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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 #3 : First And Second Derivatives Of Functions

The position of a car is given by the following function:

$f(x)=\cos(x)e^{11x}+\csc(x)$

What is the velocity function of the car?

Possible Answers:

$-\sin(x)e^{11x}+\cos(x)e^{11x}-\csc(x)\cot(x)$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}+\csc(x)\cot(x)$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

$\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

Correct answer:

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

Explanation:

The velocity function of the car is equal to the first derivative of the position function of the car, and is equal to

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

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Example Question #1 : Applications Of Derivatives

Let

$f(x)=e^{\pi x}$

Find the first and second derivative of the function.

Possible Answers:

$f'(x)=e^{\pi x }$

$f''(x)=e^{\pi x}$

$f'(x)=\pi e^{\pi x }$

$f''(x)=\pi^2 e^{\pi x}$

$f'(x)=e^{\pi }$

$f''(x)= e^{x}$

$f'(x)=\pi$

$f''(x)=\0$

Correct answer:

$f'(x)=\pi e^{\pi x }$

$f''(x)=\pi^2 e^{\pi x}$

Explanation:

In order to solve for the first and second derivative, we must use the chain rule.

The chain rule states that if

y=f(u)

and

u=g(x)

then the derivative is

$\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}$

In order to find the first derviative of the function

$f(x)=e^{\pi x}$

we set

y=e^u

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

$\frac{dy}{du}=\frac{d}{du}[e^u]$

$\frac{dy}{du}=e^u$

And differentiating we use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

As such

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]$

$\frac{du}{dx}=\pi x^{1-1}=\pi$

And so

$\frac{dy}{dx}=e^{\pi x}*\pi=\pi e^{\pi x}$

$f'(x)=\pi e^{\pi x}$

To solve for the second derivative we set

$y=\pi e^{u}$

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

$\frac{dy}{du}=\frac{d}{du}[\pi e^u]$

$\frac{dy}{du}=\pi e^u$

And differentiating we use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

As such

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]$

$\frac{du}{dx}=\pi x^{1-1}=\pi$

And so the second derivative becomes

$\frac{dy}{dx}=\pi e^{\pi x}*\pi=\pi^2 e^{\pi x}$

$f''(x)=\pi^2e^{\pi x}$

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Example Question #2 : Applications Of Derivatives

Find the velocity function of the particle if its position is given by the following function:

$s(t)=e^t\tan(t)+t^5+3$

Possible Answers:

$te^t\tan(t)+e^t\sec^2(t)+5t^4$

$e^t\tan(t)+e^t\sec^2(t)+t^5$

$e^t\tan(t)-e^t\sec^2(t)+5t^4$

$e^t\tan(t)+e^t\sec^2(t)+5t^4$

Correct answer:

$e^t\tan(t)+e^t\sec^2(t)+5t^4$

Explanation:

The velocity function is given by the first derivative of the position function:

$v(t)=s'(t)=e^t\tan(t)+e^t\sec^2(t)+5t^4$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(t)=\sec^2(t)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

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Example Question #1 : Applications Of Derivatives

Find the first and second derivatives of the function

f(x)=sec(x)

Possible Answers:

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

f'(x)=sec^2(x)

f''(x)=sec^3(x)

f'(x)=csc(x)

f''(x)=-sec(x)

f'(x)=cot(x)

f''(x)=csc^2(x)

Correct answer:

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

Explanation:

We must find the first and second derivatives.

We use the properties that

The derivative of is
The derivative of is

As such

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)]=sec(x)tan(x)$

To find the second derivative we differentiate again and use the product rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}[uv]=uv'+vu'$

Setting

u=sec(x) and v=tan(x)

we find that

$f''(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)tan(x)]=sec(x)*sec^2(x)+tan(x)*sec(x)tan(x)$

=sec^3(x)+sec(x)tan^2(x)

As such

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

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Example Question #1 : Applications Of Derivatives

Given the velocity function

$v(x)=\int_{0}^{x^2+2x+1} z^d\sin(z^5) dz$

where is real number such that $d\in [0,1]$ , find the acceleration function

a(x) .

Possible Answers:

$a(x)=(x+1)^{d}\sin (x+1)^{10}$

$a(x)=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

$a(x)=(x^2+2x+1)^{d}\sin (x^2+2x+1)^{10}$

$a(x)=(x^2+2x+1)^{2d}\sin (x^2+2x+1)^{10}$

$a(x)=(2x+2)^{d}(x+1)^{2d}\sin (x+1)^{10}$

Correct answer:

$a(x)=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

Explanation:

We can find the acceleration function a(x) from the velocity function by taking the derivative:

a(x)=v'(x)

We can view the function

$v(x)=\int_{0}^{x^2+2x+1} z^d\sin(z^5) dz$

as the composition of the following functions

$g(x)=\int_{0}^{x} z^d\sin(z^5) dz$

h(x)=x^2+2x+1=(x+1)^2

so that v(x)=g(h(x)) . This means we use the chain rule

[g(h(x))]'=g'(h(x))h'(x)

to find the derivative. We have $g'(x)=x^d\sin x^5$ and h'(x)=2x+2 , so we have

$a(x)=v'(x)=[(x+1)^2]^d\sin [(x+1)^2]^5\cdot (2x+2)$

$=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

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Example Question #1 : Velocity, Speed, Acceleration

The position of an object is given by the equation $y = \frac{1}{3} t^3 - \frac{1}{2} t^2 -2t + 4$ . What is its acceleration at t = 2?

Possible Answers:

Correct answer:

Explanation:

If this function gives the position, the first derivative will give its speed and the second derivative will give its acceleration.

y ' = t^2 - t -2

y'' = 2t - 1

Now plug in 2 for t:

y'' = 2(2) - 1 = 4- 1 = 3

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Example Question #1 : Applications Of Derivatives

The equation $y=\frac{1}{2} t^4 +10t$ models the position of an object after t seconds. What is the acceleration at 3 seconds?

Possible Answers:

Correct answer:

Explanation:

If this function gives the position, the first derivative will give its speed and the second derivative will give its acceleration.

y' = 2 t^3 +10

y'' = 6t^2

Plug in 3 for t:

6(3)^2 = 6(9) = 54

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Example Question #1 : Velocity, Speed, Acceleration

The equation $y= \sin (2t)$ models the position of an object after t seconds. What is its speed after $\pi$ seconds?

Possible Answers:

Correct answer:

Explanation:

If this function gives the position, the first derivative will give its speed.

$y' = \cos (2t ) \cdot 2$

Plug in $\pi$ for t:

$2 \cos (2 \pi ) = 2\cdot 1 = 2$

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Example Question #2 : Velocity, Speed, Acceleration

The position of an object is modeled by the equation $y= \frac{1}{2} \cos ^2 t$ What is the speed after $\frac{\pi}{4}$ seconds?

Possible Answers:

$-\frac{1}{2}$

$-\frac{1}{\sqrt2}$

$\sqrt{2}$

Correct answer:

$-\frac{1}{2}$

Explanation:

If this function gives the position, the first derivative will give its speed. To differentiate, use the chain rule: $f(g(x)) ' = f'(g(x))\cdot g'(x)$ . In this case, $f(x) = \frac{1}{2} x^2$ and $g(x) = \cos x$ . Since f'(x) = x and $g'(x ) = - \sin (x)$ , the first derivative is $\cos t \cdot - \sin t$ .

Plug in $\frac{\pi}{4}$ for t:

$\cos (\frac{\pi}{4} ) \cdot - \sin (\frac{\pi}{4} ) = \frac{1}{\sqrt{2}} \cdot - \frac{1}{\sqrt2} = -\frac{1}{2}$

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Example Question #1 : Velocity, Speed, Acceleration

A particle's position on the -axis is given by the function $f(t)=\sin(t)-t+1$ from $0 < t <\pi$ .

When does the particle change direction?

Possible Answers:

It doesn't change direction within the given bounds

$x=2\pi$

$x=\frac{\pi}{2}$

$x= \pi$

$x= \frac{\pi}{4}$

Correct answer:

It doesn't change direction within the given bounds

Explanation:

To find when the particle changes direction, we need to find the critical values of f(t) . This is done by finding the velocity function, setting it equal to , and solving for

$0=v(t) = f'(t) = \cos(t)-1$ .

Hence $\cos(t)=1$ .

The solutions to this on the unit circle are $t =0, \pi$ , so these are the values of where the particle would normally change direction. However, our given interval is $0 < t < \pi$ , which does not contain $0, \pi$ . Hence the particle does not change direction on the given interval.

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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 #3 : First And Second Derivatives Of Functions

Example Question #1 : Applications Of Derivatives

Example Question #2 : Applications Of Derivatives

Example Question #1 : Applications Of Derivatives

Example Question #1 : Applications Of Derivatives

Example Question #1 : Velocity, Speed, Acceleration

Example Question #1 : Applications Of Derivatives

Example Question #1 : Velocity, Speed, Acceleration

Example Question #2 : Velocity, Speed, Acceleration

Example Question #1 : Velocity, Speed, Acceleration

All AP Calculus BC Resources

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