Practice questions

Let ${s}$ be the side and ${A}$ the area.

We know: ${\dfrac{ds}{dt}=3}$ cm/s.

We want: ${\dfrac{dA}{dt}}$ when ${s=5}$ cm.

The equation: ${A=s^2}$.

Differentiate with respect to ${t}$:

${\dfrac{dA}{dt}=2s\dfrac{ds}{dt}}$

Now put in the numbers:

$\dfrac{dA}{dt}=2(5)(3)=30\ \text{cm}^2\text{/s}$

Let ${x}$ be the distance along the ground from the person to the point under the kite. Let ${L}$ be the length of the string.

We know: ${\dfrac{dx}{dt}=3}$ m/s.

We want: ${\dfrac{dL}{dt}}$ when ${L=50}$ m.

The equation, by the Pythagorean theorem:

${x^2+40^2=L^2}$

Differentiate with respect to ${t}$. The height ${40}$ does not change, so its derivative is ${0}$:

${2x\dfrac{dx}{dt}=2L\dfrac{dL}{dt}}$

We also need ${x}$ at this moment. When ${L=50}$:

${x=\sqrt{50^2-40^2}=\sqrt{900}=30}$ m

Now put in the numbers:

$\begin{align*}2(30)(3)&=2(50)\dfrac{dL}{dt}\\180&=100\dfrac{dL}{dt}\\\dfrac{dL}{dt}&=1.8\ \text{m/s}\end{align*}$

The string is being let out at ${1.8}$ m/s. This is slower than the kite moves, because only part of the kite's motion makes the string longer.

Write ${f(x)=x^{1/3}}$. Then ${f(8)=2}$, and

$\begin{align*}&f'(x)\\&=\tfrac{1}{3}x^{1/3-1}\\&=\tfrac{1}{3}x^{-2/3}\end{align*}$

At ${a=8}$: ${8^{2/3}=4}$, so ${f'(8)=\dfrac{1}{3\cdot 4}=\dfrac{1}{12}}$.

The linearization is

$\begin{align*}&L(x)\\&=f(8)+f'(8)(x-8)\\&=2+\dfrac{1}{12}(x-8)\end{align*}$

Now estimate:

$\begin{align*}&\sqrt[3]{8.2}\\&\approx L(8.2)\\&=2+\dfrac{0.2}{12}\\&\approx 2.0167\end{align*}$

(A calculator gives ${2.0165}$, so the estimate is very close.)

The volume is ${V=s^3}$. The differential is

${dV=3s^2\,ds}$

Use ${s=10}$ and ${ds=0.1}$:

${dV=3(10)^2(0.1)=30\ \text{cm}^3}$

So the possible error in the volume is about ${30}$ cm${^3}$.

The volume is ${10^3=1000}$ cm${^3}$. The relative error is

${\dfrac{dV}{V}=\dfrac{30}{1000}=0.03}$,   or ${3\%}$.

$\begin{align*}&f'(x)\\&=3x^2-12\\&=3(x^2-4)\\&=3(x+2)(x-2)\end{align*}$

${f'(x)=0}$ at ${x=-2}$ and ${x=2}$. The derivative exists everywhere. So the critical numbers are ${-2}$ and ${2}$.

Both critical numbers are in the interval. Find the value of ${f}$ at each critical number and at each endpoint:

${x}$	${f(x)}$
${-3}$ (endpoint)	${10}$
${-2}$ (critical)	${17}$
${2}$ (critical)	${-15}$
${3}$ (endpoint)	${-8}$

The largest value is ${17}$, and the smallest is ${-15}$.

So the absolute maximum is ${f(-2)=17}$, and the absolute minimum is ${f(2)=-15}$.

${f}$ is a polynomial, so it is continuous on ${[0,4]}$ and differentiable on ${(0,4)}$. The theorem applies.

The slope of the secant line from ${x=0}$ to ${x=4}$:

$\begin{align*}&\dfrac{f(4)-f(0)}{4-0}\\&=\dfrac{24-0}{4}\\&=6\end{align*}$

Now find ${c}$ with ${f'(c)=6}$. Here ${f'(x)=2x+2}$:

$\begin{align*}2c+2&=6\\c&=2\end{align*}$

The number ${c=2}$ is between ${0}$ and ${4}$.

Check the three conditions:

• ${\sin x}$ is continuous on ${[0,\pi]}$.
• ${\sin x}$ is differentiable on ${(0,\pi)}$.
• ${\sin 0=0}$ and ${\sin\pi=0}$, so the end values are equal.

So Rolle's Theorem applies. Find ${c}$ with ${f'(c)=0}$:

${\cos c=0}$

The only solution between ${0}$ and ${\pi}$ is ${c=\dfrac{\pi}{2}}$. There the tangent line is horizontal, at the top of the sine curve.

$\begin{align*}&f'(x)\\&=3x^2-12x+9\\&=3(x^2-4x+3)\\&=3(x-1)(x-3)\end{align*}$

The critical numbers are ${1}$ and ${3}$. Test one number in each interval. For example, on ${(1,3)}$, test ${x=2}$: ${f'(2)=-3}$, which is negative.

Interval	Sign of ${f'}$	${f}$ is
${(-\infty,1)}$	${+}$	increasing
${(1,3)}$	${-}$	decreasing
${(3,\infty)}$	${+}$	increasing

So ${f}$ is increasing on ${(-\infty,1)}$ and on ${(3,\infty)}$. It is decreasing on ${(1,3)}$.

From the table:

• At ${1}$, ${f'}$ changes from ${+}$ to ${-}$. So ${f}$ has a local maximum, ${f(1)=5}$.
• At ${3}$, ${f'}$ changes from ${-}$ to ${+}$. So ${f}$ has a local minimum, ${f(3)=1}$.

${f''(x)=6x-12=6(x-2)}$. It is ${0}$ at ${x=2}$.

Interval	Sign of ${f''}$	${f}$ is
${(-\infty,2)}$	${-}$	concave down
${(2,\infty)}$	${+}$	concave up

So ${f}$ is concave down on ${(-\infty,2)}$ and concave up on ${(2,\infty)}$.

The concavity changes at ${x=2}$. Since ${f(2)=3}$, the inflection point is ${(2,3)}$.

Domain: The denominator ${x^2+1}$ is never ${0}$. So the domain is all real numbers.

Intercepts: ${f(0)=0}$, and ${f(x)=0}$ only when ${x=0}$. So the graph crosses both axes at ${(0,0)}$.

Symmetry: ${f(-x)=\dfrac{-x}{x^2+1}=-f(x)}$. So ${f}$ is odd: the graph is symmetric about the origin.

Asymptotes: There is no vertical asymptote. Divide the top and bottom by ${x^2}$:

$\begin{align*}&\lim_{x\to\infty}\dfrac{x}{x^2+1}\\&=\lim_{x\to\infty}\dfrac{\dfrac{1}{x}}{1+\dfrac{1}{x^2}}\\&=0\end{align*}$

The same is true as ${x\to -\infty}$. So ${y=0}$ is a horizontal asymptote.

Increasing and decreasing: Use the quotient rule:

$\begin{align*}&f'(x)\\&=\dfrac{1\cdot(x^2+1)-x\cdot 2x}{(x^2+1)^2}\\&=\dfrac{1-x^2}{(x^2+1)^2}\end{align*}$

The denominator is always positive, so the sign of ${f'}$ is the sign of ${1-x^2}$. It is ${0}$ at ${x=-1}$ and ${x=1}$.

Interval	Sign of ${f'}$	${f}$ is
${(-\infty,-1)}$	${-}$	decreasing
${(-1,1)}$	${+}$	increasing
${(1,\infty)}$	${-}$	decreasing

Local extremes: ${f}$ has a local minimum ${f(-1)=-\dfrac{1}{2}}$ and a local maximum ${f(1)=\dfrac{1}{2}}$.

Concavity: Use the quotient rule again. The numerator of ${f'}$ is ${1-x^2}$. Its derivative is ${-2x}$. The denominator is ${(x^2+1)^2}$. By the chain rule, its derivative is ${2(x^2+1)\cdot 2x=4x(x^2+1)}$. So

$\begin{align*}&f''(x)\\&=\dfrac{{-2x(x^2+1)^2-(1-x^2)4x(x^2+1)}}{(x^2+1)^4}\end{align*}$

Every term has a factor ${x^2+1}$. Cancel one:

$\begin{align*}&=\dfrac{-2x(x^2+1)-4x(1-x^2)}{(x^2+1)^3}\\&=\dfrac{2x^3-6x}{(x^2+1)^3}\\&=\dfrac{2x(x^2-3)}{(x^2+1)^3}\end{align*}$

It is ${0}$ at ${x=0}$ and ${x=\pm\sqrt{3}\approx\pm 1.73}$.

