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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #2 : How To Find Position

Given the acceleration function is $a(t)=2t^2-t+3$ , intial velocity is 2 m/s, and the displacement at t=1 is x(1)=4 .

Find its initial position.

Possible Answers:

0.5

-0.5

Correct answer:

0.5

Explanation:

First, we find the velocity function by integrating the acceleration function $v(t)=\int 2t^2-t+3 = \frac{2}{3}t^3-\frac{1}{2}t^2+3t+c$

Since the initial velocity is 2 m/s, we have c=2 .

Then we integrate again to get the displacement function.

$x(t)=\int \frac{2}{3}t^3-\frac{1}{2}t^2+3t+2=\frac{1}{6}t^4-\frac{1}{6}t^3+\frac{3}{2}t^2+2t+k$

$x(1)=1.5+2+k=3.5+k=4$

This implies k=0.5

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Example Question #221 : Ap Calculus Ab

The velocity of a particle moving in the x-direction is given by the equation v(t)=3-4t . What is the maximum value of the particle's displacement, provided that its displacement is zero when t=0?

Possible Answers:

3/4

No maximum exists.

1/2

$\frac{9}{8}$

4/3

Correct answer:

$\frac{9}{8}$

Explanation:

In order to find the maximum value (if one exists) of the displacement (which we can call s), we must determine where its derivative equals zero or is undefined. Because the velocity of the particle is the derivative of its displacement with respect to time, we must set the velocity equal to zero.

v(t)=3-4t = 0

$t=\frac{3}{4}$

We must verify if this is indeed where a local maximum occurs. One way to do this is to examine the value of the second derivative of the displacement function. Because the first derivative is equal to the velocity, the second derivative of the displacement function is simply the derivative of the velocity function with respect to time.

s''(t)=v'(t)

s''(t)=-4 < 0

The second derivative of the displacement function is negative, which means that a maximum does indeed occur where t = 3/4.

However, the question asks us to find the maximum value of the displacement. Thus, we will need to determine the equation for the displacement. Because v(t)=s'(t) , if we integrate both sides of this equation, we can obtain s(t).

s'(t)=3-4t

$s(t)=\int v(t)dt=\int (3-4t)dt$

s(t)=3t-2t^2+C

We can find C by using the given information that the displacement of the particle is zero when t = 0.

s(0)=3(0)-2(0)^2+C=0

C=0

Thus, the equation of the displacement function is

s(t)=3t-2t^2 .

The problem ultimately asks us to determine the maximum value of the displacement, which we already verified will occur when t = 3/4.

$s(\frac{3}{4})=3\cdot \frac{3}{4}-2\cdot(\frac{3}{4})= \frac{9}{8}$

The answer is $\frac{9}{8}$ .

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Example Question #1 : Integrals

Write the equation of a tangent line to the given function at the point.

y = ln(x²) at (e, 3)

Possible Answers:

y = (2/e)(x – e)

y = (2/e)

y – 3 = ln(e²)(x – e)

y – 3 = (x – e)

y – 3 = (2/e)(x – e)

Correct answer:

y – 3 = (2/e)(x – e)

Explanation:

To solve this, first find the derivative of the function (otherwise known as the slope).

y = ln(x²)

y' = (2x/(x²))

Then, to find the slope in respect to the given points (e, 3), plug in e.

y' = (2e)/(e²)

Simplify.

y'=(2/e)

The question asks to find the tangent line to the function at (e, 3), so use the point-slope formula and the points (e, 3).

y – 3 = (2/e)(x – e)

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Example Question #1 : Numerical Approximations To Definite Integrals

Find the equation of the tangent line at (1,-1) when

$y=\frac{x^{3}-2x(x^{1/2})}{x}$

Possible Answers:

y=x+1

y=1

y=x-2

y=-x-2

y=x

Correct answer:

y=x-2

Explanation:

The answer is

$y=\frac{x^{3}-2x(x^{1/2})}{x}$ let's go ahead and cancel out the 's. This will simplify things.

$y=x^{2}-2(x^{1/2})$

$y'=2x-\frac{1}{(x^{1/2})}$

$y'(1)=2(1)-\frac{1}{(1^{1/2})} =1$ this is the slope so let's use the point slope formula.

y+1=1(x-1)

y+1=x-1

y=x-2

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Example Question #11 : Integrals

Differentiate $y=\frac{t+2}{t^{2}-4t-12}$

Possible Answers:

$y'=\frac{1}{(t^{2}-4t-12)^{2}}$

$y'=\frac{1}{(t-6)}$

$y'=\frac{(t+2)-1}{(t^{2}-4t-12)^{2}}$

$y'=\frac{-1}{(t-6)^{2}}$

$y'=\frac{1}{2t-4}$

Correct answer:

$y'=\frac{-1}{(t-6)^{2}}$

Explanation:

We see the answer is $y'=\frac{-1}{(t-6)^{2}}$ after we simplify and use the quotient rule.

$y=\frac{t+2}{t^{2}-4t-12}$ we could use the quotient rule immediatly but it is easier if we simplify first.

$y=\frac{t+2}{(t+2)(t-6)}$

$y=\frac{1}{(t-6)}$

$y=(t-6)^{-1}$

$y'=-(t-6)^{-2}$

$y'=\frac{-1}{(t-6)^{2}}$

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Example Question #11 : Integrals

Find $\lim_{x\rightarrow \infty }\frac{-2x^3+x^2-2}{x^2+10}$

Possible Answers:

$1$

$0$

$\infty$

$-\infty$

$-2$

Correct answer:

$-\infty$

Explanation:

When taking limits to infinity, we usually only consider the highest exponents. In this case, the numerator has $-2x^3$ and the denominator has $x^2$ . Therefore, by cancellation, it becomes $-2x$ as approaches infinity. So the answer is $-\infty$ .

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Example Question #12 : Integrals

Evaluate:

$\sum_{n=1}^{\infty }4(\frac{-1}{2})^{n-1}$

Possible Answers:

cannot be determined

$\infty$

$\frac{8}{3}$

Correct answer:

$\frac{8}{3}$

Explanation:

First, we can write out the first few terms of the sequence $a_{n}=$ $4(\frac{-1}{2})^{n-1}$ , where ranges from 1 to 3.

$a_{1}=4(\frac{-1}{2})^{1-1}=4$

$a_{2}=4(\frac{-1}{2})^{2-1}=-2$

$a_{3}=4(\frac{-1}{2})^{3-1}=1$

Notice that each term $\left ( n>1 \right )$ , is found by multiplying the previous term by $-\frac{1}{2}$ . Therefore, this sequence is a geometric sequence with a common ratio of $-\frac{1}{2}$ . We can find the sum of the terms in an infinite geometric sequence, provided that |r|<1 , where is the common ratio between the terms. Because $r=-\frac{1}{2}$ in this problem, $\left | r \right |$ is indeed less than 1. Therefore, we can use the following formula to find the sum, , of an infinite geometric series.

$S=\frac{a_{1}}{1-r}=\frac{4}{1-(\frac{-1}{2})}=\frac{4}{3/2}=\frac{8}{3}$

The answer is $\frac{8}{3}$ .

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Example Question #11 : Integrals

If $f(x)=4, f'(x)=3, g(x)=2, \ and\ g'(x)=1$ then find h'(x) .

$h(x)=\frac{2f}{g}$

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

The answer is 1.

$h(x)=\frac{2f}{g}$

$h'(x)=\frac{2(f'(x)g(x)-f(x)g'(x))}{g^{2}}$

$h'(x)=\frac{2(3*2-4*1)}{4} = 1$

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Example Question #2 : Numerical Approximations To Definite Integrals

Find the equation of the tangent line at (4,1) on graph f(x)=(x-2)(x+4)

Possible Answers:

y=3x-7

y=7x+8

y=x-11

y=x+40

y=10x-39

Correct answer:

y=10x-39

Explanation:

The answer is 10x-39

f(x)=(x-2)(x+4)

$f(x)=x^{2}+2x-8$

$f'(x)=2x+2$

f'(x) = 2(4)+2=10

(This is the slope. Now use the point-slope formula)

y-1=10(x-4)

y-1=10x-40

y=10x-39

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Example Question #13 : Integrals

Find the equation of the tangent line at (1,1) in

$f(x)=ax^{2}+bx+c$

Possible Answers:

y=2ax-b+1

y=2ab+b-1

y=2ax+bx-2a-b+1

y=2ax-bx-2a+1

y=2a+b

Correct answer:

y=2ax+bx-2a-b+1

Explanation:

The answer is y=2ax+bx-2a-b+1

$f(x)=ax^{2}+bx+c$

$f'(x)=2ax+b$

$f'(1)=2a(1)+b = 2a+b$

(This is the slope. Now use the point-slope formula.)

y-1=(2a+b)(x-1)

y-1=2ax+bx-2a-b

y=2ax+bx-2a-b+1

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #2 : How To Find Position

Example Question #221 : Ap Calculus Ab

Example Question #1 : Integrals

Example Question #1 : Numerical Approximations To Definite Integrals

Example Question #11 : Integrals

Example Question #11 : Integrals

Example Question #12 : Integrals

Example Question #11 : Integrals

Example Question #2 : Numerical Approximations To Definite Integrals

Example Question #13 : Integrals

All AP Calculus AB Resources

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