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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #552 : Derivatives

Find the eqation of the line tangent to the graph of $y=x+x^{1/2}$ at point (4,6) .

Possible Answers:

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

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

$y=\frac{1}{2x^{1/2}}$

$y=1+\frac{1}{2(6^{1/2})}$

$y=\frac{5}{4}x+1$

Correct answer:

$y=\frac{5}{4}x+1$

Explanation:

$y=x+x^{1/2}$

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

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

Now we plug in x=4 .

$y'=1+\frac{1}{2(4^{1/2})} = 1+\frac{1}{4} = \frac{5}{4}$ this is the slope.

Now use the point slope formula to find the tangent line.

$y-6=\frac{5}{4}(x-4)$

$y-6=\frac{5}{4}x-5$

$y=\frac{5}{4}x+1$

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Example Question #21 : Applications Of Derivatives

Find the rate of change of y if $y=5x^{-3}+3e^{x}$

Possible Answers:

$y'=6x^{3}+e^{3x}$

$y'=-15x^{-4}+3e^{x}$

$y'=-6x^{-2}+3e^{3x}$

$y'=-15x^{-3}+4e^{x}$

$y'=-3x^{-4}+3e^{x}$

Correct answer:

$y'=-15x^{-4}+3e^{x}$

Explanation:

The rate of change of y is also the derivative of y.

Differentiate the function given.

You should get $y'=-15x^{-4}+3e^{x}$

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Example Question #22 : Applications Of Derivatives

If p(t) gives the position of an asteroid as a function of time, find the function which models the velocity of the asteroid as a function of time.

p(t)=-365sin(t)+12t^3-3t^2+164

Possible Answers:

p'(t)=365cos(t)+36t^3-6t

p'(t)=-365cos(t)+36t^3-6t

p'(t)=-365+cos(t)+36t^3-6t

p'(t)=-365cos(t)+36t^3-6t+164

Correct answer:

p'(t)=-365cos(t)+36t^3-6t

Explanation:

If p(t) gives the position of an asteroid as a function of time, find the function which models the velocity of the asteroid as a function of time.

p(t)=-365sin(t)+12t^3-3t^2+164

Begin by recalling that velocity is the first derivative of position. So all we need to do is find the first derivative of our position function.

Recall that the derivative of sine is cosine, and that the derivative of polynomials can be found by multiplying each term by its exponent, and decreasing the exponent by 1.

Starting with:

p(t)=-365sin(t)+12t^3-3t^2+164

We get:

p'(t)=-365cos(t)+36t^3-6t

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Example Question #23 : Applications Of Derivatives

Given j(k), find the rate of change when k=5.

j(k)=-3k^3+4k+16

Possible Answers:

125

Correct answer:

Explanation:

Given j(k), find the rate of change when k=5

j(k)=-3k^3+4k+16

Let's begin by realizing that a rate of change refers to a derivative.

So, we need to find the derivative of j(k)

j(k)=-3k^3+4k+16

We find this by multiplying each term by the exponent, and decreasing the exponent by 1

j(k)=-3k^3+4k+16

j'(k)=-9k^2+4

Next, plug in 5 to find our answer:

j'(5)=-9(5)^2+4=-9*25+4=-225+4=-221

So, our rate of change is -221.

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Example Question #4 : Interpretation Of The Derivative As A Rate Of Change

If p(t) gives the position of a planet as a function of time, find the function which models the planet's velocity.

p(t)=-12t^3+6t^2-12t+sin(t)

Possible Answers:

v(t)=-36t^2+12t-12+cos(t)

v(t)=36t^2+12+cos(t)

v(t)=-72t^2+12-sin(t)

v(t)=-36t^2+12t-cos(t)

Correct answer:

v(t)=-36t^2+12t-12+cos(t)

Explanation:

If p(t) gives the position of a planet as a function of time, find the function which models the planet's velocity.

p(t)=-12t^3+6t^2-12t+sin(t)

Velocity is the first derivative of position.

Therefore, all we need to do to solve this problem is to find the first derivative.

We can do this via the power rule and the rule for differentiating sine.

1) $\frac{d}{dx}x^n=(n)x^{n-1}$

2) $\frac{d}{dx}(sin(x))=cos(x)$

So, we these rules in mind, we get:

p(t)=-12t^3+6t^2-12t+sin(t)

p'(t)=v(t)=-36t^2+12t-12+cos(t)

So our final answer is:

v(t)=-36t^2+12t-12+cos(t)

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Example Question #5 : Interpretation Of The Derivative As A Rate Of Change

If p(t) gives the position of a planet as a function of time, find the planet's velocity when t=0.

p(t)=-12t^3+6t^2-12t+sin(t)

Possible Answers:

Correct answer:

Explanation:

If p(t) gives the position of a planet as a function of time, find the planet's velocity when t=0.

p(t)=-12t^3+6t^2-12t+sin(t)

Velocity is the first derivative of position. Therefore, all we need to do to solve this problem is to find the first derivative of p(t) and then plug in 0 for t and solve.

We can do this via the power rule and the rule for differentiating sine and cosine.

1) $\frac{d}{dx}x^n=(n)x^{n-1}$

2) $\frac{d}{dx}(sin(x))=cos(x)$

So, we these rules in mind, we get:

p(t)=-12t^3+6t^2-12t+sin(t)

p'(t)=v(t)=-36t^2+12t-12+cos(t)

So our velocity function is:

v(t)=-36t^2+12t-12+cos(t)

Now, plug in 0 and simplify.

v(0)=-36(0)^2+12(0)-12+cos(0)=0+0-12+1=-11

So, our answer is -11. We are not given any units, so we do not need to worry about them.

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Example Question #4 : Interpretation Of The Derivative As A Rate Of Change

A factory producing pens wants to maximize its output; to do so, it needs to find the rate of change of pen production. Find this rate if pens are produced according to the following function:

$p(t)=5000t^2+400\cos(t)$

Possible Answers:

$1000t-400\sin(t)$

$4000t-400\sin(t)$

$10000t+400\sin(t)$

$10000t-400\sin(t)$

Correct answer:

$10000t-400\sin(t)$

Explanation:

The rate of change of a function is given by the derivative of that function. So, to find the rate of change of production, we must take the first derivative of the function for production, which is equal to

$p'(t)=10000t-400\sin(t)$

found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(t)=-\sin(t)$

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Example Question #6 : Interpretation Of The Derivative As A Rate Of Change

Concrete at a factory flows according to the following theoretical model:

$c(x)=2x^2-5\cos(x)+5\sin(x)$

What is the rate of change of the concrete flow?

Possible Answers:

$4x-5\sin(x)+5\cos(x)$

$4x+5\sin(x)+5\cos(x)$

$4x+5\sin(x)-5\cos(x)$

Correct answer:

$4x+5\sin(x)+5\cos(x)$

Explanation:

The rate of change of the concrete flow is given by the first derivative of the concrete flow function:

$c'(x)=4x+5\sin(x)+5\cos(x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

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Example Question #1 : Interpretation Of The Derivative As A Rate Of Change

Find the speed of the car at t=5 if its position is given by

$s(t)=4t^2+\frac{t}{5}$

Possible Answers:

40.2

-40.2

Correct answer:

40.2

Explanation:

To determine the speed of the car, we must take the first derivative of the position function, which gives us the rate of change of the position of the car - in other words, speed.

$s'(t)=8t+\frac{1}{5}$

The derivative was found using the following rule:

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

Evaluating the derivative function at t=5, we get our speed as

$s'(5)=8(5)+\frac{1}{5}=40.2$

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Example Question #2 : Interpretation Of The Derivative As A Rate Of Change

A group of scientists use the following code for describing velocity and acceleration of particles:

$\alpha$ , when both the velocity and acceleration are positive;
$\beta$ , when the velocity is positive, but the acceleration is negative;
$\gamma$ , when the velocity is negative, but the acceleration is positive;
$\delta$ , when both the velocity and acceleration are negative.

What code - at t=2 - will the scientists use when describing a particle moving with a position function given by the following equation:

s(t)=12t-t^2

Possible Answers:

$\alpha$

$\gamma$

$\beta$

$\delta$

We need more information to determine the code used

Correct answer:

$\beta$

Explanation:

To determine which code the scientists will use, we must find the velocity and acceleration of the particle, given by the first and second derivatives of the position function, respectively, evaluated at the given point.

The velocity and acceleration functions, therefore, are

v(t)=s'(t)=12-2t

a(t)=s''(t)=-2

The derivatives were found using the following rules:

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

Evaluated at t=2, we find that

v(2)=8 > 0, a(2)=-2 < 0

When the velocity is positive, but the acceleration is negative, the code the scientists use is $\beta$ .

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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 #552 : Derivatives

Example Question #21 : Applications Of Derivatives

Example Question #22 : Applications Of Derivatives

Example Question #23 : Applications Of Derivatives

Example Question #4 : Interpretation Of The Derivative As A Rate Of Change

Example Question #5 : Interpretation Of The Derivative As A Rate Of Change

Example Question #4 : Interpretation Of The Derivative As A Rate Of Change

Example Question #6 : Interpretation Of The Derivative As A Rate Of Change

Example Question #1 : Interpretation Of The Derivative As A Rate Of Change

Example Question #2 : Interpretation Of The Derivative As A Rate Of Change

All AP Calculus AB Resources

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