Solutions To Selected Problems Chapter L Chapter N

Chapter M

Exercises M.1

1.

Answers will vary.
2.

surface
3.

topographical
4.

T
5.

surface
6.

When level curves are close together, it means the function is changing $z$ -values rapidly. When far apart, it changes $z$ -values slowly.
7.

domain: $\mathbb{R}^{2}$

range: $z\geq 2$
8.

domain: $\mathbb{R}^{2}$

range: $\mathbb{R}$
9.

domain: $\mathbb{R}^{2}$

range: $\mathbb{R}$
10.

domain: $x\neq 2y$ ; in set notation, $\{(x,y)\,|\,x\neq 2y\}$

range: $z\neq 0$
11.

domain: $\mathbb{R}^{2}$

range: $0<z\leq 1$
12.

domain: $\mathbb{R}^{2}$

range: $-1\leq z\leq 1$
13.

domain: $\{(x,y)\,|\,x^{2}+y^{2}\leq 9\}$ , i.e., the domain is the circle and interior of a circle centered at the origin with radius 3.

range: $0\leq z\leq 3$
14.

domain: $\{(x,y)\,|\,x^{2}+y^{2}\geq 9\}$ , i.e., the domain is the exterior of the circle (not including the circle itself) centered at the origin with radius 3.

range: $0<z<\infty$ , or $(0,\infty)$
15.

Level curves are lines $y=(3/2)x-c/2$ .
16.

Level curves are hyperbolas $\frac{x^{2}}{c}-\frac{y^{2}}{c}=1$ , except for $c=0$ , where the level curve is the pair of lines $y=x$ , $y=-x$ .
17.

Level curves are parabolas $x=y^{2}+c$ .
18.

Level curves are hyperbolas $(x-c)^{2}-(y-c)^{2}=1$ , drawn in graph in different styles to differentiate the curves.
19.

Level curves are circles, centered at $(1/c,-1/c)$ with radius $\sqrt{2/c^{2}-1}$ . When $c=0$ , the level curve is the line $y=x$ .
20.

Level curves are cubics of the form $y=x^{3}+cx+1$ . Note how each curve passes through $(0,1)$ and that the function is not defined at $x=0$ .
21.

Level curves are ellipses of the form $\frac{x^{2}}{c^{2}}+\frac{y^{2}}{c^{2}/4}=1$ , i.e., $a=c$ and $b=c/2$ .
22.

Level curves are ellipses of the form $\frac{x^{2}}{c}+\frac{y^{2}}{c/4}=1$ , i.e., $a=\sqrt{c}$ and $b=\sqrt{c}/2$ .
23.

domain: $x+2y-4z\neq 0$ ; the set of points in $\mathbb{R}^{3}$ NOT in the domain form a plane through the origin.

range: $\mathbb{R}$
24.

domain: $x^{2}+y^{2}+z^{2}\neq 1$ ; the set of points in $\mathbb{R}^{3}$ NOT in the domain form a sphere of radius 1.

range: $(-\infty,0)\cup[1,\infty)$
25.

domain: $z\geq x^{2}-y^{2}$ ; the set of points in $\mathbb{R}^{3}$ above (and including) the hyperbolic paraboloid $z=x^{2}-y^{2}$ .

range: $[0,\infty)$
26.

domain: $\mathbb{R}^{3}$

range: $\mathbb{R}$
27.

The level surfaces are spheres, centered at the origin, with radius $\sqrt{c}$ .
28.

The level surfaces are hyperbolic paraboloids of the form $z=x^{2}-y^{2}+c$ ; each is shifted up/down by $c$ .
29.

The level surfaces are paraboloids of the form $z=\frac{x^{2}}{c}+\frac{y^{2}}{c}$ ; the larger $c$ , the “wider” the paraboloid.
30.

The level surfaces are planes through the origin of the form $cx-cy-z=0$ , that is, planes through the origin with normal vector $\left\langle c,-c,-1\right\rangle$ .
31.

The level curves for each surface are similar; for $z=\sqrt{x^{2}+4y^{2}}$ the level curves are ellipses of the form $\frac{x^{2}}{c^{2}}+\frac{y^{2}}{c^{2}/4}=1$ , i.e., $a=c$ and $b=c/2$ ; whereas for $z=x^{2}+4y^{2}$ the level curves are ellipses of the form $\frac{x^{2}}{c}+\frac{y^{2}}{c/4}=1$ , i.e., $a=\sqrt{c}$ and $b=\sqrt{c}/2$ . The first set of ellipses are spaced evenly apart, meaning the function grows at a constant rate; the second set of ellipses are more closely spaced together as $c$ grows, meaning the function grows faster and faster as $c$ increases.

The function $z=\sqrt{x^{2}+4y^{2}}$ can be rewritten as $z^{2}=x^{2}+4y^{2}$ , an elliptic cone; the function $z=x^{2}+4y^{2}$ is a paraboloid, each matching the description above.

Exercises M.2

1.

Answers will vary.
2.

Answers will vary. One answer is “As $(x,y)$ gets close to $(1,2)$ , $f(x,y)$ gets close to 17.”
3.

Answers will vary.

One possible answer: $\{(x,y)|x^{2}+y^{2}\leq 1\}$
4.

Answers will vary.

One possible answer: $\{(x,y)|y\geq x^{2}\}$
5.

Answers will vary.

One possible answer: $\{(x,y)|x^{2}+y^{2}<1\}$
6.

Answers will vary.

One possible answer: $\{(x,y)|y>x^{2}\}$
7.
- Answers will vary.
  interior point: $(1,3)$
  boundary point: $(3,3)$
  
  $S$ is a closed set
  
  $S$ is bounded
8.
- Answers will vary.
  interior point: $(-5,28)$
  boundary point: $(3,9)$
  
  $S$ is an open set
  
  $S$ is unbounded
9.
- Answers will vary.
  interior point: none
  boundary point: $(0,-1)$
  
  $S$ is a closed set, consisting only of boundary points
  
  $S$ is bounded
10.
- Answers will vary.
  Interior point: $(0,1)$
  Boundary point: $(0,0)$
  
  $S$ is a closed set, containing all of its boundary points.
  
  $S$ is unbounded.
11.
- $D=\left\{(x,y)\,|\,9-x^{2}-y^{2}\geq 0\right\}$ .
  
  $D$ is a closed set.
  
  $D$ is bounded.
12.
- $D=\left\{(x,y)\,|\,y\geq x^{2}\right\}$ .
  
  $D$ is a closed set.
  
  $D$ is unbounded.
13.
- $D=\left\{(x,y)\,|\,y>x^{2}\right\}$ .
  
  $D$ is an open set.
  
  $D$ is unbounded.
14.
- $D=\left\{(x,y)\,|\,(x,y)\neq(0,0)\right\}$ .
  
  $D$ is an open set.
  
  $D$ is unbounded.
15.
- $D=\mathbb{R}^{2}$ .
  
  $D$ is an open set.
  
  $D$ is unbounded.
16.
- $D=\{(x,y)\,|\,x-2y+3\neq 0\}$ . This is all points in the plane except those on the line $y=\frac{x}{2}-\frac{3}{2}$ .
  
  $D$ is an open set.
  
  $D$ is unbounded.
17.
- $D=\{(x,y)\,|\,x^{2}+y^{2}<1\}$ . This is the open disk of radius 1 centered at the origin.
  
  $D$ is an open set.
  
  $D$ is bounded.
18.
- $D=\{(x,y)\,|\,x,y>0\text{ or }x,y<0\}$ . This is the first and third quadrants of the $x y$ -plane without the axes.
  
  $D$ is an open set.
  
  $D$ is unbounded.
19.
- Along $y=0$ , the limit is 1.
  