Interval	Sign of ${f''}$	${f}$ is
${(-\infty,-\sqrt{3})}$	${-}$	concave down
${(-\sqrt{3},0)}$	${+}$	concave up
${(0,\sqrt{3})}$	${-}$	concave down
${(\sqrt{3},\infty)}$	${+}$	concave up

The concavity changes at all three, so the inflection points are

$\left(-\sqrt{3},-\dfrac{\sqrt{3}}{4}\right)$,   ${(0,0)}$,   $\left(\sqrt{3},\dfrac{\sqrt{3}}{4}\right)$.

Sketch: Put it all together:

Let ${x}$ be the side of the base, and ${h}$ the height, both in cm.

Formula: The material is the area of the base plus the four sides:

${S=x^2+4xh}$

Constraint: The volume is ${32{,}000}$:

${x^2h=32{,}000}$,   so   ${h=\dfrac{32{,}000}{x^2}}$.

One variable:

$\begin{align*}&S(x)\\&=x^2+4x\cdot\dfrac{32{,}000}{x^2}\\&=x^2+\dfrac{128{,}000}{x}\end{align*}$

Domain: ${x>0}$.

Minimize:

${S'(x)=2x-\dfrac{128{,}000}{x^2}}$

Set it equal to ${0}$:

$\begin{align*}2x&=\dfrac{128{,}000}{x^2}\\x^3&=64{,}000\\x&=40\end{align*}$

For ${x<40}$, ${S'(x)<0}$, and for ${x>40}$, ${S'(x)>0}$. So ${S}$ falls and then rises: ${x=40}$ gives the least material.

The height is ${h=\dfrac{32{,}000}{40^2}=20}$.

Answer: The base should be ${40}$ cm by ${40}$ cm, and the height ${20}$ cm.

Direct substitution gives ${\dfrac{1-1}{0}=\dfrac{0}{0}}$. Use L'Hôpital's rule: differentiate the top and the bottom.

$\begin{align*}&\lim_{x\to 0}\dfrac{e^{2x}-1}{x}\\&=\lim_{x\to 0}\dfrac{2e^{2x}}{1}\\&=2e^0\\&=2\end{align*}$

The form is ${\dfrac{\infty}{\infty}}$. Use L'Hôpital's rule:

$\begin{align*}&\lim_{x\to\infty}\dfrac{x^2}{e^x}\\&=\lim_{x\to\infty}\dfrac{2x}{e^x}\end{align*}$

This is still ${\dfrac{\infty}{\infty}}$. Use the rule again:

$\begin{align*}&=\lim_{x\to\infty}\dfrac{2}{e^x}\\&=0\end{align*}$

So ${e^x}$ grows much faster than ${x^2}$.

The form is ${0\cdot(-\infty)}$. Rewrite the product as a fraction:

${x^2\ln x=\dfrac{\ln x}{x^{-2}}}$

Now the form is ${\dfrac{-\infty}{\infty}}$. Use L'Hôpital's rule:

$\begin{align*}&\lim_{x\to 0^+}\dfrac{\ln x}{x^{-2}}\\&=\lim_{x\to 0^+}\dfrac{\dfrac{1}{x}}{-2x^{-3}}\end{align*}$

Simplify. Dividing by ${x^{-3}}$ is the same as multiplying by ${x^3}$:

$\begin{align*}&=\lim_{x\to 0^+}\left(-\dfrac{x^2}{2}\right)\\&=0\end{align*}$

The form is ${1^\infty}$. Let ${y=(1+2x)^{1/x}}$, and take ${\ln}$ of both sides:

${\ln y=\dfrac{\ln(1+2x)}{x}}$

This has the form ${\dfrac{0}{0}}$. Use L'Hôpital's rule. The derivative of ${\ln(1+2x)}$ is ${\dfrac{2}{1+2x}}$:

$\begin{align*}&\lim_{x\to 0^+}\dfrac{\ln(1+2x)}{x}\\&=\lim_{x\to 0^+}\dfrac{\dfrac{2}{1+2x}}{1}\\&=2\end{align*}$

So ${\ln y\to 2}$. Then ${y=e^{\ln y}\to e^2}$:

$\displaystyle\lim_{x\to 0^+}(1+2x)^{1/x}=e^2$