Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus BC : AP Calculus BC

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 #11 : Polar Form

Calculate the polar form hypotenuse of the following cartesian equation: y^2=100x^8

Possible Answers:

$\sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

$\frac{\tan(\theta)}{10\cos^3(\theta)}$

$\sqrt[3]{\frac{\tan(\theta)}{\cos(\theta)}}$

$\sqrt[3]{\frac{\tan(\theta)}{10\cos(\theta)}}$

Correct answer:

$\sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

Explanation:

In a cartesian form, the primary parameters are and . In polar form, they are and $\theta$

is the hypotenuse, and $\theta$ is the angle created by .

2 things to know when converting from Cartesian to polar.

$x= r\cos(\theta)$

$y= r\sin(\theta)$

You want to calculate the hypotenuse,

Solution:

y^2=100x^8

$y=10x^4=r\sin(\theta)$

$y=10(r\cos(\theta))^4$

$y=10r^4\cos^4(\theta) = r\sin(\theta)$

$r^3 = \frac{\tan(\theta)}{10\cos^3(\theta)}$

$r = \sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

Report an Error

Example Question #261 : Parametric, Polar, And Vector

Graph the equation $r=cos(\theta )$ where $0\leq \theta \leq2\pi$ .

Possible Answers:

R_sinx

R_cosx

R_cosx_1

Faker_cosx

R_cos2x

Correct answer:

R_cosx

Explanation:

At angle the graph as a radius of . As it approaches $\frac{\pi }{2}$ , the radius approaches .

As the graph approaches $\pi$ , the radius approaches .

Because this is a negative radius, the curve is drawn in the opposite quadrant between $\frac{3\pi }{2}$ and $2\pi$ .

Between $\pi$ and $\frac{3\pi }{2}$ , the radius approaches from and redraws the curve in the first quadrant.

Between $\frac{3\pi }{2}$ and $2\pi$ , the graph redraws the curve in the fourth quadrant as the radius approaches from .

Report an Error

Example Question #12 : Polar Form

Draw the graph of $r=cos(2\theta )$ from $0\leq \theta \leq2\pi$ .

Possible Answers:

R2_cos2x

R_sin2x

R_cos2x

R_sin2x

R_cosx_1

Correct answer:

R_cos2x

Explanation:

Because this function has a period of $\pi$ , the x-intercepts of the graph y=cos(2x) happen at a reference angle of $\frac{\pi }{4}$ (angles halfway between the angles of the axes).

Between and $\frac{\pi }{4}$ the radius approaches from .

Between $\frac{\pi }{4}$ and $\frac{\pi }{2}$ , the radius approaches from and is drawn in the opposite quadrant, the third quadrant because it has a negative radius.

From $\frac{\pi }{2}$ to $\frac{3\pi }{4}$ the radius approaches from , and is drawn in the fourth quadrant, the opposite quadrant.

Between $\frac{3\pi }{4}$ and $\pi$ , the radius approaches from .

From $\pi$ and $\frac{5\pi }{4}$ , the radius approaches from .

Between $\frac{5\pi }{4}$ and $\frac{3\pi }{2}$ , the radius approaches from . Because it is a negative radius, it is drawn in the opposite quadrant, the first quadrant.

Then between $\frac{3\pi }{2}$ and $\frac{7\pi }{4}$ the radius approaches from and is draw in the second quadrant.

Finally between $\frac{7\pi }{4}$ and $2\pi$ , the radius approaches from .

Report an Error

Example Question #2 : Derivatives Of Polar Form

Find the derivative of the following function:

$r=3\theta+e^\theta$

Possible Answers:

$\frac{(3+e^\theta)\sin(\theta)+(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)-(3\theta+e^\theta)\sin(\theta)}$

$\frac{(3+e^\theta)\sin(\theta)+(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)+(3\theta+e^\theta)\sin(\theta)}$

$\frac{(3+e^\theta)\sin(\theta)-(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)+(3\theta+e^\theta)\sin(\theta)}$

$3+e^\theta$

Correct answer:

$\frac{(3+e^\theta)\sin(\theta)+(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)-(3\theta+e^\theta)\sin(\theta)}$

Explanation:

The derivative of a polar function is given by the following:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin(\theta)+r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta)-r\sin(\theta)}$

First, we must find $\frac{dr}{d\theta}=3+e^\theta$

We found the derivative using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x}e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Finally, we plug in the above derivative and the original function into the above formula:

$\frac{(3+e^\theta)\sin(\theta)+(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)-(3\theta+e^\theta)\sin(\theta)}$

Report an Error

Example Question #31 : Functions, Graphs, And Limits

What is the derivative of $r=4sin\theta+2$ ?

Possible Answers:

$\frac{8cos\theta\sin\theta+2\cos\theta}{4-8\sin^{2}\theta+2sin\theta}$

$\frac{8cos\theta\sin\theta-2\cos\theta}{4-8\sin^{2}\theta-2sin\theta}$

$\frac{8cos\theta\sin\theta+2\cos\theta}{4-8\sin^{2}\theta-2sin\theta}$

$\frac{8cos\theta\sin\theta+2\cos\theta}{4+8\sin^{2}\theta-2sin\theta}$

$\frac{8cos\theta\sin\theta-2\cos\theta}{4+8\sin^{2}\theta+2sin\theta}$

Correct answer:

$\frac{8cos\theta\sin\theta+2\cos\theta}{4-8\sin^{2}\theta-2sin\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=4sin\theta+2$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=4cos\theta$

We can then swap the given values of $\frac{dr}{d\theta}$ and into the equation of the derivative of an expression into polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta}{\frac{dr}{d\theta}cos\theta-rsin\theta}$

$\frac{dy}{dx}=\frac{4cos\theta\sin\theta+(4sin\theta+2)cos\theta}{4cos\theta\cos\theta-(4sin\theta+2)sin\theta}$

$\frac{dy}{dx}=\frac{4cos\theta\sin\theta+4sin\theta\cos\theta+2\cos\theta}{4cos\theta\cos\theta-4sin\theta\sin\theta-2sin\theta}$

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4(cos^{2}\theta-sin^{2}\theta)-2sin\theta}$

Using the trigonometric identity $\sin^{2}\theta+\cos^{2}\theta=1$ , we can deduce that $\cos^{2}\theta=1-\sin^{2}\theta$ . Swapping this into the denominator, we get:

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4((1-\sin^{2}\theta)-sin^{2}\theta)-2sin\theta}$

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4(1-2\sin^{2}\theta)-2sin\theta}$

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4-8\sin^{2}\theta-2sin\theta}$

Report an Error

Example Question #32 : Functions, Graphs, And Limits

What is the derivative of $r=2+5\sin\theta$ ?

Possible Answers:

$\frac{10\cos\theta\sin\theta-2\cos\theta}{5-10\sin^{2}\theta-2\sin\theta}$

$\frac{10\cos\theta\sin\theta+2\cos\theta}{5-10\sin^{2}\theta-2\sin\theta}$

$\frac{10\cos\theta\sin\theta-2\cos\theta}{5+10\sin^{2}\theta+2\sin\theta}$

$\frac{10\cos\theta\sin\theta+2\cos\theta}{5+10\sin^{2}\theta-2\sin\theta}$

$\frac{10\cos\theta\sin\theta+2\cos\theta}{5-10\sin^{2}\theta+2\sin\theta}$

Correct answer:

$\frac{10\cos\theta\sin\theta+2\cos\theta}{5-10\sin^{2}\theta-2\sin\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=2+5\sin\theta$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=5\cos\theta$

We can then swap the given values of $\frac{dr}{d\theta}$ and into the equation of the derivative of an expression into polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}$

$\frac{dy}{dx}=\frac{(5\cos\theta)\sin\theta+(2+5\sin\theta)\cos\theta}{(5\cos\theta)\cos\theta-(2+5\sin\theta)\sin\theta}$

$\frac{dy}{dx}=\frac{5\cos\theta\sin\theta+2\cos\theta+5\sin\theta\cos\theta}{5\cos^{2}\theta-2\sin\theta-5\sin^{2}\theta}$

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5(\cos^{2}\theta-\sin^{2}\theta)-2\sin\theta}$

Using the trigonometric identity $\sin^{2}\theta+\cos^{2}\theta=1$ , we can deduce that $\cos^{2}\theta=1-\sin^{2}\theta$ . Swapping this into the denominator, we get:

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5(1-\sin^{2}\theta-\sin^{2}\theta)-2\sin\theta}$

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5(1-2\sin^{2}\theta)-2\sin\theta}$

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5-10\sin^{2}\theta-2\sin\theta}$

Report an Error

Example Question #1 : Vectors

Find the vector form of (6,3,1) to (0,1,3) .

Possible Answers:

$\overrightarrow{v}=[6,-2,-2]$

$\overrightarrow{v}=[6,2,2]$

$\overrightarrow{v}=[6,2,-2]$

$\overrightarrow{v}=[6,2,-3]$

$\overrightarrow{v}=[-6,-2,-2]$

Correct answer:

$\overrightarrow{v}=[6,2,-2]$

Explanation:

When we are trying to find the vector form we need to remember the formula which states to take the difference between the ending and starting point.

Thus we would get:

Given (a,b,c) and (d,e,f)

$\overrightarrow{v}=[d-a, e-b, f-c]$

In our case we have ending point at (6,3,1) and our starting point at (0,1,3) .

Therefore we would set up the following and simplify.

$\overrightarrow{v}=[6-0,3-1,1-3]=[6,2,-2]$

Report an Error

Example Question #33 : Parametric, Polar, And Vector Functions

Given points (2,4,5) and (9,1,0) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle 7,3,-5\right \rangle$

$\left \langle -7,-3,-5\right \rangle$

$\left \langle 7,3,5\right \rangle$

$\left \langle -7,3,-5\right \rangle$

$\left \langle 7,-3,-5\right \rangle$

Correct answer:

$\left \langle 7,-3,-5\right \rangle$

Explanation:

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points.

That is, for any point

$a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ ,

the distance is the vector

$v=\left \langle b_1-a_1,b_2-a_2,b_3-a_3\right \rangle$ .

Subbing in our original points (2,4,5) and (9,1,0) , we get:

$v=\left \langle 9-2,1-4,0-5\right \rangle$

$v=\left \langle 7,-3,-5\right \rangle$

Report an Error

Example Question #1 : Vector Form

Given points (4,2,-2) and (7,1,0) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle 3,1,-2\right \rangle$

$\left \langle -3,1,2\right \rangle$

$\left \langle 3,-1,2\right \rangle$

$\left \langle 3,1,2\right \rangle$

$\left \langle -3,1,-2\right \rangle$

Correct answer:

$\left \langle 3,-1,2\right \rangle$

Explanation:

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points.

That is, for any point $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , the distance is the vector $v=\left \langle b_1-a_1,b_2-a_2,b_3-a_3\right \rangle$ .

Subbing in our original points (4,2,-2) and (7,1,0) , we get:

$v=\left \langle 7-4,1-2,0-(-2)\right \rangle$

$v=\left \langle 3,-1,2\right \rangle$

Report an Error

Example Question #11 : Graphing Vectors

The graph of the vector function $r(t)=\left \langle \frac{1}{2}t,\sin(t)\right \rangle$ can also be represented by the graph of which of the following functions in rectangular form?

Possible Answers:

$y=\sin(\frac{x}2{})$

$y=\sin(2x)$

$y=-\sin(\frac{x}2{})$

$y=-\sin(2x)$

$y=2\sin(x)$

Correct answer:

$y=\sin(2x)$

Explanation:

We can find the graph of $r(t)=\left \langle \frac{1}{2}t,\cos(t)\right \rangle$ in rectangular form by mapping the parametric coordinates to Cartesian coordinates (x,y) :

$x=\frac{1}{2}t$

t=2x

We can now use this value to solve for :

$y=\sin(t)$

$y=\sin(2x)$

Report an Error

View AP Calculus BC Tutors

Lyndell
Certified Tutor

University of Virginia-Main Campus, Doctor of Philosophy, Mathematics.

View AP Calculus BC Tutors

Robert
Certified Tutor

Rensselaer Polytechnic Institute, Bachelors, Electrical Engineering. Brooklyn Polytechnic, Masters, Computer Science.

View AP Calculus BC Tutors

Abhishake
Certified Tutor

Delhi Technological University, Bachelor of Technology, Polymer Chemistry. National Tsing Hua University, Master of Engineeri...

All AP Calculus BC Resources

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

Popular Subjects

Statistics Tutors in Dallas Fort Worth, Chemistry Tutors in Washington DC, Calculus Tutors in Atlanta, French Tutors in Denver, SAT Tutors in Houston, ISEE Tutors in Dallas Fort Worth, ISEE Tutors in Atlanta, Calculus Tutors in Philadelphia, MCAT Tutors in Los Angeles, SAT Tutors in Atlanta

Popular Courses & Classes

GMAT Courses & Classes in Atlanta, ACT Courses & Classes in San Diego, ISEE Courses & Classes in Los Angeles, GRE Courses & Classes in Washington DC, MCAT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Boston, GRE Courses & Classes in Boston, GRE Courses & Classes in Miami, SSAT Courses & Classes in New York City, SSAT Courses & Classes in Phoenix

Popular Test Prep

SAT Test Prep in Seattle, ACT Test Prep in San Diego, GMAT Test Prep in Philadelphia, GRE Test Prep in Miami, GRE Test Prep in Seattle, GMAT Test Prep in Washington DC, ISEE Test Prep in Houston, GRE Test Prep in Washington DC, LSAT Test Prep in Denver, SSAT Test Prep in Los Angeles

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 : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #11 : Polar Form

Example Question #261 : Parametric, Polar, And Vector

Example Question #12 : Polar Form

Example Question #2 : Derivatives Of Polar Form

Example Question #31 : Functions, Graphs, And Limits

Example Question #32 : Functions, Graphs, And Limits

Example Question #1 : Vectors

Example Question #33 : Parametric, Polar, And Vector Functions

Example Question #1 : Vector Form

Example Question #11 : Graphing Vectors

All AP Calculus BC Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: