### All AP Calculus BC Resources

## Example Questions

### Example Question #10 : First And Second Derivatives Of Functions

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

What is the velocity function of the car?

**Possible Answers:**

**Correct answer:**

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

The derivative was found using the following rules:

, , ,

### Example Question #32 : Derivatives

Let

Find the first and second derivative of the function.

**Possible Answers:**

**Correct answer:**

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

The chain rule states that if

and

then the derivative is

In order to find the first derviative of the function

we set

and

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

And differentiating we use the power rule which states

As such

And so

To solve for the second derivative we set

and

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

And differentiating we use the power rule which states

As such

And so the second derivative becomes

### Example Question #13 : First And Second Derivatives Of Functions

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

**Possible Answers:**

**Correct answer:**

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

and was found using the following rules:

, , ,

### Example Question #1 : Applications Of Derivatives

Find the first and second derivatives of the function

**Possible Answers:**

**Correct answer:**

We must find the first and second derivatives.

We use the properties that

- The derivative of is
- The derivative of is

As such

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

Setting

and

we find that

As such

### Example Question #2 : Applications Of Derivatives

Given the velocity function

where is real number such that , find the acceleration function

.

**Possible Answers:**

**Correct answer:**

We can find the acceleration function from the velocity function by taking the derivative:

We can view the function

as the composition of the following functions

so that . This means we use the chain rule

to find the derivative. We have and , so we have

### Example Question #3 : Applications Of Derivatives

The position of an object is given by the equation . What is its acceleration at t = 2?

**Possible Answers:**

**Correct answer:**

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

Now plug in 2 for t:

### Example Question #4 : Applications Of Derivatives

The equation models the position of an object after t seconds. What is the acceleration at 3 seconds?

**Possible Answers:**

**Correct answer:**

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

Plug in 3 for t:

### Example Question #5 : Applications Of Derivatives

The equation models the position of an object after t seconds. What is its speed after seconds?

**Possible Answers:**

**Correct answer:**

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

Plug in for t:

### Example Question #6 : Applications Of Derivatives

The position of an object is modeled by the equation What is the speed after seconds?

**Possible Answers:**

**Correct answer:**

If this function gives the position, the first derivative will give its speed. To differentiate, use the chain rule: . In this case, and . Since and , the first derivative is .

Plug in for t:

### Example Question #7 : Applications Of Derivatives

A particle's position on the -axis is given by the function from .

When does the particle change direction?

**Possible Answers:**

It doesn't change direction within the given bounds

**Correct answer:**

It doesn't change direction within the given bounds

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

.

Hence .

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

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