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AP Calculus BC : Derivatives

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Example Questions

Example Question #1 : Applications Of Derivatives

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:

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

It doesn't change direction within the given bounds

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

$x=2\pi$

$x= \pi$

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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Example Question #191 : Derivatives

A particle moves in space with velocity given by

$v(t)=\alpha t^3+\beta \sqrt{t}$

where $\alpha, \beta$ are constant parameters.

Find the acceleration of the particle when t=4.

Possible Answers:

$3\alpha t^2+\frac{\beta}{2\sqrt{t}}$

$48\alpha +2\beta$

$48\alpha + \frac{\beta}{4}$

Correct answer:

$48\alpha + \frac{\beta}{4}$

Explanation:

To find the acceleration of the particle, we must take the first derivative of the velocity function:

$a(t)=v'(t)=3\alpha t^2+\frac{\beta}{2\sqrt{t}}$

The derivative was found using the following rule:

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

Now, we evaluate the acceleration function at the given point:

$a(4)=3\alpha(4^2)+\frac{\beta}{2\sqrt{4}}=48\alpha+\frac{\beta}{4}$

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Example Question #12 : Applications Of Derivatives

Find the velocity function from an acceleration function given by

a(t)=4t+5

and the condition v(0)=10

Possible Answers:

2t^2+15

2t^2+5t+C

2t^2+5t+10

2t^2+5t

Correct answer:

2t^2+5t+10

Explanation:

Acceleration is the rate of change of velocity, so we must integrate the acceleration function to find the velocity function:

$v(t)=\int a(t)dt=\int (4t+5 ) dt=2t^2+5t+C$

The integration was performed using the following rules:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ , $\int a dx=ax+C$

To find the integration constant C, we must use the initial condition given:

v(0)=10=2(0)^2+5(0)+C

C=10

Our final answer is

v(t)=2t^2+5t+10

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Example Question #13 : Applications Of Derivatives

The velocity of a particle is given by v(t). Find the function which models the particle's acceleration.

v(t)=3t^3-4t^2-5t+16

Possible Answers:

a(t)=9t^3+8t^2-5

a(t)=9t^2-8t-17

a(t)=6t^2-8t+5

a(t)=9t^2-8t-5

Correct answer:

a(t)=9t^2-8t-5

Explanation:

The velocity of a particle is given by v(t). Find the function which models the particle's acceleration.

v(t)=3t^3-4t^2-5t+16

To find the acceleration from a velocity function, simply take the derivative.

In this case, we are given v(t), and we need to find v'(t) because v'(t)=a(t).

To find v'(t), we need to use the power rule.

For each term, simply multiply by the exponent, and then subtract one from the exponent. Constant terms will drop out, linear terms will become constants, and so on.

v(t)=3t^3-4t^2-5t+16

$v'(t)=a(t)=(3)3t^{3-1}-(2)4t^{2-1}-5=9t^2-8t-5$

So, our answer is:

a(t)=9t^2-8t-5

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Example Question #11 : Applications Of Derivatives

The velocity of a particle is given by v(t). Find the particle's acceleration when t=3.

v(t)=3t^3-4t^2-5t+16

Possible Answers:

a=27

a=-3

a=52

a=53

Correct answer:

a=52

Explanation:

The velocity of a particle is given by v(t). Find the particle's acceleration when t=3.

v(t)=3t^3-4t^2-5t+16

To find the acceleration from a velocity function, simply take the derivative.

In this case, we are given v(t), and we need to find v'(t) because v'(t)=a(t).

To find v'(t), we need to use the power rule.

For each term, simply multiply by the exponent, and then subtract one from the exponent. Constant terms will drop out, linear terms will become constants, and so on.

v(t)=3t^3-4t^2-5t+16

$v'(t)=a(t)=(3)3t^{3-1}-(2)4t^{2-1}-5=9t^2-8t-5$

So, our acceleration function is:

a(t)=9t^2-8t-5

Now, plug in 3 for t and solve.

a(3)=9(3)^2-8(3)-5=81-24-5=52

So, our answer is 52.

