Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Spatial 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 #901 : Spatial Calculus

Which of the following vectors is perpendicular to (-3,-8) ?

Possible Answers:

(3,8)

(3,-8)

(-3,8)

(8,-4)

(-8,3)

Correct answer:

(-8,3)

Explanation:

By definition, a given vector (a,b) has a perpendicular vector (b,-a) . Therefore, the vector perpendicular to (-3,-8) is (-8,3) .

To verify that two vectors are perpendicular their dot product must equal zero.

The dot product is as follows.

$(a_1,b_1),(a_2,b_2)\rightarrow (a_1\times a_2)+(b_1\times b_2)=0$

Substituting in our vector values we get:

$(-3,-8),(-8,3)\rightarrow (-3\times -8)+(-8\times 3)=24+-24=0$

Report an Error

Example Question #902 : Spatial Calculus

Which of the following is perpendicular to the vector (7,-5) ?

Possible Answers:

(-5,-7)

(5,-7)

(7,5)

(-7,5)

(-7,-5)

Correct answer:

(-5,-7)

Explanation:

By definition, a given vector (a,b) has a perpendicular vector (b,-a) .

Given a vector (7,-5) , its perpendicular vector will be (-5,-7) .

We can further verify this result by noting that the product of two perpendicular vectors is .

Since $(7,-5)\times(-5,-7) = (7\times -5) + (-5\times -7)=-35+35=0$ , then (-5,-7) is perpendicular to (7,-5) .

Report an Error

Example Question #903 : Spatial Calculus

Which of the following is perpendicular to the vector (8, 4) ?

Possible Answers:

(-8, -4)

(4,-8)

(-8, 4)

(8, -4)

(-3, 8)

Correct answer:

(4,-8)

Explanation:

By definition, a given vector (a,b) has a perpendicular vector (b,-a) .

Given a vector (8,4) , its perpendicular vector will be (4,-8) .

We can further verify this result by noting that the product of two perpendicular vectors is .

Since $(8,4)\times (4,-8) = (8\times 4) + (4\times -8)=32-32=0$ , then (4,-8) is perpendicular to (8,4) .

Report an Error

Example Question #904 : Spatial Calculus

Which of the following is perpendicular to the vector (2, 2) ?

Possible Answers:

$\left(2,-\frac{1}{2}\right)$

(2,-2)

(-1, 2)

(-2, -2)

$\left(\frac{1}{2},\frac{1}{2}\right)$

Correct answer:

(2,-2)

Explanation:

By definition, a given vector (a,b) has a perpendicular vector (b,-a) .

Given a vector (2,2) , its perpendicular vector will be (2,-2) .

We can further verify this result by noting that the product of two perpendicular vectors is .

Since $(2,2)\times (2,-2) = (2\times 2) + (2\times -2)=4-4=0$ , then (2,-2) is perpendicular to (2,2) .

Report an Error

Example Question #51 : How To Find Position

At time t=0 a particle is at the origin at rest with no velocity. It then experiences an acceleration of

$a\left ( t\right ) = \sin{3t}$ .

After $\pi$ seconds, what is the particle's position?

Possible Answers:

$\frac{-1}{9}-\frac{1}{3}\pi$

$\frac{1}{3}\pi$

$\frac{1}{9}-\frac{1}{3}\pi$

$\frac{1}{3}\pi-\frac{1}{9}$

$-\frac{1}{3}\pi$

Correct answer:

$\frac{1}{3}\pi$

Explanation:

First, we integrate $a\left ( t\right )$ with respect to to get velocity:

$v\left ( t\right ) = -\frac{\cos{3t}}{3} + C$

We know that the particle is not moving at t=0 , so

v(0)=0 .

Solving this gives us

$v(t)=\frac{1}{3}-\frac{cos{3t}}{3}$ .

Now, we have to solve for position. To do this, we integrate v(t) with respect to . Thus, we get

$p(t)=\frac{1}{3}t-\frac{\sin{3t}}{9}+C$ .

We know that the particle starts at the origin, so we have to solve:

p(0)=0

This implies that C=0 , so we have:

$p(t)=\frac{1}{3}t-\frac{\sin{3t}}{9}$ .

As a special case:

$p(\pi)=\frac{1}{3}\pi-\frac{\sin{3\cdot \pi}}{9}=\frac{1}{3}\pi$ .

Report an Error

Example Question #905 : Spatial Calculus

Given the initial velocity, initial position and acceleration of an object, find its position function.

v(0)=-12,p(0)=32,a(t)=5.8

Possible Answers:

p(t)=-6.2t+32

p(t)=5.8t^2+20t

p(t)=5.8t^2-12t+32

p(t)=2.9t^2-12t+32

Correct answer:

p(t)=2.9t^2-12t+32

Explanation:

We begin by integrating the acceleration function and using the initial condition to find the value of the constant of integration:

$\\ \int a(t)dt=\int5.8dt=5.8t+C=v(t)\\ v(0)=C=-12\Rightarrow v(t)=5.8t-12$

Now that we have the velocity function we repeat the process to find the position function:

$\\ \int v(t)dt=\int 5.8t-12dt=2.9t^2-12t+D=p(t)\\ p(0)=32=D\Rightarrow p(t)=2.9t^2-12t+32$

Report an Error

Example Question #906 : Spatial Calculus

Which of the following is perpendicular to the vector (9,6) ?

Possible Answers:

(6,-9)

(-9,-6)

(-6,8)

(9,-6)

(-9,6)

Correct answer:

(6,-9)

Explanation:

By definition, a given vector (a,b) has a perpendicular vector (b,-a) . Given a vector (9,6) , we therefore know its perpendicular vector is (6,-9) .

We can further verify this by noting that the product of a vector and its perpendicular vector is .

Since $(9,6)\times (6,-9)=(9\times 6)+(6\times -9)=54+ (-54)=0$ , the two vectors are perpendicular to each other.

Report an Error

Example Question #907 : Spatial Calculus

Which of the following is perpendicular to the vector (-5,-1) ?

Possible Answers:

(-5,1)

(-1,5)

(5,-1)

(1,-4)

(5,1)

Correct answer:

(-1,5)

Explanation:

By definition, a given vector (a,b) has a perpendicular vector (b,-a) . Given a vector (-5,-1) , we therefore know its perpendicular vector is (-1,5) .

We can further verify this by noting that the product of a vector and its perpendicular vector is .

Since $(-5,-1)\times (-1,5)=(-5\times -1)+(-1\times 5)=5+(-5)=0$ , the two vectors are perpendicular to each other.

Report an Error

Example Question #908 : Spatial Calculus

Which of the following is perpendicular to the vector (9,7) ?

Possible Answers:

(-7,8)

(-9,7)

(9,-7)

(-9,-7)

(7,-9)

Correct answer:

(7,-9)

Explanation:

By definition, a given vector (a,b) has a perpendicular vector (b,-a) . Given a vector (9,7) , we therefore know its perpendicular vector is (-7,9) .

We can further verify this by noting that the product of a vector and its perpendicular vector is .

Since $(9,7)\times (7,-9)=(9\times 7)+(7\times -9)=63+(-63)=0$ , the two vectors are perpendicular to each other.

Report an Error

Example Question #909 : Spatial Calculus

Which of the following is perpendicular to the vector (7,-6) ?

Possible Answers:

(-6,-7)

(-7,-6)

(-6,7)

(7,6)

(-7,6)

Correct answer:

(-6,-7)

Explanation:

By definition, a given vector (a,b) has a perpendicular vector (b,-a) . Given a vector (7,-6) , its perpendicular vector will be (-6,-7) . We can further verify this result by noting that the product of two perpendicular vectors is ; since $(7,-6)\times (-6,-7)=(7\times -6)+(-6\times -7)=-42+42=0$ , we know the two vectors are perpendicular to each other.

Report an Error

View Calculus Tutors

Ainsley
Certified Tutor

Rice University, Bachelor of Science, Physics.

View Calculus Tutors

Mike Bahaa
Certified Tutor

Concordia University-Ann Arbor, Master of Arts Teaching, Mathematics Teacher Education.

View Calculus Tutors

Jeremy
Certified Tutor

University of Illinois at Chicago, Bachelor of Science, Physics.

All Calculus 1 Resources

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

Popular Subjects

Chemistry Tutors in Washington DC, SAT Tutors in Washington DC, English Tutors in Chicago, Calculus Tutors in Seattle, Computer Science Tutors in San Francisco-Bay Area, LSAT Tutors in Boston, Math Tutors in Los Angeles, SAT Tutors in Denver, Statistics Tutors in Boston, ACT Tutors in Philadelphia

Popular Courses & Classes

SSAT Courses & Classes in Washington DC, MCAT Courses & Classes in Los Angeles, GRE Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Washington DC, SAT Courses & Classes in San Diego, LSAT Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Chicago, MCAT Courses & Classes in Phoenix, ACT Courses & Classes in Philadelphia, GMAT Courses & Classes in Philadelphia

Popular Test Prep

LSAT Test Prep in Atlanta, LSAT Test Prep in San Diego, MCAT Test Prep in Los Angeles, MCAT Test Prep in Seattle, MCAT Test Prep in Miami, ACT Test Prep in Boston, ISEE Test Prep in Dallas Fort Worth, SAT Test Prep in Atlanta, ACT Test Prep in Philadelphia, SSAT Test Prep in Atlanta

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 : Spatial Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #901 : Spatial Calculus

Example Question #902 : Spatial Calculus

Example Question #903 : Spatial Calculus

Example Question #904 : Spatial Calculus

Example Question #51 : How To Find Position

Example Question #905 : Spatial Calculus

Example Question #906 : Spatial Calculus

Example Question #907 : Spatial Calculus

Example Question #908 : Spatial Calculus

Example Question #909 : Spatial Calculus

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: