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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

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2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #131 : Integral Applications

Give the arclength of the graph of the function $f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ on the interval $\left [ 0, \frac{\pi}{8} \right ]$ .

Possible Answers:

$1+\frac{1 }{2} \ln 2$

$\ln \left ( 1 + \sqrt{2} \right )$

$1 + \sqrt{2}$

$\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right )$

$\frac{1 }{2}+\frac{1 }{2} \sqrt{2}$

Correct answer:

$\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right )$

Explanation:

The length of the curve of f(x) on the interval [a,b] can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x$ .

$f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ , so

$f'(x) = \frac{1}{2} \cdot \frac{\frac{\mathrm{d} }{\mathrm{d} x}\cos 2 x}{\cos 2 x} = \frac{1}{2} \cdot \frac{-2 \sin 2x}{\cos 2 x} = - \tan 2x$

The integral becomes

$\int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \left ( - \tan 2x \right )^{2}} \; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \tan ^{2} 2x } \; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{\sec ^{2} 2x } \; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sec 2x \; \mathrm{d}x$

Use substitution - set u = 2x . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 2$ , and $\mathrm{d} x = \frac{1}{2}\mathrm{d} u$ . The bounds of integration become $2 \cdot 0 = 0$ and $2 \cdot \frac{\pi}{8} = \frac{\pi}{4}$ , and the integral becomes

$\frac{1}{2} \int_{0}^{ \frac{\pi}{4}} \sec u \; \mathrm{d}u$

$=\left.\begin{matrix}\frac{1}{2} \ln | \sec u + \tan u | \\ \\ \end{matrix}\right|\begin{matrix} \frac{\pi}{4}\\ \\ 0 \end{matrix}$

$=\frac{1}{2} \ln \left | \sec \frac{\pi}{4} + \tan \frac{\pi}{4} \right | - \frac{1}{2} \ln | \sec 0 + \tan 0 |$

$=\frac{1}{2} \ln \left | \sqrt{2} + 1 \right | -\frac{1}{2} \ln | 1 + 0 |$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) -\frac{1}{2} \ln 1$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) - 0$

$=\frac{1}{2} \ln \left (1 + \sqrt{2} \right )$

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Example Question #1 : Average Values And Lengths Of Functions

What is the length of the curve $g(x)=x^{\frac{3}{2}}$ over the interval $0\le x\le4$ ?

Possible Answers:

$\frac{56}{27}$

$\frac{8}{27}(10\sqrt{10}-1)$

$4\sqrt2-1$

$\frac{4\pi}{3}$

Correct answer:

$\frac{8}{27}(10\sqrt{10}-1)$

Explanation:

The general formula for finding the length of a curve f(x) over an interval [a,b] is $\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx$

In this example, the arc length can be found by computing the integral

$\int_{0}^{4}\sqrt{1+(\frac{d}{dx}x^{\frac{3}{2}})^2}\ dx$ .

The derivative of $x^{\frac{3}{2}}$ can be found using the power rule, $\frac{d}{dx} x^n=nx^{n-1}$ , which leads to

$\int_{0}^{4}\sqrt{1+\left(\frac{3}{2}x^{\frac{1}{2}}\right)^2}\ dx=\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx$ .

At this point, a substitution is useful.

Let

$u=1+\frac{9}{4}x,\ du=\frac{9}{4}dx,\frac{4}{9}du=dx$ .

We can also express the limits of integration in terms of to simplify computation. When x=0, u=1 , and when x = 4, u=10 .

Making these substitutions leads to

$\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx=\int_{1}^{10}\frac{4}{9}\sqrt{u}du=\frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du$ .

Now use the power rule, which in general is $\int x^n=\frac{x^{n+1}}{n+1}+c$ , to evaluate the integral.

$\frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du=\frac{4}{9}\frac{u^{\frac{3}{2}}}{\frac{3}{2}}|_{1}^{10}$

$=\frac{8}{27}(10^{\frac{3}{2}}-1^\frac{3}{2})=\frac{8}{27}(10\sqrt{10}-1)$

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Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Find the total distance traveled by a particle along the curve y=x^2 from x=0 to x=2 .

