We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #201 : Calculus

The position of a particle in two dimensions is given by the functions x(t)=t^2+3t-4 , y(t)=3x^2(t) . What is the particle's velocity in the (x,y) directions at time t=3 ?

Possible Answers:

(9,81)

(3,243)

(9,756)

(6,588)

(9,486)

Correct answer:

(9,756)

Explanation:

Velocity is the time derivative of position:

$v(t)=\frac{d}{dt}p(t)$

In the case of two position functions for the description of two-dimensional location, there will simply be two corresponding velocity functions:

For the position functions

x(t)=t^2+3t-4

y(t)=3x^2(t)

The velocity functions are then:

v_x(t)=2t+3

$v_y(t)=6x(t)\frac{dx(t)}{dt}$

v_y(t)=6x(t)v_x(t)

v_y(t)=6(t^2+3t-4)(2t+3)

At time t=3

v_x(t)=6+3=9

v_y(t)=6(9+9-4)(6+3)=6(14)(9)=756

Report an Error

Example Question #201 : Calculus

The position of a particle in three-dimensional space is given by the functions

$x(t)=3cos(\frac{1}{2}t)$

y(t)=2sin(t)cos(t)

z(t)=t^2+t

What is the magnitude of the particle's velocity at time $t=\pi$ ?

Possible Answers:

$\sqrt{8\pi+\frac{29}{4}}$

$\sqrt[3]{2\pi+\frac{3}{2}}$

$\sqrt{4\pi^2+4\pi+\frac{29}{4}}$

$\sqrt{2\pi+\frac{3}{2}}$

$\sqrt[3]{4\pi^2+4\pi+\frac{29}{4}}$

Correct answer:

$\sqrt{4\pi^2+4\pi+\frac{29}{4}}$

Explanation:

Velocity can be found as the time derivative of position:

$v(t)=\frac{d}{dt}p(t)$

In the case of three-dimensional descriptions, that simply means there are three velocity functions. For the positional functions

$x(t)=3cos(\frac{1}{2}t)$

y(t)=2sin(t)cos(t)

z(t)=t^2+t

The resultant velocity functions are:

$v_x(t)=-\frac{3}{2}sin(\frac{1}{2}t)$

v_y(t)=2cos^2(t)-2sin^2(t)

v_z(t)=2t+1

At time $t=\pi$

$v_x(\pi)=-\frac{3}{2}sin(\frac{1}{2}\pi)=-\frac{3}{2}$

$v_y(\pi)=2cos^2(\pi)-2sin^2(\pi)=2$

$v_z(\pi)=2\pi+1$

The magnitude of the velocity is then the root of the sum of the squares:

$v(\pi)=\sqrt{(-\frac{3}{2})^2+(2)^2+(2\pi+1)^2}$

$v(\pi)=\sqrt{\frac{9}{4}+4+4\pi^2+4\pi+1}$

$v(\pi)=\sqrt{4\pi^2+4\pi+\frac{29}{4}}$

Report an Error

Example Question #201 : Spatial Calculus

The acceleration of a particle is given by the function $a(t)=9t^2+15e^{-3t}$ . If the particle has an initial velocity of , what is it's velocity after five seconds?

Possible Answers:

$390-5e^{-15}$

$390-15e^{-15}$

$375-5e^{-15}$

$360-5e^{-15}$

$360+5e^{-15}$

Correct answer:

$390-5e^{-15}$

Explanation:

Velocity can be found by integrating acceleration with respect to time:

$v(t)=\int a(t)dt$

To find the integral we will need to apply a few different rules.

$\int x^n=\frac{x^{n+1}}{n+1}$

$\int e^u=e^u\cdot \frac{1}{\frac{du}{dx}}$

For the acceleration function

$a(t)=9t^2+15e^{-3t}$ ,

the velocity function is thus:

$v(t)=3t^3-5e^{-3t}+C$

To find the constant of integration, use the initial velocity condition:

$v(0)=3(0^3)-5e^{-3(0)}+C=10$

C=15

From here, the full velocity function can be found:

$v(t)=3t^3-5e^{-3t}+15$

$v(5)=3(5^3)-5e^{-3(5)}+15$

$v(5)=390-5e^{-15}$

Report an Error

Example Question #201 : Spatial Calculus

The position of a particle in three dimensions is given by the functions:

x(t)=2t+4

y(t)=x^3(t)

z(t)=2y^2(t)

What is the velocity at time t=2 ?

