2} h$ b. $\frac{\partial V}{\partial r} = 2 π r h$ c. $\frac{\partial V}{\partial h} = π r^{2}$

135.

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

137.

$\frac{\partial^{2} z}{\partial x^{2}} = 2, \frac{\partial^{2} z}{\partial y^{2}} = 4$

139.

$f_{x y y} = f_{y x y} = f_{y y x} = 0$

141.

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

143.

$f_{x y z} = 6 y^{2} x - 18 y z^{2}$

145.

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

147.

$(0, 0), (0, 2), (\sqrt{3}, −1), (− \sqrt{3}, −1)$

149.

$\frac{\partial^{2} z}{\partial x^{2}} + \frac{\partial^{2} z}{\partial y^{2}} = e^{x} sin (y) - e^{x} sin y = 0$

151.

$c^{2} \frac{\partial^{2} z}{\partial x^{2}} = e^{− t} cos (\frac{x}{c})$

153.

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

155.

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

159.

$\frac{\partial F}{\partial θ} = 6, \frac{\partial F}{\partial x} = 4 - 3 \sqrt{3}$

161.

$\frac{\partial f}{\partial x}$ at $(500, 1000) = 172.36,$ $\frac{\partial f}{\partial y}$ at $(500, 1000) = 36.93$

Section 4.4 Exercises

163.

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

165.

Normal vector: $i + j,$ tangent vector: $i - j$

167.

Normal vector: $7 i - 17 j,$ tangent vector: $17 i + 7 j$

169.

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

171.

$−36 x - 6 y - z = −39$

173.

$z = 0$

175.

$5 x + 4 y + 3 z - 22 = 0$

177.

$4 x - 5 y + 4 z = 0$

179.

$2 x + 2 y - z = 0$

181.

$−2 (x - 1) + 2 (y - 2) - (z - 1) = 0$

183.

$x = 20 t + 2, y = −4 t + 1, z = − t + 18$

185.

$x = 0, y = 0, z = t$

187.

$x - 1 = 2 t; y - 2 = −2 t; z - 1 = t$

189.

The differential of the function $z (x, y) = d z = f_{x} d x + f_{y} d y$

191.

Using the definition of differentiability, we have $e^{x y} x \approx x + y .$

193.

$Δ z = 2 x Δ x + 3 Δ y + {(Δ x)}^{2} .$ ${(Δ x)}^{2} \to 0$ for small $Δ x$ and $z$ satisfies the definition of differentiability.

195.

$Δ z \approx 1.185422$ and $d z \approx 1.108 .$ They are relatively close.

197.

$16$ cm³

199.

$Δ z =$ exact change $= 0.6449,$ approximate change is $d z = 0.65 .$ The two values are close.

201.

$13 % or 0.13$

203.

$0.025$

205.

$0.3 %$

207.

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

209.

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

211.

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

213.

$z = 0$

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.

Section 4.5 Exercises

215.

$\frac{d w}{d t} = y cos z + x cos z (2 t) - \frac{x y sin z}{\sqrt{1 - t^{2}}}$

217.

$\frac{\partial w}{\partial s} = −30 x + 4 y,$ $\frac{\partial w}{\partial t} = 10 x - 16 y$

219.

$\frac{\partial f}{\partial r} = r sin (2 θ)$

221.

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

223.

$\frac{d f}{d t} = −1$

225.

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

227.

$\frac{d w}{d t} = 2 e^{2 t}$ in both cases

229.

$\frac{d u}{d t} = \sqrt{2} (π -4$

231.

$\frac{d y}{d x} = - \frac{3 x^{2} + y^{2}}{2 x y}$

233.

$\frac{d y}{d x} = \frac{y - x}{− x + 2 y^{3}}$

235.

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

237.

$\frac{d y}{d x} = - \frac{y e^{x y}}{x e^{x y} + e^{y} (1 + y)}$

239.

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

241.

$\frac{d z}{d t} = - \frac{10}{3} t^{7 / 3} \times e^{1 - t^{10 / 3}}$

243.

$\frac{\partial z}{\partial u} = \frac{−2 sin u}{3 sin v}$ and $\frac{\partial z}{\partial v} = \frac{−2 cos u cos v}{3 {sin}^{2} v}$

245.

$\frac{\partial z}{\partial r} = \sqrt{3} e^{\sqrt{3}},$ $\frac{\partial z}{\partial θ} = (2 - 4 \sqrt{3}) e^{\sqrt{3}}$

247.

$\frac{\partial w}{\partial t} = cos (x y z) \times y z \times (−3) - cos (x y z) x z e^{1 - t} + cos (x y z) x y \times 4$

249.

$f (t x, t y) = \sqrt{t^{2} x^{2} + t^{2} y^{2}} = t^{1} f (x, y),$ $\frac{\partial f}{\partial y} = x \frac{1}{2} {(x^{2} + y^{2})}^{− 1 / 2} \times 2 x + y \frac{1}{2} {(x^{2} + y^{2})}^{− 1 / 2} \times 2 y = 1 f (x, y)$

251.

$V' = 4 π$

253.

$\frac{d V}{d t} = \frac{1066 π}{3} {cm}^{3} / min$

255.

$\frac{d A}{d t} = 12 in .^{2} / min$

257.

$2 ° C/sec$

259.

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

Section 4.6 Exercises

261.

$- \frac{4 \sqrt{3} + 3}{2}$

263.

$−1$

265.

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

267.

$\sqrt{3}$

269.

$−1.0$

271.

$\frac{22}{25}$

273.

$\frac{2}{3}$

275.

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

277.

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

279.

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

281.

$〈 5, 4, 3 〉$

283.