Possible Answers:

$\int_0^2\sqrt{1+4x^2}dx$

$\int_0^2\sqrt{1+2x}dx$

$\int_0^2\sqrt{1+2x^2}dx$

$\int_0^2\sqrt{1+x^2}dx$

$\int_0^2\sqrt{1+4x^2}dx$

Correct answer:

$\int_0^2\sqrt{1+4x^2}dx$

Explanation:

To find the required distance, we can use the arc length expression given by $\int_{x_0}^{x_1}\sqrt{1+f'(x)^2}dx$ .

Taking the derivative of our function, we have y'=f'(x)=2x . Plugging in our values for our integral bounds, we have

$\int_0^2\sqrt{1+(2x)^2}dx = \int_0^2 \sqrt{1+4x^2}dx$ .

As with most arc length integrals, this integral is too difficult (if not, outright impossible) to evaluate explicitly by hand. So we will just leave it this form, or evaluate it with some computer software.

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Example Question #1 : Modeling By Solving Separable Differential Equations

Solve the separable differential equation

$\frac{\mathrm{d} y}{\mathrm{d} x}= yx^2$

with the condition y(0)=4 .

Possible Answers:

$y=Ce^{\frac{x^3}{3}}$

$y=4e^{\frac{x^3}{3}}$

$y=Ce^{\frac{x^2}{2}}$

$y=4e^{\frac{x^2}{3}}$

$y=Ce^{\frac{64}{3}}$

Correct answer:

$y=4e^{\frac{x^3}{3}}$

Explanation:

To solve the separable differential equation, we must separate x and y, dx and dy respectively to opposite sides:

$\frac{dy}{y}=x^2dx$

Integrating both sides, we get

$\ln\left | y\right |=\frac{ x^3}{3}+C$

The rules of integration used were

$\int \frac{dx}{x}=\ln\left | x\right |+C$ , $\int x^ndx= \frac{x^{n+1}}{n+1}+C$

The constants of integration merged into one.

Now, we exponentiate both sides of the equation to solve for y, and use the properties of exponents to simplify:

$y=e^{\frac{x^3}{3}}+e^C=e^{{\frac{x^3}{3}}\cdot C}=Ce^{\frac{x^3}{3}}$

To solve for C, we use our given condition:

4=Ce^0

4=C

Our final answer is

$y=4e^{\frac{x^3}{3}}$

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Example Question #1 : Parametric Form

Rewrite as a Cartesian equation:

$x = t^{2} + 2t + 1, y = t^{2} - 2t + 1, t \in [-1, 1]$

Possible Answers:

y = x+2

y = x - 2

$y = x - 4 - 4\sqrt{x}$

$y = x + 4 + 4\sqrt{x}$

$y = x + 4 - 4\sqrt{x}$

Correct answer:

$y = x + 4 - 4\sqrt{x}$

Explanation:

$x = t^{2} + 2t + 1$

$x = \left ( t + 1\right )^{2}$

$\pm \sqrt{x} = t + 1$

$\sqrt{x} = t + 1$ or $-\sqrt{x} = t + 1$

We are restricting to values on $\left [ -1, 1\right ]$ , so t + 1 is nonnegative; we choose

$\sqrt{x} = t + 1$ .

Also,

$y = t^{2} - 2t + 1$

$y= \left ( t - 1\right )^{2}$

$\sqrt{y} = \pm \left ( t - 1\right )$

$\sqrt{y} = t- 1$ or $-\sqrt{y} = t- 1$

We are restricting to values on $\left [ -1, 1\right ]$ , so t - 1 is nonpositive; we choose

$-\sqrt{y} = t- 1$

or equivalently,

$\sqrt{y} = -t+ 1$

to make t - 1 nonpositive.

Then,

$\sqrt{x}+ \sqrt{y} = (t + 1) + (-t + 1)$

and

$\sqrt{x}+ \sqrt{y} = 2$

$\sqrt{y} = 2 - \sqrt{x}$

$y =\left ( 2 - \sqrt{x} \right )^2$

$y = 4 - 4\sqrt{x} + \left (\sqrt{x} \right )^{2}$

$y = x + 4 - 4\sqrt{x}$

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Example Question #1 : Parametric, Polar, And Vector Functions

Rewrite as a Cartesian equation:

$x =10^ {t}, y = \sinh t, t \in (0,\infty )$

Possible Answers:

$y = \frac{ x ^{2 \log e} -1}{ x ^{2 \log e} + 1}$

$y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$

$y = \frac{ x ^{2 \log e} +1}{2 x ^{\log e}}$

$y = \frac{ 2x ^{2 \log e} -1}{2 x ^{\log e}}$

$y = \frac{ x ^{2 \log e} }{2 x ^{\log e}}$

Correct answer:

$y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$

Explanation:

$y = \sinh t = \frac{e ^{2t} -1}{2e ^{t}} = \frac{\left (e ^{ t} \right) ^{2} -1}{2e ^{t}}$

$e^{t} = 10 ^{\log e \cdot t} =\left ( 10 ^{ t} \right )^{\log e} = x ^{\log e}$ , so

$y = \frac{\left ( x ^{\log e} \right) ^{2} -1}{2 x ^{\log e}}= \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$

This makes the Cartesian equation

$y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$ .

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Example Question #1 : Parametric, Polar, And Vector Functions

If x=3+t and y=4t+7 , what is in terms of (rectangular form)?

Possible Answers:

y=4x-5

y=4x-7

y=4x-3

y=4x-12

y=4x-2

Correct answer:

y=4x-5

Explanation:

Given x=3+t and y=4t+7 , we can find in terms of by isolating in both equations:

$x=3+t\rightarrow t=x-3$

$y=4t+7\rightarrow t=\frac{y-7}{4}$

Since both of these transformations equal , we can set them equal to each other:

$\frac{y-7}{4}=x-3$

y-7=4(x-3)

y-7=4x-12

y=4x-5

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Example Question #191 : Ap Calculus Bc

Given x=t+10 and y=2t-9 , what is the arc length between $0\leqt\leq t\leq4$ ?

Possible Answers:

$2\sqrt{5}$

$4\sqrt{5}$

$\sqrt{5}$

$3\sqrt{5}$

$5\sqrt{5}$

Correct answer:

$4\sqrt{5}$

Explanation:

In order to find the arc length, we must use the arc length formula for parametric curves:

$L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt$ .

Given x=t+10 and y=2t-9 , we can use using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , to derive

$\frac{dx}{dt}=(1)t^{1-1}+(0)10=1$ and

$\frac{dy}{dt}=(1)2t^{1-1}-(0)9=2$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$L=\int_{0}^{4}\sqrt{(1)^{2}+(2)^{2}}dt$

$L=\int_{0}^{4}(\sqrt{1+4})dt$

$L=\int_{0}^{4}(\sqrt{5})dt$

Now, using the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

we can determine that:

$L=[t\sqrt{5}]_{0}^{4}\textrm{}dt$

$L=(4)\sqrt{5}-(0)\sqrt{5}$

$L=4\sqrt{5}$

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Example Question #4 : Parametric, Polar, And Vector Functions

Given x=4t-5 and y=6t+2 , what is the length of the arc from $0\leq t\leq2$ ?

Possible Answers:

$3\sqrt{13}$

$4\sqrt{13}$

$5\sqrt{13}$

$6\sqrt{13}$

$7\sqrt{13}$

Correct answer:

$4\sqrt{13}$

Explanation:

In order to find the arc length, we must use the arc length formula for parametric curves:

$L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt$ .

Given x=4t-5 and y=6t+2 , we can use using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , to derive

$\frac{dx}{dt}=(1)4t^{1-1}-(0)5=4$ and

$\frac{dy}{dt}=(1)6t^{1-1}+(0)2=6$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$L=\int_{0}^{2}(\sqrt{(4)^{2}+(6)^{2}}dt$

$L=\int_{0}^{2}(\sqrt{16+36})dt$

$L=\int_{0}^{2}(\sqrt{52})dt$

$L=\int_{0}^{2}(\sqrt{4\times13})dt$

$L=\int_{0}^{2}(2\sqrt{13})dt$

Now, using the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

we can determine that:

$L=[2t\sqrt{13}]_{0}^{2}\textrm{}$

$L=2(2)\sqrt{13}-2(0)\sqrt{13}$

$L=4\sqrt{13}$

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Example Question #151 : Parametric, Polar, And Vector

Find the length of the following parametric curve

$x=e^{t}cos(t)$ , $y=e^{t}sin(t)$ , $0\leq t \leq 2 \pi$ .

Possible Answers:

sin(t)+cos(t)

$e^{2 \pi}$

$e^{t}cos(t)$

$\sqrt{2}\left ( e^{2 \pi}-1}\right )$

Correct answer:

$\sqrt{2}\left ( e^{2 \pi}-1}\right )$

Explanation:

The length of a curve is found using the equation $L=\int_{\alpha }^{\beta }\sqrt{\left ( \frac{dx}{dt} \right )^{2}+\left ( \frac{dy}{dt} \right )^{2}}dt$

We use the product rule,

$\frac{d}{dt}(a*b)=a\frac{db}{dt}+b\frac{da}{dt}$ , when and are functions of ,

the trigonometric rule,

$\frac{d}{dt}[cos(t)]=-sin(t)$ and $\frac{d}{dt}[sin(t)]=cos(t)$

and exponential rule,

$\frac{d}{dt}[e^{t}]=e^{t}$ to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ .

In this case

$x=e^{t}cos(t)$ , $y=e^{t}sin(t)$

$\frac{dx}{dt}=\frac{d}{dt}e^{t}cos(t)=e^{t}\frac{d}{dt}cos(t)+cos(t)\frac{d}{dt}e^{t}$

$=e^{t}*(-sin(t))+cos(t)*e^{t}=e^{t}cos(t)-e^{t}sin(t)$

$\frac{dy}{dt}=\frac{d}{dt}e^{t}sin(t)=e^{t}\frac{d}{dt}sin(t)+sin(t)\frac{d}{dt}e^{t}$

$=e^{t}cos(t)+sin(t)*e^{t}=e^{t}cos(t)+e^{t}sin(t)$

$\alpha=0, \beta=2\pi$

The length of this curve is

$L=\int_{0}^{2\pi }\sqrt{\left (e^{t}cos(t)-e^{t}sin(t)\right )^{2}+\left (e^{t}cos(t)+e^{t}sin(t)\right )^{2}}dt$

Using the identity $(a\pm b)^{2}=a^{2}\pm2ab+b^{2}$

$L=\int_{0}^{2\pi }\sqrt{\binom{\left (e^{2t}cos^{2}(t)-2e^{t}cos(t)e^{t}sin(t)+e^{2t}sin^{2}(t)\right )}{\left +(e^{2t}cos^{2}(t)+2e^{t}cos(t)e^{t}sin(t)+e^{2t}sin^{2}(t)\right)}dt}$

$=\int_{0}^{2\pi }\sqrt{(2e^{2t}cos^{2}(t)+2e^{2t}sin^{2}(t))dt}$

$=\int_{0}^{2\pi }\sqrt{2e^{2t}(cos^{2}(t)+sin^{2}(t))dt}$

Using the identity $\sqrt{a*b}=\sqrt{a}\sqrt{b}$

$L=\int_{0}^{2\pi }\sqrt{2e^{2t}} \sqrt{cos^{2}(t)+sin^{2}(t)dt}$

Using the trigonometric identity $sin^{2}(at)+cos^{2}(at)=1$ where is a constant and $e^{2t}=(e^{t})^2$

$=\int_{0}^{2\pi }\sqrt{2}e^{t}dt}=\sqrt{2}\int_{0}^{2\pi }e^{t}dt}$

Using the exponential rule, $\int e^{t}dt}=e^{t}$

$\sqrt{2}\int_{0}^{2\pi }e^{t}dt}=\sqrt{2}e^{t}|_{0}^{2\pi}=\sqrt{2}\left ( e^{2 \pi}-e^{0}}\right )$

Using the exponential rule, $e^{0}=1$ , gives us the final solution

$L=\sqrt{2}\left ( e^{2 \pi}-1}\right )$

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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #131 : Integral Applications

Example Question #1 : Average Values And Lengths Of Functions

Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Example Question #1 : Modeling By Solving Separable Differential Equations

Example Question #1 : Parametric Form

Example Question #1 : Parametric, Polar, And Vector Functions

Example Question #1 : Parametric, Polar, And Vector Functions

Example Question #191 : Ap Calculus Bc

Example Question #4 : Parametric, Polar, And Vector Functions

Example Question #151 : Parametric, Polar, And Vector

All AP Calculus BC Resources

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