Possible Answers:

(4,5260,984213)

(2,384,786432)

(2,716,432)

(2,298,842352)

(32,384,1384681)

Correct answer:

(2,384,786432)

Explanation:

Velocity can be found as the time derivative of position:

$v(t)=\frac{d}{dt}p(t)$

For the three position functions

x(t)=2t+4

y(t)=x^3(t)

z(t)=2y^2(t)

There can be found three velocity funcitons:

v_x(t)=2

$v_y(t)=3x^2(t)\frac{dx(t)}{dt}$

$v_z(t)=4y(t)\frac{dy(t)}{dt}$

Now note that these derivative terms correspond to velocities, i.e. $v_x(t)=\frac{dx(t)}{dt}$ and $v_y(t)=\frac{dy(t)}{dt}$ .

Knowing this, we can replace terms in each function so that each velocity is in terms of time:

v_x(t)=2

v_y(t)=3(2t+4)^2(2)

v_z(t)=4(2t+4)^3[3(2t+4)^2(2)]

Therefore at time t=2

v_x(2)=2

v_y(2)=3(4+4)^2(2)=384

v_z(2)=4(4+4)^3(384)=786432

Report an Error

Example Question #205 : Velocity

The position of a particle in two dimensions is given by the functions:

$x(t)=4cos(\frac{1}{2}t)+sin(3t)$

$y(t)=2sin(\frac{1}{2}t)cos(\frac{3}{2}t)$

What is the magnitude of the particle's velocity at time $t=\pi$ ?

Possible Answers:

$\sqrt2$

$\sqrt{34}$

$2\sqrt{2}$

Correct answer:

$\sqrt{34}$

Explanation:

Velocity can be found as the time derivative of position:

$v(t)=\frac{d}{dt}p(t)$

For the two position functions

$x(t)=4cos(\frac{1}{2}t)+sin(3t)$

$y(t)=2sin(\frac{1}{2}t)cos(\frac{3}{2}t)$

The velocity functions are found to be:

$v_x(t)=-2sin(\frac{1}{2}t)+3cos(3t)$

$v_y(t)=cos(\frac{1}{2}t)cos(\frac{3}{2}t)-3sin(\frac{1}{2}t)sin(\frac{3}{2}t)$

The velocities at time $t=\pi$ are found to be:

$v_x(\pi)=-2sin(\frac{1}{2}\pi)+3cos(3\pi)=-2-3=-5$

$v_y(\pi)=cos(\frac{1}{2}\pi)cos(\frac{3}{2}\pi)-3sin(\frac{1}{2}\pi)sin(\frac{3}{2}\pi)$

$v_y(\pi)=0(0)-3(1)(-1)=3$

The magnitude of the velocity can be found by taking the root of the sum of squares:

$v(\pi)=\sqrt{(-5)^2+(3)^2}$

$v(\pi)=\sqrt{34}$

Report an Error

Example Question #202 : Spatial Calculus

The acceleration of a football is given by $a(t)=3t^{2}+2t$ . What is the velocity function v(t) of the football if the initial velocity is $3 \frac{meters}{sec}$ ?

Possible Answers:

$v(t)=t^{3}+4t^{2}-3$

$v(t)=t^{3}+t^{2}+3$

$v(t)=-t^{3}+t^{2}+3$

$v(t)=t^{3}-t^{2}+3$

$v(t)=3t^{3}+2t$

Correct answer:

$v(t)=t^{3}+t^{2}+3$

Explanation:

To find the velocity function v(t) , we need to integrate the acceleration function a(t) .

To integrate this function, we need the following formulae:

$\int x^{n}dx=\frac{1}{n+1}x^{n+1}, \int cf(x)dx=c\int f(x)dx$

$v(t)=\int a(t) = \int 3t^{2}+2t = \frac{3t^{3}}{3}+\frac{2t^2}{2}+C=t^{3}+t^{2}+C$

Now, to solve for the constant, we use the initial conditions:

Since v(0)=3 , we can plug in for in the velocity function and set it equal to :

$v(0)=3(0)^{2}+2(0)+C=0+0+C=3$

Therefore, C=3

So, the specific solution for the velocity is $v(t)=t^{3}+t^{2}+3$ .

Report an Error

Example Question #203 : Spatial Calculus

A particle's velocity is given by the function v(t)=3t^2+4t+6 . What is the average velocity of the particle over the interval of time t=[2,10] ?

Possible Answers:

154

123.2

157.5

126

Correct answer:

154

Explanation:

Average velocity can be found as the total distance travelled over the total time expended.

Distance travelled can be found by integrating the velocity function over the specified interval of time t=[a,b] .

Therefore

$v_{avg}=\frac{1}{b-a}\int_a^b v(t)dt$

For the velocity function

v(t)=3t^2+4t+6

$v_{avg}=\frac{1}{10-2}\int_2^{10} (3t^2+4t+6)dt$

$v_{avg}=\frac{1}{8}(t^3+2t^2+6t)|_2^{10}$

$v_{avg}=\frac{1}{8}[(1000+200+60)-(8+8+12)]$

$v_{avg}=\frac{1}{8}[(1260)-(28)]$

$v_{avg}=\frac{1}{8}(1232)$

$v_{avg}=154$

Report an Error

Example Question #204 : Spatial Calculus

The velocity of a particle is given by the function v(t)=t^2+t+1 . What is the particle's average velocity over the period of time t=[0,3] ?

