### All AP Calculus AB Resources

## Example Questions

### Example Question #22 : Calculus Ii — Integrals

Evaluate the following limit:

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**Correct answer:**

When approaches 0 both and will approach . Therefore, L’Hopital’s Rule can be applied here. Take the derivatives of the numerator and denominator and try the limit again:

### Example Question #1 : Derivative Interpreted As An Instantaneous Rate Of Change

Find the instantaneous rate of change for the function,

at the point .

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**Correct answer:**

Find the instantaneous rate of change for the function,

at the point .

**1) **First compute the derivative of the function, since this will give us the instantaneous rate of change of the function as a function of .

**2) **Now evaluate the derivative at the value ,

Therefore, is the instantaneous rate of change of the function

at the point .

### Example Question #2 : Derivative Interpreted As An Instantaneous Rate Of Change

A particle is traveling in a straight line along the x-axis with position function . What is the instantaneous rate of change in the particle's position at time seconds?

**Possible Answers:**

**Correct answer:**

To find the instantaneous rate of change of the particle at time , we have to find the derivative of and plug into it.

.

And

.

Hence the instantaneous rate of change in position (or just 'velocity') of the particle at is . (At that very instant, the particle is not moving.)

### Example Question #3 : Derivative Interpreted As An Instantaneous Rate Of Change

Find the function values and as well as the instantaneous rate of change for the function corresponding to the following values of

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**Correct answer:**

Find the instantaneous rate of change for the function corresponding to the following values of

Evaluate the function at each value of

The instantaneous rate of change at any point will be given by the derivative at that point. First compute the derivative of the function:

Apply the product rule:

Therefore,

Now evaluate the derivative for each given value of :

Therefore, the instantaneous rate of change of the function at the corresponding values of are:

### Example Question #4 : Derivative Interpreted As An Instantaneous Rate Of Change

Given that v(t) is the velocity of a particle, find the particle's acceleration when t=3.

**Possible Answers:**

Not enough information provided

**Correct answer:**

Given that v(t) is the velocity of a particle, find the particle's acceleration when t=3.

We are given velocity and asked to find acceleration. Our first step should be to find the derivative.

We can use our standard power rule for our 1st and 3rd terms, but we need to remember something else for our second term. Namely, that the derivative of is simply

With that in mind, let;s find v'(t)

Now, for the final push, we need to find the acceleration when t=3. We do this by plugging in 3 for t and simplifying.

So, our answer is 123.5

### Example Question #5 : Derivative Interpreted As An Instantaneous Rate Of Change

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### Example Question #6 : Derivative Interpreted As An Instantaneous Rate Of Change

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### Example Question #7 : Derivative Interpreted As An Instantaneous Rate Of Change

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### Example Question #8 : Derivative Interpreted As An Instantaneous Rate Of Change

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### Example Question #9 : Derivative Interpreted As An Instantaneous Rate Of Change

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