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Example Question #1 : Limits

Evaluate:

$\lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1}$

Possible Answers:

The limit does not exist.

$\infty$

$\frac{1}{2}$

Correct answer:

Explanation:

Let's examine the limit

$\lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1}$

first.

$\lim_{x\rightarrow \infty }x= \lim_{x\rightarrow \infty } \left ( 2^{x} - 1 \right )= \infty$

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

and

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{ x \ln 2} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x}e^{ x \ln 2} - \frac{\mathrm{d} }{\mathrm{d} x} 1$

$=\ln 2 \cdot e^{ x \ln 2} - 0 = \ln 2 \cdot 2^{x}$ ,

so by L'Hospital's Rule,

$\lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1} = \lim_{x\rightarrow \infty } \frac{1 }{\ln 2 \cdot 2^{x}}$

Since $\lim_{x\rightarrow \infty } \ln 2 \cdot 2^{x} = \infty$ ,

$\lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1} = \lim_{x\rightarrow \infty } \frac{1 }{\ln 2 \cdot 2^{x}} = 0$

Now, for each x > 0 , $0 \leq \sin^{2} x \leq 1$ ; therefore,

$0 \leq \frac{x \sin^{2} x}{2^{x} - 1} \leq \frac{x }{2^{x} - 1}$

By the Squeeze Theorem,

$0 \leq \lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1} \leq \lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1} = 0$

and

$\lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1} = 0$

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Example Question #1 : L'hospital's Rule

Evaluate:

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}$

Possible Answers:

$\frac{5}{3}$

$\infty$

$\ln \frac{5}{3}$

The limit does not exist.

$\frac{\ln 5}{\ln 3}$

Correct answer:

$\infty$

Explanation:

$\lim_{x\rightarrow \infty }\left ( 5^{x} - 1 \right ) = \lim_{x\rightarrow \infty }\left ( 3^{x} - 1 \right ) = \infty$

Therefore, by L'Hospital's Rule, we can find $\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}$ by taking the derivatives of the expressions in both the numerator and the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 5^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x \ln 5} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 5 } - \frac{\mathrm{d} }{\mathrm{d} x}1$

$=\ln 5\cdot e^{x \ln 5 } - 0 = \ln 5\cdot 5^{x }$

Similarly,

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 3^{x} - 1 \right ) = \ln 3 \cdot 3 ^{x}$

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1} = \lim_{x\rightarrow \infty } \frac{\ln 5 \cdot 5^{x} }{\ln 3 \cdot 3^{x} }$

$=\frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty } \frac{ 5^{x} }{ 3^{x} }= \frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty }\left ( \frac{ 5 }{ 3 } \right )^{x}$

But $\lim_{x\rightarrow \infty }a^{x} = \infty$ for any a > 1 , so

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}= \frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty }\left ( \frac{ 5 }{ 3 } \right )^{x} = \infty$

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Example Question #3 : Limits

Evaluate:

$\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1}$

Possible Answers:

$\infty$

$\ln \frac{7}{2}$

$\frac{1}{2}\ln 7$

The limit does not exist.

$\log_{2}7$

Correct answer:

$\log_{2}7$

Explanation:

$\lim_{x\rightarrow 0 }\left ( 7^{x} - 1 \right ) = 7^{0} - 1 = 1- 1 = 0$

and

$\lim_{x\rightarrow 0 }\left ( 2^{x} - 1 \right ) = 2^{0} - 1 = 1- 1 = 0$

Therefore, by L'Hospital's Rule, we can find $\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1}$ by taking the derivatives of the expressions in both the numerator and the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 7^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x \ln 7} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 7 } - \frac{\mathrm{d} }{\mathrm{d} x}1$

$=\ln 7\cdot e^{x \ln 7 } - 0 = \ln 7\cdot 7^{x }$

Similarly,

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2^{x} - 1 \right ) = \ln 2 \cdot 2 ^{x}$

$\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1} = \lim_{x\rightarrow 0 } \frac{ \ln 7\cdot 7^{x }}{ \ln 2\cdot 2^{x }}$

$= \frac{ \ln 7 \cdot 7^{0 }}{ \ln 2\cdot 2^{0 }} = \frac{ \ln 7\cdot 1}{ \ln 2\cdot 1} = \frac{ \ln 7}{ \ln 2 }= \log_{2}7$

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Example Question #1 : L'hospital's Rule

Evaluate:

$\lim_{x\rightarrow 0^{+} } \frac{x \sin x}{2^{x} - 1}$

Possible Answers:

$\frac{1}{\ln2}$

$\ln 2$

Correct answer:

Explanation:

$\lim_{x\rightarrow 0^{+} } x \sin x= 0 \cdot \sin 0 = 0$

and

$\lim_{x\rightarrow 0^{+} } \left ( 2 ^{x} - 1 \right )= 2^{0} -1 = 0$

Therefore, by L'Hospital's Rule, we can find $\lim_{x\rightarrow 0^{+} } \frac{x \sin x}{2^{x} - 1}$ by taking the derivatives of the expressions in both the numerator and the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}x \sin x = \frac{\mathrm{d} }{\mathrm{d} x} x \cdot \sin x + \frac{\mathrm{d} }{\mathrm{d} x}\sin x \cdot x$

$= 1 \cdot \sin x + \cos x \cdot x = \sin x + x \cos x$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x \ln 2} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 2} - \frac{\mathrm{d} }{\mathrm{d} x} 1$

$= \ln 2\cdot e^{x \ln 2} - 0 = \ln 2\cdot 2^{x }$

$\lim_{x\rightarrow 0^{+} } \frac{x \sin x}{2^{x} - 1} = \lim_{x\rightarrow 0^{+} } \frac{\sin x + x \cos x}{ \ln 2\cdot 2^{x }}$

$= \frac{\sin 0 + 0 \cos 0}{ \ln 2\cdot 2^{0 }} = \frac{ 0 + 0 }{ \ln 2\cdot 1} = 0$

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Example Question #1 : Euler's Method

Suppose we have the following differential equation with the initial condition:

$\frac{\partial p }{\partial x} = 0.5x(1-x), \ p(0) = 2$

Use Euler's method to approximate p(2) , using a step size of .

Possible Answers:

Correct answer:

Explanation:

We start at x = 0 and move to x=2, with a step size of 1. Essentially, we approximate the next step by using the formula:

$p(x + \Delta x) = p(x) + p'(x)\cdot (\Delta x)$ .

So applying Euler's method, we evaluate using derivative:

$p'(x) = 0.5x(1-x), \ p(0) = 2$

And two step sizes, at x = 1 and x = 2.

${}p(x=1) = p(0) + (1 - 0)p'(0) = 2 + 1\cdot p'(0) = 2 + 1 \cdot 0.5(0)(1-0) = 2 + 0 = 2$

${}p(x=2) = p(1) + (2-1)\cdot p'(1) = 2 + 1\cdot p'(1) = 2 + 0.5(1)(1-1) = 2 + 0 = 2$

And thus the evaluation of p at x = 2, using Euler's method, gives us p(2) = 2.

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AP Calculus BC : Derivatives

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Applications Of Derivatives

Example Question #191 : Derivatives

Example Question #12 : Applications Of Derivatives

Example Question #13 : Applications Of Derivatives

Example Question #11 : Applications Of Derivatives

Example Question #1 : Limits

Example Question #1 : L'hospital's Rule

Example Question #3 : Limits

Example Question #1 : L'hospital's Rule

Example Question #1 : Euler's Method

All AP Calculus BC Resources

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