AP Calculus AB : Interpretation of the derivative as a rate of change

Study concepts, example questions & explanations for AP Calculus AB

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Example Questions

Example Question #11 : Interpretation Of The Derivative As A Rate Of Change

The velocity of a particle in feet per second can be modeled by the function

What is the acceleration of this particle at 6 seconds?

Possible Answers:

Correct answer:

Explanation:

The acceleration of a particle is its change in velocity; therefore, to find its acceleration at a particular point in time, it is necessary to take the derivative of the velocity function and evaluate it for that value.  In short, we want 

Differentiate :

Apply the sum rule:

Apply the chain rule by setting :

This is the acceleration function; substitute 6 for :

Example Question #12 : Interpretation Of The Derivative As A Rate Of Change

The velocity of a particle in feet after  seconds can be modeled by the equation 

.

At  seconds, give the acceleration of this particle in feet per second squared.

Possible Answers:

Correct answer:

Explanation:

The acceleration function of a particle with respect to time is the derivative  of the velocity function of the particle. 

Find this derivative by first using the product rule:

Apply the product rule:

Evaluate :

 feet per second squared.

Example Question #31 : Applications Of Derivatives

Determine the velocity of a particle at x=10 when the position function is given by

Possible Answers:

Correct answer:

Explanation:

To determine the velocity function - the rate of change of position - we must take the derivative of the position function:

The derivative was found using the following rules:

To find the velocity at a certain value of x - the instantaneous rate of change of position - we simply plug in the given value of x into the velocity function:

 

Example Question #31 : Applications Of Derivatives

A crystal forms at a rate given by the following equation:

What is the rate of change of the crystal growth rate at t=5?

Possible Answers:

None of the other answers

Correct answer:

Explanation:

To find the rate of change of crystal growth at a given time, we must take the derivative of the growth rate function and evaluate it at a specific time.

The derivative of the function is

and was found using the following rules:

Now, to find the growth rate at t=5, simply plug in this value for t into the derivative function:

Example Question #571 : Derivatives

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

 

Possible Answers:

Correct answer:

Explanation:

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

Velocity is the first derivative of position. Acceleration is the first derivative of velocity.

Therefore, all we need to do to solve this problem is to find the second derivative of p(t).

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

1) 

2) 

So, we these rules in mind, we get:

So our velocity function is:

Next, differentiate v(t) to get a(t).

So, our acceleration function is

Example Question #33 : Applications Of Derivatives

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

Possible Answers:

Correct answer:

Explanation:

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

Velocity is the first derivative of position. Acceleration is the first derivative of velocity.

Therefore, all we need to do to solve this problem is to find the second 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) 

2) 

So, we these rules in mind, we get:

So our velocity function is:

Next, differentiate v(t) to get a(t).

So, our acceleration function is

Next, plug in 0 and simplify.

Our answer is 12. Since we are not given any units, we can leave it as 12

Example Question #31 : Applications Of Derivatives

A car is moving with a velocity that can be modeled by the equation

What is the cars acceleration at 

Possible Answers:

Correct answer:

Explanation:

The car's acceleration is the instantaneous rate of change in its velocity with respect to time (), so we can find the value of the cars acceleration at any time by taking the derivative of the velocity equation

Evaluating at , we get

Example Question #32 : Applications Of Derivatives

A body's position "s" is given by the equation:

    ,   

 

a) Find the body's speed at the endpoints of the given interval

b) Find the body's acceleration at the endpoints of the given interval

Possible Answers:

Correct answer:

Explanation:

We are given the function describing the position of the body given a time "t":

We are also given the interval that the function can be applied over: 

First, we are tasked to find the speed of the body at each of the endpoints (0 and 2 seconds, respectively). 

To figure this, we must understand that speed is the absolute value of velocity

To find the velocity function, we must take the derivative of the position function with respect to time:

                                  

                                  

Now that we have the function of velocity given a time "t", we can find the speed of the body given a time "t" by simply taking the absolute value of the velocity function. 

Now, to find the speed of the endpoints:

 

We can repeat the same process to find the speed at 2 seconds.

 

For part a, the speeds at 0 and 2 seconds are 3 and 1 [m/s], respectively. 

Part b asks us to find the acceleration at the endpoint times (0 and 2 seconds). To do this, we must first understand that to find acceleration at a time "t", we must take the derivative, with respect to time, of the velocity function

From part a, we found the velocity function:

Thus, to find acceleration, we derive this function with respect to time.

                                 

                                  

                                  

The result of our derivation tells us that no matter what time is plugged into the function, the acceleration shall always return 2[m/s^2].

So, for part b, the acceleration at 0 and 2 seconds are 2 and 2 [m/s^2], respectively. 

 

 

 

 

Example Question #19 : Interpretation Of The Derivative As A Rate Of Change

At any time t, the position of a body is given by the equation:

 

Find the body's acceleration at each time the velocity is zero. 

 

Possible Answers:

Correct answer:

Explanation:

We are given the position equation:

To find the acceleration when the velocity is equal to zero, we must first find the function to describe the velocity given a time "t".

We know that the velocity function is found by deriving the position function with respect to time

                                  

                                  

Thus, the velocity function is given by the equation: 

To find the acceleration when the velocity is equal to zero, we must set the velocity function equal to zero and solve for the times. 

 (multiply c factor by the a coefficient)

 (divide out the original a coefficient)

 

Thus, the velocity is zero at both 3 and 1 seconds. The question asks for the acceleration at these times. 

To do the next part, we must understand that the acceleration function is the derivative of the velocity function, with respect to time. We already know the velocity function:

So, if we derive this function with respect to time and plug in our times, we will know the acceleration when the velocity is equal to zero. 

                                  

                                  


Our derivation tells us that given any time "t", the acceleration of the body can be found by the function:

So, to find the acceleration at times 3 and 1, we simply plug in the values to our acceleration function above. 

 

So, the acceleration at time 1 is -6[m/s^2] and the acceleration at time 3 is 6[m/s^2].

 

Example Question #20 : Interpretation Of The Derivative As A Rate Of Change

Given the function describing an object's position with respect to time:

 

a) Find the function describing the object's velocity with respect to time

 

and

 

b) Find the function describing the object's acceleration with respect to time

Possible Answers:

Correct answer:

Explanation:

This problem, although not technically complex, requires some understanding of the derivative and how it relates to elementary physics. 

 

For part a)

Given a function that returns an objects position at time t, we can take the derivative of the position function to find the velocity function

We are given the position function:

 

 

 

 

We can apply the general power rule for derivatives to find the derivative of this function:

And from our understanding that the derivative of the position function is the velocity function, we can make the substitution

Thus, our part a answer is:

 

For part b)

Furthermore, we can take the derivative of the velocity function to find our acceleration function.

We found the velocity function to be:

Thus, the derivative of the velocity function (the acceleration function) is simply:

 

The full answer should be:

 

 

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