Chapter 4

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3 ft/min

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2500; the two numbers are each 50.

There is no maximum sum; the fundamental equation has only 1 critical value that corresponds to a minimum.

Area = 1/4, with sides of length $1/\sqrt{2}$.

The radius should be about $3.84$cm and the height should be $2r=7.67$cm. No, this is not the size of the standard can.

The height and width should be $18$ and the length should be $36$, giving a volume of $11,664$in${}^3$.

$5-10/\sqrt{39}\approx 3.4$ miles should be run underground, giving a minimum cost of $374,899.96.

The dog should run about 19 feet along the shore before starting to swim.

The largest area is 2 formed by a square with sides of length $\sqrt{2}$.

A length of 2 in and height of 2.5 will give a cost of 60 ¢.

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F

Answers will vary.

Use $y=x^2$; $dy=2x\cdot dx$ with $x=2$ and $dx=0.05$. Thus $dy=.2$; knowing $2^2=4$, we have ${2.05}^2\approx 4.2$.

Use $y=x^3$; $dy=3x^2\cdot dx$ with $x=5$ and $dx=0.1$. Thus $dy=7.5$; knowing $5^3=125$, we have ${5.1}^3\approx 132.5$.

Use $y=\sqrt{x}$; $dy=1/\left(2\sqrt{x}\right)\cdot dx$ with $x=16$ and $dx=0.5$. Thus $dy=.0625$; knowing $\sqrt{16}=4$, we have $\sqrt{16.5}\approx 4.0625$.

Use $y=\sqrt[3]{x}$; $dy=1/\left(3\sqrt[3]{x^2}\right)\cdot dx$ with $x=64$ and $dx=-1$. Thus $dy=-1/48\approx 0.0208$; we could use $-1/48\approx -1/50=-0.02$; knowing $\sqrt[3]{64}=4$, we have $\sqrt[3]{63}\approx 3.98$.

Use $y=\sin x$; $dy=\cos x\cdot dx$ with $x=\pi$ and $dx\approx -0.14$. Thus $dy=0.14$; knowing $\sin \pi =0$, we have $\sin 3\approx 0.14$.

$dy=\left(2x+3\right)dx$

$dy=\frac{-2}{4x^3}dx$

$dy=\left(2x{e}^{3x}+3x^2{e}^{3x}\right)dx$

$dy=\frac{2\left(\tan x+1\right)-2x{\sec }^{2}x}{{\left(\tan x+1\right)}^2}dx$

$dy=\left(e^x\sin x+e^x\cos x\right)dx$

$dy=\frac{1}{{\left(x+2\right)}^2}dx$

$dy=\left(\ln x\right)dx$

$1-6x$

$1-\frac{x}{2}$

$2^{2/3}+\frac{x}{2^{4/3}}$

$dV=\pm 0.157$

$\pm 15\pi /8\approx \pm 5.89$in${}^2$

F

$x_0=1.5$, $x_1=1.5709148$, $x_2=1.5707963$, $x_3=1.5707963$, $x_4=1.5707963$, $x_5=1.5707963$

$x_0=0$, $x_1=2$, $x_2=1.2$, $x_3=1.0117647$, $x_4=1.0000458$, $x_5=1$

$x_0=2$, $x_1=0.6137056389$, $x_2=0.9133412072$, $x_3=0.9961317034$, $x_4=0.9999925085$, $x_5=1$

roots are: $x=-5.156$, $x=-0.369$ and $x=0.525$

roots are: $x=-1.013$, $x=0.988$, and $x=1.393$

$x=\pm 0.824$,

$x=\pm 0.743$

The approximations alternate between $x=1$ and $x=2$.

$f\left(x\right)=x^2-16.5$ and $x_0=4$ yield $x_1=\frac{65}{16}=4.0625$ and $x_2=\frac{8449}{2080}\approx 4.0620192$.

$f\left(x\right)=x^3-63$ and $x_0=4$ yield $x_1=\frac{191}{48}\approx 3.97916667$ and $x_2\approx 3.9790572$.