  Along $x=0$ , the limit is $-1$ .
Since the above limits are not equal, the limit does not exist.
20.
- Along $y=mx$ , the limit is $\frac{m+1}{m-1}$ .
Since the above limit varies according to what $m$ is used, each limit is different, meaning the overall limit does not exist.
21.
- Along $y=mx$ , the limit is $0$ .
  
  Along $x=0$ , the limit is $-1$ .
Since the above limits are not equal, the limit does not exist.
22.
- Along $y=mx$ , the limit is:
  
  $\displaystyle\lim_{(x,y)\to(0,0)}\frac{\sin(x^{2})}{y}$ $\displaystyle=\lim_{x\to 0}\frac{\sin(x^{2})}{mx}$
  
  apply L’Hôpital’s Rule
  
  $\displaystyle=\lim_{x\to 0}\frac{2x\cos(x^{2})}{m}$
  
  $\displaystyle=0.$
  
  Along $x=0$ , the limit is:
  
  $\lim_{(x,y)\to(0,0)}\frac{\sin(x^{2})}{y}=\lim_{x\to 0}\frac{\sin(x^{2})}{x^{2% }}.$
  
  This can be evaluated with L’Hôpital’s Rule or from known limits; it is 1.
Since the limits along the lines $y=mx$ are not the same as the limit along $y=x^{2}$ , the overall limit does not exist.
23.
- Along $y=2$ , the limit is:
  
  $\displaystyle\lim_{(x,y)\to(1,2)}\frac{x+y-3}{x^{2}-1}$ $\displaystyle=\lim_{x\to 1}\frac{x-1}{x^{2}-1}$
  
  $\displaystyle=\lim_{x\to 1}\frac{1}{x+1}$
  
  $\displaystyle=1/2.$
  
  Along $y=x+1$ , the limit is:
  
  $\displaystyle\lim_{(x,y)\to(1,2)}\frac{x+y-3}{x^{2}-1}$ $\displaystyle=\lim_{x\to 1}\frac{2(x-1)}{x^{2}-1}$
  
  $\displaystyle=\lim_{x\to 1}\frac{2}{x+1}$
  
  $\displaystyle=1.$
Since the limits along the lines $y=2$ and $y=x+1$ differ, the overall limit does not exist.
24.
- Along $x=\pi$ , the limit is:
  
  $\displaystyle\lim_{(x,y)\to(\pi,\pi/2)}\frac{\sin x}{\cos y}$ $\displaystyle=\lim_{y\to\pi/2}\frac{0}{\cos y}$
  
  $\displaystyle=0.$
  
  Along $y=x-\pi/2$ , the limit is:
  
  $\displaystyle\lim_{(x,y)\to(\pi,\pi/2)}\frac{\sin x}{\cos y}$ $\displaystyle=\lim_{x\to\pi}\frac{\sin x}{\cos(x-\pi/2)}$
  
  Apply L’Hôpital’s Rule:
  
  $\displaystyle=\lim_{x\to\pi}\frac{\cos x}{\sin(x-\pi/2)}$
  
  $\displaystyle=-1.$
Since the limits along the lines $x=\pi$ and $y=x-\pi$ differ, the overall limit does not exist.
25.

Hint: Consider $-\left\lvert\sin(xy)\right\rvert\leq\sin(xy)\cos\left(\frac{1}{x^{2}+y^{2}}% \right)\leq\left\lvert\sin(xy)\right\rvert$ .
26.

Hint: For the second part, consider the path $y=x^{2}$ .
27.

Hint: $x^{4}-9y^{4}=(x^{2}+3y^{2})(x^{2}-3y^{2})$ .

Exercises M.3

1.

A constant is a number that is added or subtracted in an expression; a coefficient is a number that is being multiplied by a nonconstant function.
2.

Answers will vary; each should include something about treating $y$ as a constant or a coefficient.
3.

$f_{x}$
4.

$f_{y}$
5.

$f_{x}=2xy-1$ , $f_{y}=x^{2}+2$

$f_{x}(1,2)=3$ , $f_{y}(1,2)=3$
6.

$f_{x}=3x^{2}-3$ , $f_{y}=2y-6$

$f_{x}(-1,3)=0$ , $f_{y}(-1,3)=0$
7.

$f_{x}=-\sin x\sin y$ , $f_{y}=\cos x\cos y$

$f_{x}(\pi/3,\pi/3)=-3/4$ , $f_{y}(\pi/3,\pi/3)=1/4$
8.

$f_{x}=1/x$ , $f_{y}=1/y$

$f_{x}(-2,-3)=-1/2$ , $f_{y}(-2,-3)=-1/3$
9.

$f_{x}=2xy+6x$ , $f_{y}=x^{2}+4$

$f_{xx}=2y+6$ , $f_{yy}=0$

$f_{xy}=2x$ , $f_{yx}=2x$
10.

$f_{x}=3x^{2}+6xy+3y^{2}$ , $f_{y}=3x^{2}+6xy+3y^{2}$

$f_{xx}=6x+6y$ , $f_{yy}=6x+6y$

$f_{xy}=6x+6y$ , $f_{yx}=6x+6y$
11.

$f_{x}=1/y$ , $f_{y}=-x/y^{2}$

$f_{xx}=0$ , $f_{yy}=2x/y^{3}$

$f_{xy}=-1/y^{2}$ , $f_{yx}=-1/y^{2}$
12.

$f_{x}=-4/(x^{2}y)$ , $f_{y}=-4/(xy^{2})$

$f_{xx}=8/(x^{3}y)$ , $f_{yy}=8/(xy^{3})$

$f_{xy}=4/(x^{2}y^{2})$ , $f_{yx}=4/(x^{2}y^{2})$
13.

$f_{x}=2xe^{x^{2}+y^{2}}$ , $f_{y}=2ye^{x^{2}+y^{2}}$

$f_{xx}=2e^{x^{2}+y^{2}}+4x^{2}e^{x^{2}+y^{2}}$ , $f_{yy}=2e^{x^{2}+y^{2}}+4y^{2}e^{x^{2}+y^{2}}$

$f_{xy}=4xye^{x^{2}+y^{2}}$ , $f_{yx}=4xye^{x^{2}+y^{2}}$
14.

$f_{x}=e^{x+2y}$ , $f_{y}=2e^{x+2y}$

$f_{xx}=e^{x+2y}$ , $f_{yy}=4e^{x+2y}$

$f_{xy}=2e^{x+2y}$ , $f_{yx}=2e^{x+2y}$
15.

$f_{x}=\cos x\cos y$ , $f_{y}=-\sin x\sin y$

$f_{xx}=-\sin x\cos y$ , $f_{yy}=-\sin x\cos y$

$f_{xy}=-\sin y\cos x$ , $f_{yx}=-\sin y\cos x$
16.

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

$f_{xx}=6(x+y)$ , $f_{yy}=6(x+y)$

$f_{xy}=6(x+y)$ , $f_{yx}=6(x+y)$
17.

$f_{x}=-5y^{3}\sin(5xy^{3})$ , $f_{y}=-15xy^{2}\sin(5xy^{3})$

$f_{xx}=-25y^{6}\cos(5xy^{3})$ , $f_{yy}=-225x^{2}y^{4}\cos(5xy^{3})-30xy\sin(5xy^{3})$

$f_{xy}=-75xy^{5}\cos(5xy^{3})-15y^{2}\sin(5xy^{3})$ , $f_{yx}=-75xy^{5}\cos(5xy^{3})-15y^{2}\sin(5xy^{3})$
18.

$f_{x}=10x\cos(5x^{2}+2y^{3})$ , $f_{y}=6y^{2}\cos(5x^{2}+2y^{3})$

$f_{xx}=10\cos(5x^{2}+2y^{3})-100x^{2}\sin(5x^{2}+2y^{3})$ , $f_{yy}=12y\cos(5x^{2}+2y^{3})-36y^{4}\sin(5x^{2}+2y^{3})$

$f_{xy}=-60xy^{2}\sin(5x^{2}+2y^{3})$ , $f_{yx}=-60xy^{2}\sin(5x^{2}+2y^{3})$
19.

$f_{x}=\frac{2y^{2}}{\sqrt{4xy^{2}+1}}$ , $f_{y}=\frac{4xy}{\sqrt{4xy^{2}+1}}$

$f_{xx}=-\frac{4y^{4}}{\sqrt{4xy^{2}+1}^{3}}$ , $f_{yy}=-\frac{16x^{2}y^{2}}{\sqrt{4xy^{2}+1}^{3}}+\frac{4x}{\sqrt{4xy^{2}+1}}$

$f_{xy}=-\frac{8xy^{3}}{\sqrt{4xy^{2}+1}^{3}}+\frac{4y}{\sqrt{4xy^{2}+1}}$ , $f_{yx}=-\frac{8xy^{3}}{\sqrt{4xy^{2}+1}^{3}}+\frac{4y}{\sqrt{4xy^{2}+1}}$
20.

$f_{x}=2\sqrt{y}$ , $f_{y}=5\sqrt{y}+\frac{2x+5y}{2\sqrt{y}}$

$f_{xx}=0$ , $f_{yy}=\frac{5}{\sqrt{y}}-\frac{2x+5y}{4y^{3/2}}$

$f_{xy}=\frac{1}{\sqrt{y}}$ , $f_{yx}=\frac{1}{\sqrt{y}}$
21.

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

$f_{xx}=\frac{8x^{2}}{(x^{2}+y^{2}+1)^{3}}-\frac{2}{(x^{2}+y^{2}+1)^{2}}$ , $f_{yy}=\frac{8y^{2}}{(x^{2}+y^{2}+1)^{3}}-\frac{2}{(x^{2}+y^{2}+1)^{2}}$

$f_{xy}=\frac{8xy}{(x^{2}+y^{2}+1)^{3}}$ , $f_{yx}=\frac{8xy}{(x^{2}+y^{2}+1)^{3}}$
22.

$f_{x}=5$ , $f_{y}=-17$

$f_{xx}=0$ , $f_{yy}=0$

$f_{xy}=0$ , $f_{yx}=0$
23.

$f_{x}=6x$ , $f_{y}=0$

$f_{xx}=6$ , $f_{yy}=0$

$f_{xy}=0$ , $f_{yx}=0$
24.

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

$f_{xx}=-\frac{4x^{2}}{(x^{2}+y)^{2}}+\frac{2}{(x^{2}+y)}$ , $f_{yy}=-\frac{1}{(x^{2}+y)^{2}}$

$f_{xy}=-\frac{2x}{(x^{2}+y)^{2}}$ , $f_{yx}=-\frac{2x}{(x^{2}+y)^{2}}$
25.

$f_{x}=\frac{1}{4xy}$ , $f_{y}=-\frac{\ln x}{4y^{2}}$

$f_{xx}=-\frac{1}{4x^{2}y}$ , $f_{yy}=\frac{\ln x}{2y^{3}}$

$f_{xy}=-\frac{1}{4xy^{2}}$ , $f_{yx}=-\frac{1}{4xy^{2}}$
26.

$f_{x}=5e^{x}\sin y$ , $f_{y}=5e^{x}\cos y$

$f_{xx}=5e^{x}\sin y$ , $f_{yy}=-5e^{x}\sin y$

$f_{xy}=5e^{x}\cos y$ , $f_{yx}=5e^{x}\cos y$
27.

$f(x,y)=x\sin y+x+C$ , where $C$ is any constant.
28.

$f(x,y)=\frac{1}{2}x^{2}+xy+\frac{1}{2}y^{2}+C$ , where $C$ is any constant.
29.

$f(x,y)=3x^{2}y-4xy^{2}+2y+C$ , where $C$ is any constant.
30.

$f(x,y)=\ln(x^{2}+y^{2})+C$ , where $C$ is any constant.
31.

$f_{x}=2xe^{2y-3z}$ , $f_{y}=2x^{2}e^{2y-3z}$ , $f_{z}=-3x^{2}e^{2y-3z}$

$f_{yz}=-6x^{2}e^{2y-3z}$ , $f_{zy}=-6x^{2}e^{2y-3z}$
32.

$f_{x}=3x^{2}y^{2}+3x^{2}z$ , $f_{y}=2x^{3}y+2yz$ , $f_{z}=x^{3}+y^{2}$

$f_{yz}=2y$ , $f_{zy}=2y$
33.

$f_{x}=\frac{3}{7y^{2}z}$ , $f_{y}=-\frac{6x}{7y^{3}z}$ , $f_{z}=-\frac{3x}{7y^{2}z^{2}}$

$f_{yz}=\frac{6x}{7y^{3}z^{2}}$ , $f_{zy}=\frac{6x}{7y^{3}z^{2}}$
34.

$f_{x}=\frac{1}{x}$ , $f_{y}=\frac{1}{y}$ , $f_{z}=\frac{1}{z}$

$f_{yz}=0$ , $f_{zy}=0$

Exercises M.4

1.

T
2.

T
3.

T
4.

amount of change
5.

$\operatorname{d}\!z=(\sin y+2x)\operatorname{d}\!x+(x\cos y)\operatorname{d}\!y$
6.

$\operatorname{d}\!z=8x(2x^{2}+3y)\operatorname{d}\!x+6(2x^{2}+3y)\operatorname% {d}\!y$
7.

$\operatorname{d}\!z=5\operatorname{d}\!x-7\operatorname{d}\!y$
8.

$\operatorname{d}\!z=(e^{x+y}+xe^{x+y})\operatorname{d}\!x+xe^{x+y}% \operatorname{d}\!y$
9.

$\operatorname{d}\!z=\frac{x}{\sqrt{x^{2}+y}}\operatorname{d}\!x+\frac{1}{2% \sqrt{x^{2}+y}}\operatorname{d}\!y$ , with $\operatorname{d}\!x=-0.05$ and $\operatorname{d}\!y=.1$ . At $(3,7)$ , $\operatorname{d}\!z=3/4(-0.05)+1/8(.1)=-0.025$ , so $f(2.95,7.1)\approx-0.025+4=3.975$ .
10.

$\operatorname{d}\!z=(\cos x\cos y)\operatorname{d}\!x-(\sin x\sin y)% \operatorname{d}\!y$ , with $\operatorname{d}\!x=0.1$ and $\operatorname{d}\!y=-0.1$ . At $(0,0)$ , $\operatorname{d}\!z=1(.1)-(0)(-0.1)=0.1$ , so $f(0.1,-0.1)\approx 0.1+0=0.1$ .
11.

$\operatorname{d}\!z=(2xy-y^{2})\operatorname{d}\!x+(x^{2}-2xy)\operatorname{d}\!y$ , with $\operatorname{d}\!x=0.04$ and $\operatorname{d}\!y=0.06$ . At $(2,3)$ , $\operatorname{d}\!z=3(0.04)+(-8)(0.06)=-0.36$ , so $f(2.04,3.06)\approx-0.36-6=-6.36$ .
12.

$\operatorname{d}\!z=\frac{1}{x-y}\operatorname{d}\!x-\frac{1}{x-y}% \operatorname{d}\!y$ , with $\operatorname{d}\!x=0.1$ and $\operatorname{d}\!y=-0.02$ . At $(5,4)$ , $\operatorname{d}\!z=1(0.1)+(-1)(-0.02)=0.12$ , so $f(5.1,3.98)\approx 0.12+0=0.12$ .
13.

The total differential of volume is $\operatorname{d}\!V=4\pi\operatorname{d}\!r+\pi\operatorname{d}\!h$ . The coefficient of $\operatorname{d}\!r$ is greater than the coefficient of $\operatorname{d}\!h$ , so the volume is more sensitive to changes in the radius.
14.

Distance of the projectile is a function of two variables (leaving $t=3$ ): $D(v_{0},\theta)=3v_{0}\cos\theta$ . The total differential of $D$ is $\operatorname{d}\!D=3\cos\theta\operatorname{d}\!v_{0}-3v_{0}\sin\theta% \operatorname{d}\!\theta$ . The coefficient of $\operatorname{d}\!\theta$ has a much greater magnitude than the coefficient of $\operatorname{d}\!v_{0}$ , so a small change in the angle of elevation has a much greater effect on distance traveled than a small change in initial velocity.
15.

Using trigonometry, $\ell=x\tan\theta$ , so $\operatorname{d}\!\ell=\tan\theta\operatorname{d}\!x+x\sec^{2}\theta% \operatorname{d}\!\theta$ . With $\theta=85^{\circ}$ and $x=30$ , we have $\operatorname{d}\!\ell=11.43\operatorname{d}\!x+3949.38\operatorname{d}\!\theta$ . The measured length of the wall is much more sensitive to errors in $\theta$ than in $x$ . While it can be difficult to compare sensitivities between measuring feet and measuring degrees (it is somewhat like “comparing apples to oranges”), here the coefficients are so different that the result is clear: a small error in degree has a much greater impact than a small error in distance.
16.

With $D=n\ell$ , the total differential is $\operatorname{d}\!D=\ell\operatorname{d}\!n+n\operatorname{d}\!\ell.$ If one measures with a short tape, $n$ must be large and hence $n\operatorname{d}\!\ell$ is going to be greater than when a large tape is used (wherein $n$ will be small).
17.

$\operatorname{d}\!w=2xyz^{3}\operatorname{d}\!x+x^{2}z^{3}\operatorname{d}\!y+% 3x^{2}yz^{2}\operatorname{d}\!z$
18.

$\operatorname{d}\!w=e^{x}\sin y\ln z\operatorname{d}\!x+e^{x}\cos y\ln z% \operatorname{d}\!y+e^{x}\sin y\frac{1}{z}\operatorname{d}\!z$
19.

$\operatorname{d}\!x=0.05$ , $\operatorname{d}\!y=-0.1$ . $\operatorname{d}\!z=9(.05)+(-2)(-0.1)=0.65$ . So $f(3.05,0.9)\approx 7+0.65=7.65$ .
20.

$\operatorname{d}\!x=-0.12$ , $\operatorname{d}\!y=0.07$ . $\operatorname{d}\!z=2.6(-.12)+(5.1)(0.07)=0.045$ . So $f(-4.12,2.07)\approx 13+0.045=13.045$ .
21.

$\operatorname{d}\!x=0.5$ , $\operatorname{d}\!y=0.1$ , $\operatorname{d}\!z=-0.2$ .

$\operatorname{d}\!w=2(0.5)+(-3)(0.1)+3.7(-0.2)=-0.04$ , so $f(2.5,4.1,4.8)\approx-1-0.04=-1.04$ .
22.

$\operatorname{d}\!x=0.1$ , $\operatorname{d}\!y=0.1$ , $\operatorname{d}\!z=0.1$ .

$\operatorname{d}\!w=2(0.1)+(0)(0.1)+(-2)(.1)=0$ , so $f(3.1,3.1,3.1)\approx 5+0=5$ .
23.

Everywhere except the origin.

Exercises M.5

1.

Because the parametric equations describe a level curve, $z$ is constant for all $t$ . Therefore $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=0$ .
2.

$g^{\prime}(x)$
3.

$\frac{\operatorname{d}\!x}{\operatorname{d}\!t}$ , and $\frac{\partial f}{\partial y}$
4.

T
5.

F
6.

partial
7.
- $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=3(2t)+4(2)=6t+8$ .
  
  At $t=1$ , $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=14$ .
8.
- $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=2x(1)-2y(2t)=2x-4yt$
  
  At $t=1$ , $x=1$ , $y=0$ and $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=2$ .
9.
- $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=5(-2\sin t)+2(\cos t)=-10\sin t% +2\cos t$
  
  At $t=\pi/4$ , $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=-4\sqrt{2}$ .
10.
- $\displaystyle\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=\frac{1}{1+y^{2}}% (-\sin t)-\frac{2xy}{(y^{2}+1)^{2}}(\cos t)$ .
  
  At $t=\pi/2$ , $x=0$ , $y=1$ , and $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=-1/2$ .
11.
- $\displaystyle\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=2x(\cos t)+4y(3% \cos t)$ .
  
  At $t=\pi/4$ , $x=\sqrt{2}/2$ , $y=3\sqrt{2}/2$ , and $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=19$ .
12.
- $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=-\sin x\sin y(\pi)+\cos x\cos y% (2\pi)$ .
  
  At $t=3$ , $x=3\pi$ , $y=13\pi/2$ , and $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=0$ .
13.

$t=-4/3$ ; this corresponds to a minimum
14.

$t=0,\pm\sqrt{3/2}$
15.

$t=\tan^{-1}(1/5)+n\pi$ , where $n$ is an integer
16.

We find that

$\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=-\frac{2\cos^{2}t\sin t}{(1+% \sin^{2}t)^{2}}-\frac{\sin t}{1+\sin^{2}t}.$

Setting this equal to 0, finding a common denominator and factoring out $\sin t$ , we get

$\sin t\left(\frac{2\cos^{2}t+\sin^{2}t+1}{(1+\sin^{2}t)^{2}}\right)=0.$

We have $\sin t=0$ when $t=\pi n$ , where $n$ is an integer. The expression in the parenthesis above is always positive, and hence never equal 0. So all solutions are $t=\pi n$ , n is an integer.
17.

We find that

$\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=38\cos t\sin t.$

Thus $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=0$ when $t=\pi n$ or $\pi n+\pi/2$ , where $n$ is any integer.
18.

We find that $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=$

$-\pi\sin(\pi t)\sin(2\pi t+\pi/2)+2\pi\cos(\pi t)\cos(2\pi t+\pi/2).$

One can “easily” see that when $t$ is an integer, $\sin(\pi t)=0$ and $\cos(2\pi t+\pi/2)=0$ , hence $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=0$ when $t$ is an integer. There are other places where $z$ has a relative max/min that require more work. First, verify that $\sin(2\pi t+\pi/2)=\cos(2\pi t)$ , and $\cos(2\pi t+\pi/2)=-\sin(2\pi t)$ . This lets us rewrite $\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=0$ as

$-\sin(\pi t)\cos(2\pi t)-2\cos(\pi t)\sin(2\pi t)=0.$

By bringing one term to the other side of the equality then dividing, we can get

$2\tan(2\pi t)=-\tan(\pi t).$

Using the angle sum/difference formulas found in the back of the book, we know

$\tan(2\pi t)=\tan(\pi t)+\tan(\pi t)=\frac{\tan(\pi t)+\tan(\pi t)}{1-\tan^{2}% (\pi t)}.$

Thus we write

$2\frac{\tan(\pi t)+\tan(\pi t)}{1-\tan^{2}(\pi t)}=-\tan(\pi t).$

Solving for $\tan^{2}(\pi t)$ , we find

$\tan^{2}(\pi t)=5\quad\Rightarrow\quad\tan(\pi t)=\pm\sqrt{5},$

and so

$\pi t=\tan^{-1}(\pm\sqrt{5})=\pm\tan^{-1}(\sqrt{5}).$

Since the period of tangent is $\pi$ , we can adjust our answer to be

$\pi t=\pm\tan^{-1}(\sqrt{5})+n\pi,\text{ where $n$ is an integer.}$

Dividing by $\pi$ , we find

$t=\pm\frac{1}{\pi}\tan^{-1}(\sqrt{5})+n,\text{ where $n$ is an integer.}$
19.
- $\frac{\partial z}{\partial s}=2xy(1)+x^{2}(2)=2xy+2x^{2}$ ;
  $\frac{\partial z}{\partial t}=2xy(-1)+x^{2}(4)=-2xy+4x^{2}$
  
  With $s=1$ , $t=0$ , $x=1$ and $y=2$ . Thus $\frac{\partial z}{\partial s}=6$ and $\frac{\partial z}{\partial t}=0$
20.
- $\frac{\partial z}{\partial s}=-\pi\sin(\pi x+\pi y/2)(t^{2})-\frac{1}{2}\pi% \sin(\pi x+\pi y/2)(2st)=-\pi\Big(t^{2}\sin(\pi x+\pi y/2)+st\sin(\pi x+\pi y/% 2)\Big)$ ;
  $\frac{\partial z}{\partial t}=-\pi\sin(\pi x+\pi y/2)(2st)-\frac{1}{2}\pi\sin(% \pi x+\pi y/2)(s^{2})=-\pi\Big(2st\sin(\pi x+\pi y/2)+\frac{1}{2}s^{2}\sin(\pi x% +\pi y/2)\Big)$
  
  With $s=1$ , $t=1$ , $x=1$ and $y=1$ . Thus $\frac{\partial z}{\partial s}=2\pi$ and $\frac{\partial z}{\partial t}=5\pi/2$
21.
- $\frac{\partial z}{\partial s}=2x(\cos t)+2y(\sin t)=2x\cos t+2y\sin t$ ;
  $\frac{\partial z}{\partial t}=2x(-s\sin t)+2y(s\cos t)=-2xs\sin t+2ys\cos t$
  
  With $s=2$ , $t=\pi/4$ , $x=\sqrt{2}$ and $y=\sqrt{2}$ . Thus $\frac{\partial z}{\partial s}=4$ and $\frac{\partial z}{\partial t}=0$
22.
- $\frac{\partial z}{\partial s}=-2xe^{-(x^{2}+y^{2})}(0)-2ye^{-(x^{2}+y^{2})}(t^% {2})=-2yt^{2}e^{-(x^{2}+y^{2})}$ ;
  $\frac{\partial z}{\partial t}=-2xe^{-(x^{2}+y^{2})}(1)-2ye^{-(x^{2}+y^{2})}(2% st)=-2xe^{-(x^{2}+y^{2})}-4stye^{-(x^{2}+y^{2})}$
  
  With $s=1$ , $t=1$ , $x=1$ and $y=1$ . Thus $\frac{\partial z}{\partial s}=-2/e^{2}$ and $\frac{\partial z}{\partial t}=-6/e^{2}$
23.

$f_{x}=2x\tan y$ , $f_{y}=x^{2}\sec^{2}y$ ;

$\displaystyle\frac{\operatorname{d}\!y}{\operatorname{d}\!x}=-\frac{2\tan y}{x% \sec^{2}y}$
24.

$f_{x}=4(3x^{2}+2y^{3})^{3}(6x)$ , $f_{y}=4(3x^{2}+2y^{3})^{3}(6y^{2})$ ;

$\displaystyle\frac{\operatorname{d}\!y}{\operatorname{d}\!x}=-\frac{x}{y^{2}}$
25.

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

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

$\displaystyle\frac{\operatorname{d}\!y}{\operatorname{d}\!x}=-\frac{2x(x+y^{2}% )-(x^{2}+y)}{x+y^{2}-2y(x^{2}+y)}$
26.

$\displaystyle f_{x}=\frac{2x+y}{x^{2}+xy+y^{2}}$ , $\displaystyle f_{y}=\frac{x+2y}{x^{2}+xy+y^{2}}$ ;

$\displaystyle\frac{\operatorname{d}\!y}{\operatorname{d}\!x}=-\frac{2x+y}{2y+x}$
27.

$\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=2(4)+1(-5)=3$ .
28.

$\frac{\operatorname{d}\!z}{\operatorname{d}\!t}=1(6)+(-3)(2)=0$ .
29.

$\frac{\partial z}{\partial s}=-4(5)+9(-2)=-38$ ,

$\frac{\partial z}{\partial t}=-4(7)+9(6)=26$ .
30.

$\frac{\partial z}{\partial s}=2(-2)+1(2)=-2$ ,

$\frac{\partial z}{\partial t}=2(3)+1(-1)=5$ .
31.

$z_{\theta}=-f_{x}(x,y)r\sin\theta+f_{y}(x,y)r\cos\theta$
32.

$w_{\phi}=f_{x}(x,y,z)\rho\cos\phi\cos\theta+f_{y}(x,y,z)\rho\cos\phi\sin\theta% -f_{z}(x,y,z)\rho\sin\phi$
33.

It is increasing at $104\pi/3$ cm³/sec

Exercises M.6

1.

A partial derivative is essentially a special case of a directional derivative; it is the directional derivative in the direction of $x$ or $y$ , i.e., $\left\langle 1,0\right\rangle$ or $\left\langle 0,1\right\rangle$ .
2.

$\vec{u}=\left\langle 1,0\right\rangle$
3.

$\vec{u}=\left\langle 0,1\right\rangle$
4.

orthogonal
5.

maximal, or greatest
6.

dot
7.

$\nabla f=\left\langle-2xy+y^{2}+y,-x^{2}+2xy+x\right\rangle$
8.

$\nabla f=\left\langle\cos x\cos y,-\sin x\sin y\right\rangle$
9.

$\nabla f=\left\langle\frac{-2x}{(x^{2}+y^{2}+1)^{2}},\frac{-2y}{(x^{2}+y^{2}+1% )^{2}}\right\rangle$
10.

$\nabla f=\left\langle-4,3\right\rangle$
11.

$\nabla f=\left\langle 2x-y-7,4y-x\right\rangle$
12.

$\nabla f=\left\langle 2xy^{3}-2,3x^{2}y^{2}\right\rangle$
13.
$\nabla f=\left\langle-2xy+y^{2}+y,-x^{2}+2xy+x\right\rangle$ ; $\nabla f(2,1)=\left\langle-2,2\right\rangle$ . Be sure to change all directions to unit vectors.
- $2/5$ ( $\vec{u}=\left\langle 3/5,4/5\right\rangle$ )
  
  $-2/\sqrt{5}$ ( $\vec{u}=\left\langle-1/\sqrt{5},-2/\sqrt{5}\right\rangle$ )
14.
$\nabla f=\left\langle\cos x\cos y,-\sin x\sin y\right\rangle$ ;

$\nabla f(\frac{\pi}{4},\frac{\pi}{3})=\left\langle\frac{1}{2\sqrt{2}},-\frac{1% }{2}\sqrt{\frac{3}{2}}\right\rangle$ .

Be sure to change all directions to unit vectors.
- $\frac{1}{4}(1-\sqrt{3})$ ( $\vec{u}=\left\langle 1/\sqrt{2},1/\sqrt{2}\right\rangle$ )
  
  $\frac{4\sqrt{3}-1}{10\sqrt{2}}$ ( $\vec{u}=\left\langle-3/5,-4/5\right\rangle$ )
15.
$\nabla f=\left\langle\frac{-2x}{(x^{2}+y^{2}+1)^{2}},\frac{-2y}{(x^{2}+y^{2}+1% )^{2}}\right\rangle$ ;

$\nabla f(1,1)=\left\langle-2/9,-2/9\right\rangle$ .

Be sure to change all directions to unit vectors.
- 0 ( $\vec{u}=\left\langle 1/\sqrt{2},-1/\sqrt{2}\right\rangle$ )
  
  $2\sqrt{2}/9$ ( $\vec{u}=\left\langle-1/\sqrt{2},-1/\sqrt{2}\right\rangle$ )
16.
$\nabla f=\left\langle-4,3\right\rangle$ ; $\nabla f(5,2)=\left\langle-4,3\right\rangle$ . Be sure to change all directions into unit vectors.
- $-9/\sqrt{10}$ ( $\vec{u}=\left\langle 3/\sqrt{10},1/\sqrt{10}\right\rangle$ )
  
  $27/\sqrt{34}$ ( $\vec{u}=\left\langle-3/\sqrt{34},5/\sqrt{34}\right\rangle$ )
17.
$\nabla f=\left\langle 2x-y-7,4y-x\right\rangle$ ; $\nabla f(4,1)=\left\langle 0,0\right\rangle$ .
- 0
  
  0
18.
$\nabla f=\left\langle 2xy^{3}-2,3x^{2}y^{2}\right\rangle$ ; $\nabla f(1,1)=\left\langle 0,3\right\rangle$ Be sure to change all directions to unit vectors.
- $3/\sqrt{2}$ ; ( $\vec{u}=\left\langle 1/\sqrt{2},1/\sqrt{2}\right\rangle$ )
  
  $3$
19.
$\nabla f=\left\langle-2xy+y^{2}+y,-x^{2}+2xy+x\right\rangle$
- $\nabla f(2,1)=\left\langle-2,2\right\rangle$
  
  $\left\lVert\nabla f(2,1)\right\rVert=\left\lVert\left\langle-2,2\right\rangle% \right\rVert=\sqrt{8}$
  
  $\left\langle 2,-2\right\rangle$
  
  $\left\langle 1/\sqrt{2},1/\sqrt{2}\right\rangle$
20.
$\nabla f=\left\langle\cos x\cos y,-\sin x\sin y\right\rangle$
- $\nabla f(\frac{\pi}{4},\frac{\pi}{3})=\left\langle\frac{1}{2\sqrt{2}},-\frac{1% }{2}\sqrt{\frac{3}{2}}\right\rangle$
  
  $\left\lVert\nabla f(\frac{\pi}{4},\frac{\pi}{3})\right\rVert=\left\lVert\left% \langle\frac{1}{2\sqrt{2}},-\frac{1}{2}\sqrt{\frac{3}{2}}\right\rangle\right% \rVert=1/\sqrt{2}$
  
  $\left\langle-\frac{1}{2\sqrt{2}},\frac{1}{2}\sqrt{\frac{3}{2}}\right\rangle$
  
  $\left\langle\frac{1}{2}\sqrt{\frac{3}{2}},\frac{1}{2\sqrt{2}}\right\rangle$
21.
$\nabla f=\left\langle\frac{-2x}{(x^{2}+y^{2}+1)^{2}},\frac{-2y}{(x^{2}+y^{2}+1% )^{2}}\right\rangle$
- $\nabla f(1,1)=\left\langle-2/9,-2/9\right\rangle$ .
  
  $\left\lVert\nabla f(1,1)\right\rVert=\left\lVert\left\langle-2/9,-2/9\right% \rangle\right\rVert=2\sqrt{2}/9$
  
  $\left\langle 2/9,2/9\right\rangle$
  
  $\left\langle 1/\sqrt{2},-1/\sqrt{2}\right\rangle$
22.
$\nabla f=\left\langle-4,3\right\rangle$
- $\nabla f(5,4)=\left\langle-4,3\right\rangle$ .
  
  $\left\lVert\nabla f(5,4)\right\rVert=\left\lVert\left\langle-4,3\right\rangle% \right\rVert=5$
  
  $\left\langle 4,-3\right\rangle$
  
  $\left\langle 3/5,4/5\right\rangle$
23.
$\nabla f=\left\langle 2x-y-7,4y-x\right\rangle$
- $\nabla f(4,1)=\left\langle 0,0\right\rangle$
  
  0
  
  $\left\langle 0,0\right\rangle$
  
  All directions give a directional derivative of 0.
24.
$\nabla f=\left\langle 2xy^{3}-2,3x^{2}y^{2}\right\rangle$
- $\nabla f(1,1)=\left\langle 0,3\right\rangle$
  
  3
  
  $\left\langle 0,-3\right\rangle$
  
  $\vec{u}=\left\langle 1,0\right\rangle$
25.
- $\nabla F(x,y,z)=\left\langle 6xz^{3}+4y,4x,9x^{2}z^{2}-6z\right\rangle$
  
  $113/\sqrt{3}$
26.
- $\nabla F(x,y,z)=\\ \left\langle\cos x\cos ye^{z},-\sin x\sin ye^{z},\sin x\cos ye^{z}\right\rangle$
  
  $2/3$
27.
- $\nabla F(x,y,z)=\left\langle 2xy^{2},2y(x^{2}-z^{2}),-2y^{2}z\right\rangle$
  
  $0$
28.
- $\nabla F(x,y,z)=\\ \left\langle-\frac{4x}{(x^{2}+y^{2}+z^{2})^{2}},-\frac{4y}{(x^{2}+y^{2}+z^{2})% ^{2}},-\frac{4z}{(x^{2}+y^{2}+z^{2})^{2}}\right\rangle$
  
  $0$
29.

In the direction $\left\langle 7,8,-4\right\rangle$ with maximal value $\sqrt{129}$ .
30.

In the direction $\left\langle-1,0,-3\right\rangle$ with maximal value $\frac{\sqrt{10}}{8}$ .

Exercises M.7

1.

Answers will vary. The displacement of the vector is one unit in the $x$ -direction and 3 units in the $z$ -direction, with no change in $y$ . Thus along a line parallel to $\vec{v}$ , the change in $z$ is 3 times the change in $x$ — i.e., a “slope” of 3. Specifically, the line in the $x$ - $z$ plane parallel to $z$ has a slope of 3.
2.

Answers will vary. Let $\vec{u}=\left\langle 0.6,0.8\right\rangle$ ; this is a unit vector. The displacement of the vector is one unit in the $\vec{u}$ -direction and $-2$ units in the $z$ -direction. In the plane containing the $z$ -axis and the vector $\vec{u}$ , the line parallel to $\vec{v}$ has slope $-2$ .
3.

T
4.

On a surface through a point, there are many different smooth curves, each with a tangent line at the point. Each of these tangent lines is also “tangent” to the surface. There is not just one tangent line, but many, each in a different direction. Therefore we refer to directional tangent lines, not just the tangent line.
5.
- $\ell_{x}(t)=\begin{cases}x=2+t\\ y=3\\ z=-48-12t\end{cases}$
  
  $\ell_{y}(t)=\begin{cases}x=2\\ y=3+t\\ z=-48-40t\end{cases}$
  
  $\ell_{\vec{u}\,}(t)=\begin{cases}x=2+t/\sqrt{10}\\ y=3+3t/\sqrt{10}\\ z=-48-66\sqrt{2/5}t\end{cases}$
6.
- $\ell_{x}(t)=\begin{cases}x=\pi/3+t\\ y=\pi/6\\ z=3/4-\frac{3\sqrt{3}}{4}t\end{cases}$
  
  $\ell_{y}(t)=\begin{cases}x=\pi/3\\ y=\pi/6+t\\ z=3/4+\frac{3\sqrt{3}}{4}t\end{cases}$
  
  $\ell_{\vec{u}\,}(t)=\begin{cases}x=\pi/3+t/\sqrt{5}\\ y=\pi/6+2t/\sqrt{5}\\ z=3/4+\frac{3\sqrt{3/5}}{4}t\end{cases}$
7.
- $\ell_{x}(t)=\begin{cases}x=4+t\\ y=2\\ z=2+3t\end{cases}$
  
  $\ell_{y}(t)=\begin{cases}x=4\\ y=2+t\\ z=2-5t\end{cases}$
  
  $\ell_{\vec{u}\,}(t)=\begin{cases}x=4+t/\sqrt{2}\\ y=2+t/\sqrt{2}\\ z=2-\sqrt{2}t\end{cases}$
8.
- $\ell_{x}(t)=\begin{cases}x=1+t\\ y=2\\ z=3\end{cases}$
  
  $\ell_{y}(t)=\begin{cases}x=1\\ y=2+t\\ z=3\end{cases}$
  
  $\ell_{\vec{u}\,}(t)=\begin{cases}x=1+t/\sqrt{2}\\ y=2+t/\sqrt{2}\\ z=3\end{cases}$
9.

$\ell_{\vec{n}}(t)=\begin{cases}x=2-12t\\ y=3-40t\\ z=-48-t\end{cases}$
10.

$\ell_{\vec{n}}(t)=\begin{cases}x=\pi/3-\frac{3\sqrt{3}}{4}t\\ y=\pi/6+\frac{3\sqrt{3}}{4}t\\ z=3/4-t\end{cases}$
11.

$\ell_{\vec{n}}(t)=\begin{cases}x=4+3t\\ y=2-5t\\ z=2-t\end{cases}$
12.

$\ell_{\vec{n}}(t)=\begin{cases}x=1\\ y=2\\ z=3-t\end{cases}$
13.

$(1.425,1.085,-48.078)$ , $(2.575,4.915,-47.952)$
14.

$(-0.195,1.766,-0.206)$ and $(2.289,-0.719,1.706)$
15.

$(5.014,0.31,1.662)$ and $(2.986,3.690,2.338)$
16.

$(1,2,1)$ and $(1,2,5)$
17.

$-12(x-2)-40(y-3)-(z+48)=0$
18.

$-\frac{3\sqrt{3}}{4}(x-\pi/3)+\frac{3\sqrt{3}}{4}(y-\pi/6)-(z-3/4)=0$
19.

$3(x-4)-5(y-2)-(z-2)=0$ (Note that this tangent plane is the same as the original function, a plane.)
20.

$-(z-3)=0$ , or $z=3$
21.
$\nabla F=\left\langle x/4,y/2,z/8\right\rangle$ ; at $P$ , $\nabla F=\left\langle 1/4,\sqrt{2}/2,\sqrt{6}/8\right\rangle$
- $\ell_{\vec{n}}(t)=\begin{cases}x=1+t/4\\ y=\sqrt{2}+\sqrt{2}t/2\\ z=\sqrt{6}+\sqrt{6}t/8\end{cases}$
  
  $\frac{1}{4}(x-1)+\frac{\sqrt{2}}{2}(y-\sqrt{2})+\frac{\sqrt{6}}{8}(z-\sqrt{6})=0$ .
22.
$\nabla F=\left\langle-\frac{x}{2},-\frac{2y}{9},2z\right\rangle$ ; at $P$ , $\nabla F=\left\langle-2,2/3,2\sqrt{5}\right\rangle$
- $\ell_{\vec{n}}(t)=\begin{cases}x=4-2t\\ y=-3+2t/3\\ z=\sqrt{5}+2\sqrt{5}t\end{cases}$
  
  $-2(x-4)+\frac{2}{3}(y+3)+2\sqrt{5}(z-\sqrt{5})=0$ .
23.
$\nabla F=\left\langle y^{2}-z^{2},2xy,-2xz\right\rangle$ ; at $P$ , $\nabla F=\left\langle 0,4,4\right\rangle$
- $\ell_{\vec{n}}(t)=\begin{cases}x=2\\ y=1+4t\\ z=-1+4t\end{cases}$
  
  $4(y-1)+4(z+1)=0$ .
24.
$\nabla F=\left\langle y\cos(xy),x\cos(xy)-z\sin(yz),-y\sin(yz)\right\rangle$ ; at $P$ , $\nabla F=\left\langle\frac{\pi}{8\sqrt{3}},-\sqrt{3},-\frac{\pi}{8\sqrt{3}}\right\rangle$
- $\ell_{\vec{n}}(t)=\begin{cases}x=2+\frac{\pi}{8\sqrt{3}}t\\ y=\frac{\pi}{12}-\sqrt{3}t\\ z=4-\frac{\pi}{8\sqrt{3}}t\end{cases}$
  
  $\frac{\pi}{8\sqrt{3}}(x-2)-\sqrt{3}(y-\frac{\pi}{12})-\frac{\pi}{8\sqrt{3}}(z-% 4)=0$ .

Exercises M.8

1.

F; it is the “other way around.”
2.

T
3.

T
4.

Answers will vary. A good answer will state that we are optimizing a function subject to a constraint, or limit, on the domain of the function. We are looking to maximize/minimize the function while “looking” at only a certain part of the domain.
5.

One critical point at $(-4,2)$ ; $f_{xx}=1$ and $D=4$ , so this point corresponds to a relative minimum.
6.

One critical point at $(7,-6)$ ; $D=-5$ , so this point corresponds to a saddle point.
7.

One critical point at $(6,-3)$ ; $D=-4$ , so this point corresponds to a saddle point.
8.

One critical point at $(0,0)$ ; $f_{xx}=-2$ and $D=4$ , so this point corresponds to a relative maximum.
9.

Two critical points: at $(0,-1)$ ; $f_{xx}=2$ and $D=-12$ , so this point corresponds to a saddle point;

at $(0,1)$ , $f_{xx}=2$ and $D=12$ , so this corresponds to a relative minimum.
10.

There are 4 critical points:

$(-1,-2)$ , rel. max; $(1,-2)$ , saddle point;

$(-1,2)$ , saddle point; $(1,2)$ , rel. min.,

where $f_{xx}=2x$ and $D=4xy$ .
11.

Critical points when $x$ or $y$ are $0$ . $D=-12x^{2}y^{2}$ , so the test is inconclusive. (Some elementary thought shows that these are absolute minima.)
12.

Six critical points: $f_{x}=0$ when $x=-1,0$ and 1; $f_{y}=0$ when $y=-3,3$ . Together, we get the points:

$(-1,-3)$ saddle point; $(-1,3)$ rel. min

$(0,-3)$ rel. max; $(0,3)$ saddle point

$(1,-3)$ saddle point; $(1,3)$ relative min

where $f_{xx}=12x^{2}-4$ and $D=24y(3x^{2}-1)$ .
13.

One critical point: $f_{x}=0$ when $x=3$ ; $f_{y}=0$ when $y=0$ , so one critical point at $(3,0)$ , which is a relative maximum, where $f_{xx}=\frac{y^{2}-16}{(16-(x-3)^{2}-y^{2})^{3/2}}$ and $D=\frac{16}{(16-(x-3)^{2}-y^{2})^{2}}$ .

Both $f_{x}$ and $f_{y}$ are undefined along the circle $(x-3)^{2}+y^{2}=16$ ; at any point along this curve, $f(x,y)=0$ , the absolute minimum of the function.
14.

One critical point: $f_{x}=0$ when $x=0$ ; $f_{y}=0$ when $y=0$ , so one critical point at $(0,0)$ (although it should be noted that at $(0,0)$ , both $f_{x}$ and $f_{y}$ are undefined.) The Second Derivative Test fails at $(0,0)$ , with $D=0$ . A graph, or simple calculation, shows that $(0,0)$ is the absolute minimum of $f$ .
15.

rel. max at $(0,0)$ ; rel. min at $(2,0)$ ; saddle points at $(1,\pm 1)$ .
16.

rel. min at $(27/2,5)$ ; saddle point at $(3/2,1)$ .
17.

saddle points at $(1,2/3)$ and $(-1,-4/3)$ .
18.

rel. min at $(0,0)$ , saddle point at $(1,0)$ .
19.

The triangle is bound by the lines $y=-1$ , $y=2x+1$ and $y=-2x+1$ .

Along $y=-1$ , there is a critical point at $(0,-1)$ .

Along $y=2x+1$ , there is a critical point at $(-3/5,-1/5)$ .

Along $y=-2x+1$ , there is a critical point at $(3/5,-1/5)$ .

The function $f$ has one critical point, irrespective of the constraint, at $(0,-1/2)$ .

Checking the value of $f$ at these four points, along with the three vertices of the triangle, we find the absolute maximum is at $(0,1,3)$ and the absolute minimum is at $(0,-1/2,3/4)$ .
20.

The region has two “corners” at $(1,1)$ and $(-1,1)$ .

Along $y=1$ , there is no critical point.

Along $y=x^{2}$ , there is a critical point at $(5/14,25/196)\approx(0.357,0.128)$ .

The function $f$ itself has no critical points.

Checking the value of $f$ at the corners $(1,1)$ , $(-1,1)$ and the critical point $(5/14,25/196)$ , we find the absolute maximum is at $(5/14,25/196,25/28)\approx(0.357,0.128,0.893)$ and the absolute minimum is at $(-1,1,-12)$ .
21.

The region has no “corners” or “vertices,” just a smooth edge.

To find critical points along the circle $x^{2}+y^{2}=4$ , we solve for $y^{2}$ : $y^{2}=4-x^{2}$ . We can go further and state $y=\pm\sqrt{4-x^{2}}$ .

We can rewrite $f$ as $f(x)=x^{2}+2x+(4-x^{2})+2\sqrt{4-x^{2}}=2x+4+2\sqrt{4-x^{2}}$ . (We will return and use $-\sqrt{4-x^{2}}$ later.) Solving $f\,^{\prime}(x)=0$ , we get $x=\sqrt{2}\Rightarrow y=\sqrt{2}$ . $f\,^{\prime}(x)$ is also undefined at $x=\pm 2$ , where $y=0$ .

Using $y=-\sqrt{4-x^{2}}$ , we rewrite $f(x,y)$ as $f(x)=2x+4-2\sqrt{4-x^{2}}$ . Solving $f\,^{\prime}(x)=0$ , we get $x=-\sqrt{2},\ y=-\sqrt{2}$ .

The function $f$ itself has a critical point at $(-1,-1)$ .

Checking the value of $f$ at $(-1,-1)$ , $(\sqrt{2},\sqrt{2})$ , $(-\sqrt{2},-\sqrt{2})$ , $(2,0)$ and $(-2,0)$ , we find the absolute maximum is at $(\sqrt{2},\sqrt{2},4+4\sqrt{2})$ and the absolute minimum is at $(-1,-1,-2)$ .
22.

The region has two “corners” at $(-1,-1)$ and $(1,1)$ .

Along the line $y=x$ , $f(x,y)$ becomes $f(x)=3x-2x^{2}$ . Along this line, we have a critical point at $(3/4,3/4)$ .

Along the curve $y=x^{2}+x-1$ , $f(x,y)$ becomes $f(x)=x^{2}+3x-3$ . There is a critical point along this curve at $(-3/2,-1/4)$ . Since $x=-3/2$ lies outside our bounded region, we ignore this critical point.

The function $f$ itself has no critical points.

Checking the value of $f$ at $(-1,-1)$ , $(1,1)$ , $(3/4,3/4)$ , we find the absolute maximum is at $(3/4,3/4,9/8)$ and the absolute minimum is at $(-1,-1,-5)$ .
23.

abs max is $(1,2,6)$ , abs min is $(3,0,-6)$ .
24.

abs max is $(\sqrt{2}/2,\sqrt{2}/2,3/2)$ and $(-\sqrt{2}/2,-\sqrt{2}/2,3/2)$ , abs min is $(0,0,0)$ .
25.

$10\times 10\times 5$
26.

For any $a>0$ consider the box with dimensions $a\times a\times(500/a^{2})$ .

Exercises M.9

1.

$\pm 2\sqrt{5}$ at $(\pm 4/\sqrt{5},\pm 2/\sqrt{5})$
2.

min. $-\sqrt{3}$ at $(\pm\sqrt{3},\mp 1)$ ; max. $\sqrt{3}$ at $(\pm\sqrt{3},\pm 1)$
3.

$(\pm 20/\sqrt{13},\pm 30/\sqrt{13})$
4.

min. $(-9/\sqrt{5},0,2/\sqrt{5})$ ; max. $(9/8,\sqrt{59}/4,-1/4)$
5.

$(2/\sqrt{3})^{3}=8/3\sqrt{3}$
6.

$6\sqrt[3]{4}$ square meters.
7.

Length 130/3, height and width 65/3.
8.

$\left\lvert ax_{0}+by_{0}+cz_{0}-d\right\rvert/\sqrt{a^{2}+b^{2}+c^{2}}$
9.

$(0,\pm 1,0)$
10.

$20^{3}=8000$
11.

$(2,1,2)$
12.

$(0,0,\pm\sqrt{5})$
13.

Max: 5 at $\pm(2,2)$ , min: $-9/2$ at $\pm(3/\sqrt{2},-3/\sqrt{2})$ .
14.

$3\cdot 4^{2}=48$ .
15.

$8abc/3\sqrt{3}$
16.

Closest: $(1,0,0)$ and $(0,1,0)$ .

Furthest: $(-1/\sqrt{2},-1/\sqrt{2},-1-\sqrt{2})$ .
17.

$f(x,x^{2}-1)\to\infty$ as $x\to\infty$ .

Minimum is $(-3/4,-7/16,-155/128)$ .
18.

$416/15$ at $(8/5,-28/15,64/15)$ .

	$\displaystyle\lim_{(x,y)\to(0,0)}\frac{\sin(x^{2})}{y}$	$\displaystyle=\lim_{x\to 0}\frac{\sin(x^{2})}{mx}$
apply L’Hôpital’s Rule
		$\displaystyle=\lim_{x\to 0}\frac{2x\cos(x^{2})}{m}$
		$\displaystyle=0.$

	$\displaystyle\lim_{(x,y)\to(1,2)}\frac{x+y-3}{x^{2}-1}$	$\displaystyle=\lim_{x\to 1}\frac{x-1}{x^{2}-1}$
		$\displaystyle=\lim_{x\to 1}\frac{1}{x+1}$
		$\displaystyle=1/2.$

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

	$\displaystyle\lim_{(x,y)\to(\pi,\pi/2)}\frac{\sin x}{\cos y}$	$\displaystyle=\lim_{y\to\pi/2}\frac{0}{\cos y}$
		$\displaystyle=0.$

	$\displaystyle\lim_{(x,y)\to(\pi,\pi/2)}\frac{\sin x}{\cos y}$	$\displaystyle=\lim_{x\to\pi}\frac{\sin x}{\cos(x-\pi/2)}$
Apply L’Hôpital’s Rule:
		$\displaystyle=\lim_{x\to\pi}\frac{\cos x}{\sin(x-\pi/2)}$
		$\displaystyle=-1.$