Chapter D

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3 ft/min

$63.14$mph

The cone is rising at a rate of $0.003$ft/s.

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2500; the two numbers are each 50.

The minimum sum is $2\sqrt{500}$; the two numbers are each $\sqrt{500}$.

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 $1/\sqrt{2}$.

Each pen should be $500/3\approx 166.67$ feet by 125 feet.

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 radius should be about $3.2$in and the height should be $2r=6.4$in. As the #10 is not a perfect cylinder (with extra material to aid in stacking, etc.), the dimensions are close enough to assume that minimizing surface area was a consideration.

The height and width should be $18$ and the length should be $36$, giving a volume of $11,664$in³.

$w=4\sqrt{3}$, $h=4\sqrt{6}$

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

The power line should be run directly to the off shore facility, skipping any underground, giving a cost of about $430,813.

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

The dog should run about 13 feet along the shore before starting to swim.

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

The largest volume is 62.5 in³ formed by cutting 2.5 in squares at each corner.

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

A box that is 1 in wide, 2 in long and $4/3$ in high will have a volume of $8/3{\text{ in}}^3$.

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Answers will vary.

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

Use $y=x^2$; $\mathrm{d}y=2x\cdot \mathrm{d}x$ with $x=6$ and $\mathrm{d}x=-0.07$. Thus $\mathrm{d}y=-0.84$; knowing $6^2=36$, we have ${5.93}^2\approx 35.16$.

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

Use $y=x^3$; $\mathrm{d}y=3x^2\cdot \mathrm{d}x$ with $x=7$ and $\mathrm{d}x=-0.2$. Thus $\mathrm{d}y=-29.4$; knowing $7^3=343$, we have ${6.8}^3\approx 313.6$.

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

Use $y=\sqrt{x}$; $\mathrm{d}y=1/\left(2\sqrt{x}\right)\cdot \mathrm{d}x$ with $x=25$ and $\mathrm{d}x=-1$. Thus $\mathrm{d}y=-0.1$; knowing $\sqrt{25}=5$, we have $\sqrt{24}\approx 4.9$.

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

Use $y=\sqrt[3]{x}$; $\mathrm{d}y=1/\left(3\sqrt[3]{x^2}\right)\cdot \mathrm{d}x$ with $x=8$ and $\mathrm{d}x=0.5$. Thus $\mathrm{d}y=1/24\approx 1/25=0.04$; knowing $\sqrt[3]{8}=2$, we have $\sqrt[3]{8.5}\approx 2.04$.

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

Use $y=e^x$; $\mathrm{d}y=e^x\cdot \mathrm{d}x$ with $x=0$ and $\mathrm{d}x=0.1$. Thus $\mathrm{d}y=0.1$; knowing $e^0=1$, we have $e^{0.1}\approx 1.1$.

$\mathrm{d}y=\left(2x+3\right)\mathrm{d}x$

$\mathrm{d}y=\left(7x^6-5x^4\right)\mathrm{d}x$

$\mathrm{d}y=\frac{-2}{4x^3}\mathrm{d}x$

$\mathrm{d}y=2\left(2x+\sin x\right)\left(2+\cos x\right)\mathrm{d}x$

$\mathrm{d}y=\left(2x{e}^{3x}+3x^2{e}^{3x}\right)\mathrm{d}x$

$\mathrm{d}y=\frac{-16}{x^5}\mathrm{d}x$

$\mathrm{d}y=\frac{2\left(\tan x+1\right)-2x{\sec }^{2}x}{{\left(\tan x+1\right)}^2}\mathrm{d}x$

$\mathrm{d}y=\frac{1}{x}\mathrm{d}x$

$\mathrm{d}y=\left(e^x\sin x+e^x\cos x\right)\mathrm{d}x$

$\mathrm{d}y=\left(-\sin \left(\sin x\right)\cos x\right)\mathrm{d}x$

$\mathrm{d}y=\frac{1}{{\left(x+2\right)}^2}\mathrm{d}x$

$\mathrm{d}y=\left(\left(\ln 3\right)3^x\ln x+\frac{3^x}{x}\right)\mathrm{d}x$

$\mathrm{d}y=\left(\ln x\right)\mathrm{d}x$

$\mathrm{d}y=\left(\tan x\right)\mathrm{d}x$

$1-6x$

$2+2x$

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

$\sqrt{2}$

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

$\frac{1}{2^{2/3}}+\frac{x}{3\cdot 2^{2/3}}$

$\mathrm{d}V=\pm 0.157$

$\pm 15\pi /8\approx \pm 5.89$in²

$\pm$48in², or 1/3ft²

1%

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

$x_0=1$, $x_1=-0.55740772$, $x_2=0.065936452$, $x_3=-0.000095721919$, $x_4=2.9235662\ast {10}^{-13}$, $x_5=0$

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

$x_0=1.5$, $x_1=1.4166667$, $x_2=1.4142157$, $x_3=1.4142136$, $x_4=1.4142136$, $x_5=1.4142136$

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

$x_0=1$, $x_1=1$, $x_2=1$, $x_3=1$, $x_4=1$, $x_5=1$

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

roots are: $x=-3.714$, $x=-0.857$, $x=1$ and $x=1.571$

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

roots are: $x=-2.165$, $x=0$, $x=0.525$ and $x=1.813$

$x=\pm 0.824$,

$x=-0.637$, $x=1.410$

$x=\pm 0.743$

$x=\pm 4.493$, $x=0$

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

The approximations alternate between $x=1$, $x=2$ and $x=3$.

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

$f\left(x\right)=x^2-24$ and $x_0=5$ yield $x_1=\frac{49}{10}=4.9$ and $x_2=\frac{4801/980}{\approx }4.898980$.

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

$f\left(x\right)=x^3-8.5$ and $x_0=2$ yield $x_1=\frac{49}{24}\approx 2.0416667$ and $x_2\approx 2.0408279$.