$−320$

285.

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

287.

$\frac{31}{255}$

289.

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.

291.

$\frac{4}{3} i - 3 j$

293.

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

295.

$1.6 (10^{19})$

297.

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

299.

$\sqrt{2}, 〈 1, -1 〉$

301.

$\sqrt{\frac{13}{2}}, 〈 −3, −2 〉$

303.

a. $x + y + z = 3,$ b. $x - 1 = y - 1 = z - 1$

305.

a. $x + y - z = 1,$ b. $x - 1 = y = − z$

307.

a. $\frac{32}{\sqrt{3}},$ b. $〈 38, 6, 12 〉,$ c. $2 \sqrt{406}$

309.

$〈 u, v 〉 = 〈 π cos (π x) sin (2 π y), 2 π sin (π x) cos (2 π y) 〉$

Section 4.7 Exercises

311.

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

313.

$(0, 0)$ $(\frac{1}{15}, \frac{1}{15})$

315.

Maximum at $(4, −1, 8)$

317.

Relative minimum at $(0, 0, 1)$

319.

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 $(0, 0) .$

321.

$f (−2, - \frac{3}{2}) = −6$ is a saddle point.

323.

$f (0, 0) = 0;$ $(0, 0, 0)$ is a saddle point.

325.

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.

327.

Relative minimum located at $(2, 6) .$

329.

$(1, −2)$ is a saddle point.

331.

$(2, 1)$ and $(−2, 1)$ are saddle points; $(0, 0)$ is a relative minimum.

333.

$(−1, 0)$ is a relative maximum.

335.

$(0, 0)$ is a saddle point.

337.

The relative maximum is at $(40, 40) .$

339.

$(\frac{1}{4}, \frac{1}{2})$ is a saddle point and $(1, 1)$ is the relative minimum.

341.

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.

343.

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.

345.

$(0, 1, 0)$ is the absolute minimum and $(0, −2, 9)$ is the absolute maximum.

347.

There is an absolute minimum at $(0, 1, −1)$ and an absolute maximum at $(0, −1, 1) .$

349.

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

351.

$18 by 36 by 18 in .$

353.

$(\frac{47}{24}, \frac{47}{12}, \frac{235}{24})$

355.

$x = 3$ and $y = 6$

357.

$V = \frac{64,000}{π} \approx 20, 372$ cm³

Section 4.8 Exercises

359.

maximum: $2 \frac{\sqrt{3}}{3},$ minimum: $\frac{−2 \sqrt{3}}{3}$

361.

maximum: $(\frac{\sqrt{2}}{2}, 0, \sqrt{2}),$ minimum: $(\frac{− \sqrt{2}}{2}, 0, − \sqrt{2})$

363.

maximum: $\frac{3}{2},$ minimum = $\frac{1}{2}$

365.

maxima: $f (\frac{3 \sqrt{2}}{2}, 2 \sqrt{2}) = 24,$ $f (- \frac{3 \sqrt{2}}{2}, −2 \sqrt{2}) = 24;$ minima: $f (- \frac{3 \sqrt{2}}{2}, 2 \sqrt{2}) = −24,$ $f (\frac{3 \sqrt{2}}{2}, −2 \sqrt{2}) = −24$

367.

maximum: $2 \sqrt{11}$ at $f (\frac{2}{\sqrt{11}}, \frac{6}{\sqrt{11}}, \frac{−2}{\sqrt{11}});$ minimum: $−2 \sqrt{11}$ at $f (\frac{−2}{\sqrt{11}}, \frac{−6}{\sqrt{11}}, \frac{2}{\sqrt{11}})$

369.

$2.0$

371.

$19 \sqrt{2}$

373.

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

375.

$f (1, 2) = 5$

377.

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

379.

minimum: $f (2, 3, 4) = 29$

381.

The maximum volume is $4$ ft³. The dimensions are $1 \times 2 \times 2$ ft.

383.

The point on the surface $x^{2} - 2 x y + y^{2} - x + y = z$ closest to the point $(1, 2, - 3)$ is $(\frac{3}{2}, \frac{3}{2}, 0)$ .

385.

$1.0$

387.

$\sqrt{3}$

389.

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

391.

$\frac{1}{2}$

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

393.

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).

Review Exercises

395.

True, by Clairaut’s theorem

397.

False

399.

Answers may vary

401.

Does not exist

403.

Continuous at all points on the $x, y -plane,$ except where $x^{2} + y^{2} > 4 .$

405.

$\frac{\partial u}{\partial x} = 4 x^{3} - 3 y, \frac{\partial u}{\partial y} = −3 x, \frac{d x}{d t} = 2, \frac{d y}{d t} = 3 t^{2}, \frac{d u}{d t} = 40 t^{3}$

407.

$h_{x x} (x, y, z) = \frac{6 x e^{2 y}}{z},$ $h_{x y} (x, y, z) = \frac{6 x^{2} e^{2 y}}{z},$ $h_{x z} (x, y, z) = - \frac{3 x^{2} e^{2 y}}{z^{2}},$ $h_{y x} (x, y, z) = \frac{6 x^{2} e^{2 y}}{z},$ $h_{y y} (x, y, z) = \frac{4 x^{3} e^{2 y}}{z},$ $h_{y z} (x, y, z) = - \frac{2 x^{3} e^{2 y}}{z^{2}},$ $h_{z x} (x, y, z) = - \frac{3 x^{2} e^{2 y}}{z^{2}},$ $h_{z y} (x, y, z) = - \frac{2 x^{3} e^{2 y}}{z^{2}},$ $h_{z z} (x, y, z) = \frac{2 x^{3} e^{2 y}}{z^{3}}$