Possible Answers:

5.5

16.5

Correct answer:

5.5

Explanation:

Average velocity can be found by dividing total distance traveled over total time elapsed.

When given a velocity function, distance traveled over an interval of time t=[a,b] can be found by integrating the velocity over this interval. By dividing by the length of this interval, an average velocity can be found:

$v_{avg}=\frac{1}{b-a}\int_a^b v(t)dt$

For the velocity function

v(t)=t^2+t+1

The average velocity over the interval t=[0,3] can be found as

$v_{avg}=\frac{1}{3-0}\int_0^3(t^2+t+1)dt$

$v_{avg}=\frac{1}{3}(\frac{t^3}{3}+\frac{t^2}{2}+t)|_0^3$

$v_{avg}=\frac{1}{3}[(9+\frac{9}{2}+3)-(0+0+0)]$

$v_{avg}=\frac{11}{2}=5.5$

Report an Error

Example Question #204 : Calculus

The position of a particle is given by the function x(t)=t^2-10t+24 . At what time does the particle change directions in its movement?

Possible Answers:

Correct answer:

Explanation:

The particle changes directions when its velocity undergoes a sign change. Velocity can be found as the time derivative of position, so for the position function

x(t)=t^2-10t+24

The velocity can be found as

v(t)=2t-10

The particle initially travels in the negative direction, since v(0)=-10 . The velocity will change in sign after the point that the function reaches zero, if such a point exists for a positive time value:

2t-10=0

2t=10

t=5

This indicates the turning point. All later times, the particle will move in a positive direction.

Report an Error

Example Question #205 : Calculus

A particle's position is given by the function x(t)=-t^3+9t^2+5t+20 . What is the maximum velocity achieved by the particle?

Possible Answers:

Correct answer:

Explanation:

Velocity can be found by taking the derivative of position with respect to time. For the position function

x(t)=-t^3+9t^2+5t+20

The velocity function is

v(t)=-3t^2+18t+5

Now to find where the velocity has minima or maxima, take the derivative again with respect to time to find acceleration, and find when the accerlation is zero:

a(t)=-6t+18

-6t+18=0

t=3

Prior to this time, the acceleration is postive, and after this time the acceleration is negative, so this indicates a maximum.

The maximum velocity is then

$v_{max}=v(3)=-27+54+5$

v_max=32

Report an Error

View Calculus Tutors

Jacob
Certified Tutor

Ohio State University-Main Campus, Bachelor of Science, Political Science and Government.

View Calculus Tutors

Ben
Certified Tutor

Brown University, Bachelor in Arts, International Relations. Bridge Education Group, Certificate, Teaching French as a Second...

View Calculus Tutors

Dr. Derrick
Certified Tutor

Huston-Tillotson University, Bachelor of Science, Mathematics. Grand Canyon University, Master of Arts, Education.

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Algebra Tutors in Philadelphia, MCAT Tutors in Boston, Physics Tutors in Atlanta, Chemistry Tutors in Phoenix, SSAT Tutors in Philadelphia, French Tutors in Dallas Fort Worth, Physics Tutors in Los Angeles, SSAT Tutors in Chicago, GRE Tutors in Denver, MCAT Tutors in Dallas Fort Worth

Popular Courses & Classes

ISEE Courses & Classes in Boston, SSAT Courses & Classes in San Diego, ACT Courses & Classes in San Francisco-Bay Area, SAT Courses & Classes in Phoenix, ACT Courses & Classes in San Diego, Spanish Courses & Classes in San Diego, LSAT Courses & Classes in San Francisco-Bay Area, GRE Courses & Classes in Houston, SSAT Courses & Classes in Chicago, SAT Courses & Classes in San Diego

Popular Test Prep

GMAT Test Prep in San Diego, ISEE Test Prep in Los Angeles, ACT Test Prep in San Diego, MCAT Test Prep in Houston, ACT Test Prep in San Francisco-Bay Area, GRE Test Prep in Dallas Fort Worth, GMAT Test Prep in Denver, GMAT Test Prep in Boston, ACT Test Prep in Dallas Fort Worth, LSAT Test Prep in San Francisco-Bay Area

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

Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #201 : Calculus

Example Question #201 : Calculus

Example Question #201 : Spatial Calculus

Example Question #201 : Spatial Calculus

Example Question #205 : Velocity

Example Question #202 : Spatial Calculus

Example Question #203 : Spatial Calculus

Example Question #204 : Spatial Calculus

Example Question #204 : Calculus

Example Question #205 : Calculus

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: