We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus BC : Applications of Derivatives

Study concepts, example questions & explanations for AP Calculus BC

Create An Account

All AP Calculus BC Resources

2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : First And Second Derivatives Of Functions

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

$f(x)=\cos(x)e^{11x}+\csc(x)$

What is the velocity function of the car?

Possible Answers:

$-\sin(x)e^{11x}+\cos(x)e^{11x}-\csc(x)\cot(x)$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}+\csc(x)\cot(x)$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}$

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

$\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

Correct answer:

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

Explanation:

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

$-\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

Report an Error

Example Question #1 : Applications Of Derivatives

Let

$f(x)=e^{\pi x}$

Find the first and second derivative of the function.

Possible Answers:

$f'(x)=e^{\pi x }$

$f''(x)=e^{\pi x}$

$f'(x)=\pi e^{\pi x }$

$f''(x)=\pi^2 e^{\pi x}$

$f'(x)=e^{\pi }$

$f''(x)= e^{x}$

$f'(x)=\pi$

$f''(x)=\0$

Correct answer:

$f'(x)=\pi e^{\pi x }$

$f''(x)=\pi^2 e^{\pi x}$

Explanation:

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

The chain rule states that if

y=f(u)

and

u=g(x)

then the derivative is

$\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}$

In order to find the first derviative of the function

$f(x)=e^{\pi x}$

we set

y=e^u

and

$u=\pi x$

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

$\frac{dy}{du}=\frac{d}{du}[e^u]$

$\frac{dy}{du}=e^u$

And differentiating we use the power rule which states

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

As such

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]$

$\frac{du}{dx}=\pi x^{1-1}=\pi$

And so

$\frac{dy}{dx}=e^{\pi x}*\pi=\pi e^{\pi x}$

$f'(x)=\pi e^{\pi x}$

To solve for the second derivative we set

$y=\pi e^{u}$

and

$u=\pi x$

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

$\frac{dy}{du}=\frac{d}{du}[\pi e^u]$

$\frac{dy}{du}=\pi e^u$

And differentiating we use the power rule which states

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

As such

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]$

$\frac{du}{dx}=\pi x^{1-1}=\pi$

And so the second derivative becomes

$\frac{dy}{dx}=\pi e^{\pi x}*\pi=\pi^2 e^{\pi x}$

$f''(x)=\pi^2e^{\pi x}$

Report an Error

Example Question #2 : Applications Of Derivatives

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

$s(t)=e^t\tan(t)+t^5+3$

Possible Answers:

$te^t\tan(t)+e^t\sec^2(t)+5t^4$

$e^t\tan(t)+e^t\sec^2(t)+t^5$

$e^t\tan(t)-e^t\sec^2(t)+5t^4$

$e^t\tan(t)+e^t\sec^2(t)+5t^4$

Correct answer:

$e^t\tan(t)+e^t\sec^2(t)+5t^4$

Explanation:

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

$v(t)=s'(t)=e^t\tan(t)+e^t\sec^2(t)+5t^4$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(t)=\sec^2(t)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

Report an Error

Example Question #1 : Applications Of Derivatives

Find the first and second derivatives of the function

f(x)=sec(x)

Possible Answers:

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

f'(x)=sec^2(x)

f''(x)=sec^3(x)

f'(x)=csc(x)

f''(x)=-sec(x)

f'(x)=cot(x)

f''(x)=csc^2(x)

Correct answer:

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

Explanation:

We must find the first and second derivatives.

We use the properties that

The derivative of is
The derivative of is

As such

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)]=sec(x)tan(x)$

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

$\frac{\mathrm{d} }{\mathrm{d} x}[uv]=uv'+vu'$

Setting

u=sec(x) and v=tan(x)

we find that

$f''(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)tan(x)]=sec(x)*sec^2(x)+tan(x)*sec(x)tan(x)$

=sec^3(x)+sec(x)tan^2(x)

As such

f'(x)=sec(x)tan(x)

f''(x)=sec^3(x)+sec(x)tan^2(x)

Report an Error

Example Question #4 : Applications Of Derivatives

Given the velocity function

$v(x)=\int_{0}^{x^2+2x+1} z^d\sin(z^5) dz$

where is real number such that $d\in [0,1]$ , find the acceleration function

a(x) .

Possible Answers:

$a(x)=(x^2+2x+1)^{d}\sin (x^2+2x+1)^{10}$

$a(x)=(x+1)^{d}\sin (x+1)^{10}$

$a(x)=(2x+2)^{d}(x+1)^{2d}\sin (x+1)^{10}$

$a(x)=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

$a(x)=(x^2+2x+1)^{2d}\sin (x^2+2x+1)^{10}$

Correct answer:

$a(x)=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

Explanation:

We can find the acceleration function a(x) from the velocity function by taking the derivative:

a(x)=v'(x)

We can view the function

$v(x)=\int_{0}^{x^2+2x+1} z^d\sin(z^5) dz$

as the composition of the following functions

$g(x)=\int_{0}^{x} z^d\sin(z^5) dz$

h(x)=x^2+2x+1=(x+1)^2

so that v(x)=g(h(x)) . This means we use the chain rule

[g(h(x))]'=g'(h(x))h'(x)

to find the derivative. We have $g'(x)=x^d\sin x^5$ and h'(x)=2x+2 , so we have

$a(x)=v'(x)=[(x+1)^2]^d\sin [(x+1)^2]^5\cdot (2x+2)$

$=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

Report an Error

Example Question #1 : Velocity, Speed, Acceleration

The position of an object is given by the equation $y = \frac{1}{3} t^3 - \frac{1}{2} t^2 -2t + 4$ . What is its acceleration at t = 2?

Possible Answers:

Correct answer:

Explanation:

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

y ' = t^2 - t -2

y'' = 2t - 1

Now plug in 2 for t:

y'' = 2(2) - 1 = 4- 1 = 3

Report an Error

Example Question #1 : Applications Of Derivatives

The equation $y=\frac{1}{2} t^4 +10t$ models the position of an object after t seconds. What is the acceleration at 3 seconds?

Possible Answers:

Correct answer:

Explanation:

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

y' = 2 t^3 +10

y'' = 6t^2

Plug in 3 for t:

6(3)^2 = 6(9) = 54

Report an Error

Example Question #1 : Velocity, Speed, Acceleration

The equation $y= \sin (2t)$ models the position of an object after t seconds. What is its speed after $\pi$ seconds?

Possible Answers:

Correct answer:

Explanation:

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

$y' = \cos (2t ) \cdot 2$

Plug in $\pi$ for t:

$2 \cos (2 \pi ) = 2\cdot 1 = 2$

Report an Error

Example Question #2 : Velocity, Speed, Acceleration

The position of an object is modeled by the equation $y= \frac{1}{2} \cos ^2 t$ What is the speed after $\frac{\pi}{4}$ seconds?

Possible Answers:

$-\frac{1}{2}$

$-\frac{1}{\sqrt2}$

$\sqrt{2}$

Correct answer:

$-\frac{1}{2}$

Explanation:

If this function gives the position, the first derivative will give its speed. To differentiate, use the chain rule: $f(g(x)) ' = f'(g(x))\cdot g'(x)$ . In this case, $f(x) = \frac{1}{2} x^2$ and $g(x) = \cos x$ . Since f'(x) = x and $g'(x ) = - \sin (x)$ , the first derivative is $\cos t \cdot - \sin t$ .

Plug in $\frac{\pi}{4}$ for t:

$\cos (\frac{\pi}{4} ) \cdot - \sin (\frac{\pi}{4} ) = \frac{1}{\sqrt{2}} \cdot - \frac{1}{\sqrt2} = -\frac{1}{2}$

Report an Error

Example Question #1 : Velocity, Speed, Acceleration

A particle's position on the -axis is given by the function $f(t)=\sin(t)-t+1$ from $0 < t <\pi$ .

When does the particle change direction?

Possible Answers:

It doesn't change direction within the given bounds

$x=2\pi$

$x=\frac{\pi}{2}$

$x= \pi$

$x= \frac{\pi}{4}$

Correct answer:

It doesn't change direction within the given bounds

Explanation:

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

$0=v(t) = f'(t) = \cos(t)-1$ .

Hence $\cos(t)=1$ .

The solutions to this on the unit circle are $t =0, \pi$ , so these are the values of where the particle would normally change direction. However, our given interval is $0 < t < \pi$ , which does not contain $0, \pi$ . Hence the particle does not change direction on the given interval.

Report an Error

View AP Calculus BC Tutors

James
Certified Tutor

University of Georgia, Bachelor in Arts, Political Science and Government.

View AP Calculus BC Tutors

Leo
Certified Tutor

Columbia International University, Bachelor of Science, Chemistry.

View AP Calculus BC Tutors

Reza
Certified Tutor

University of Sherbrooke, Doctor of Philosophy, Mathematics. University of Manitoba, Master of Science, Mathematics.

All AP Calculus BC Resources

2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

ISEE Tutors in Dallas Fort Worth, SAT Tutors in Denver, English Tutors in New York City, GRE Tutors in Denver, Physics Tutors in Atlanta, Biology Tutors in Washington DC, LSAT Tutors in San Francisco-Bay Area, ACT Tutors in Atlanta, Reading Tutors in Dallas Fort Worth, Reading Tutors in Seattle

Popular Courses & Classes

GRE Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in New York City, GRE Courses & Classes in Philadelphia, ISEE Courses & Classes in Houston, SSAT Courses & Classes in Houston, ISEE Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in Atlanta, Spanish Courses & Classes in Atlanta, GRE Courses & Classes in Los Angeles

Popular Test Prep

ISEE Test Prep in Philadelphia, GMAT Test Prep in New York City, GMAT Test Prep in Chicago, ACT Test Prep in Atlanta, MCAT Test Prep in Miami, ISEE Test Prep in Atlanta, ACT Test Prep in San Francisco-Bay Area, SAT Test Prep in Atlanta, MCAT Test Prep in New York City, LSAT Test Prep in New York City

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

AP Calculus BC : Applications of Derivatives

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #3 : First And Second Derivatives Of Functions

Example Question #1 : Applications Of Derivatives

Example Question #2 : Applications Of Derivatives

Example Question #1 : Applications Of Derivatives

Example Question #4 : Applications Of Derivatives

Example Question #1 : Velocity, Speed, Acceleration

Example Question #1 : Applications Of Derivatives

Example Question #1 : Velocity, Speed, Acceleration

Example Question #2 : Velocity, Speed, Acceleration

Example Question #1 : Velocity, Speed, Acceleration

All AP Calculus BC Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: