Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Derivatives of Vectors

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 4 Next →

Example Question #1 : Derivatives Of Vectors

Given the following vector:

$\vec{u}=(x^n,arctan(x),3)$

Find the second derivative of $\vec{u}$ .

Possible Answers:

$\left(n(n-1)x^{n-1},\frac{-2x}{(1+x^2)^2},0\right)$

$\left(n(n-1)x^{n-2},\frac{-2x}{(1+x^2)^2},0\right)$

$\left(n(n-1)x^{n-2},\frac{2x}{(1+x^2)^2},0\right)$

$\left(n(n-1)x^{n-2},\frac{-2x}{(1+x^2)^4},0\right)$

$\left(n(n-1)x^{n-2},\frac{-2x}{(1+x^2)^2},3\right)$

Correct answer:

$\left(n(n-1)x^{n-2},\frac{-2x}{(1+x^2)^2},0\right)$

Explanation:

In order to obtain the second derivative, we will have to differentiate each component twice. We know how to differentiate the following:

$(x^n)''=n(n-1)x^{n-2}$ .

We also have :

$(arctan(x))'=\frac{1}{1+x^2}$

and this gives:

$\{arctan(x)\}''=\frac{-2x}{(1+x^2)^2}$

For the constant component , we know that it is derivative is zero.

This gives us the solution that we are looking for:

$\vec{u}''=\left(n(n-1)x^{n-2},\frac{-2x}{(1+x^2)^2},0\right)$

Report an Error

Example Question #5 : Derivatives Of Parametric, Polar, And Vector Functions

Given that $f: \mathbb{R}^{n}}\rightarrow \mathbb{R}$ . We define its gradient as :

$\nabla f =\left[\frac{\partial f}{\partial x_{1}} \cdots \frac{\partial f}{\partial x_{n}}\right]$

Let $f: \mathbb{R}^{n}}\rightarrow \mathbb{R}$ be given by:

$f(x_{1},x_{2},\cdots, x_{n}})=\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

What is the gradient of ?

Possible Answers:

$\left[\frac{x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\cdots \frac{x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}} \right]$

$\left[\frac{x_{1}}{2\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\cdots \frac{x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}} \right]$

$\left[\frac{2x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\cdots \frac{x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}} \right]$

$\left[\frac{x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\cdots \frac{2x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}} \right]$

$\left[\frac{x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\cdots \frac{x_{n}}{\sqrt{{x_{1}^2}+2x_{2}^2+\cdots+x_{n}^2}} \right]$

Correct answer:

$\left[\frac{x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\cdots \frac{x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}} \right]$

Explanation:

By definition, to find the gradient vector , we will have to find the gradient components. We know that the gradient components are that partial derivatives.

We know that in our case we have :

$\frac{\partial f}{\partial x_{i}}=\frac{x_{i}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

To see this, fix all other variables and assume that you have only $x_{i}$ as the only variable.

Now we apply the given defintion , i.e,

$\nabla f =\left[\frac{\partial f}{\partial x_{1}} \cdots \frac{\partial f}{\partial x_{n}}\right]$

with :

$\frac{\partial f}{\partial x_{1}}=\frac{x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

$\frac{\partial f}{\partial x_{2}}=\frac{x_{2}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

$\frac{\partial f}{\partial x_{3}}=\frac{x_{3}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

$\vdots$

$\frac{\partial f}{\partial x_{n}}=\frac{x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

this gives us the solution .

Report an Error

Example Question #1 : Derivatives Of Parametric, Polar, And Vector Functions

Let $f:\mathbb{R}^{n}\mapsto \mathbb{R}$ .

We define the gradient of as:

$\nabla f =\left[\frac{\partial f}{\partial x_{1}} \cdots \frac{\partial f}{\partial x_{n}}\right]$

Let $f=sin(x_{1}+2x_{2}+\cdots nx_{n})$ .

Find the vector gradient.

Possible Answers:

$[cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad -ncos(x_{1}+2x_{2}+\cdots nx_{n})\]$

$[cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad cos(x_{1}+2x_{2}+\cdots nx_{n})\]$

$[x_{1}cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad ncos(x_{1}+2x_{2}+\cdots nx_{n})\]$

$[cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+x_{2}+\cdots nx_{n})\quad \cdots\quad ncos(x_{1}+2x_{2}+\cdots nx_{n})\]$

$[cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad ncos(x_{1}+2x_{2}+\cdots nx_{n})\]$

Correct answer:

$[cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad ncos(x_{1}+2x_{2}+\cdots nx_{n})\]$

Explanation:

We note first that :

$\frac{\partial f}{\partial x_{1}}=cos(x_{1}+2x_{2}+\cdots nx_{n})$

Using the Chain Rule where $x_{1}$ is the only variable here.

$\frac{\partial f}{\partial x_{2}}=2 cos(x_{1}+2x_{2}+\cdots nx_{n})$

Using the Chain Rule where $x_{2}$ is the only variable here.

Continuing in this fashion we have:

$\frac{\partial f}{\partial x_{n}}=ncos(x_{1}+2x_{2}+\cdots nx_{n})$

Again using the Chain Rule and assuming that $x_{n}$ is the variable and all the others are constant.

Now applying the given definition of the gradient we have the required result.

$[cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad ncos(x_{1}+2x_{2}+\cdots nx_{n})\]$

Report an Error

Example Question #1 : Derivatives Of Parametric, Polar, And Vector Functions

Let

$\vec{u}=(\sqrt{1+t^2},2\sqrt{1+t^2},3\sqrt{1+t^2},\cdots n\sqrt{1+t^2})$

What is the derivative of $\vec{u}$ ?

Possible Answers:

$\left(\frac{t}{\sqrt{t^2+1}},\frac{t}{\sqrt{t^2+1}},\cdots \frac{nt}{\sqrt{t^2+1}}\right)$

$\left(\frac{t}{\sqrt{t^2+1}},\frac{2t}{\sqrt{t^2+1}},\cdots \frac{nt}{\sqrt{t+1}}\right)$

$\left(\frac{t}{\sqrt{t^2+1}},\frac{2t}{\sqrt{t^2+1}},\cdots \frac{nt}{\sqrt{t^2+1}}\right)$

$\left(\frac{t}{\sqrt{t^2+1}},\frac{t}{\sqrt{t^2+1}},\cdots \frac{t}{\sqrt{t^2+1}}\right)$

$\left(\frac{t}{\sqrt{t^2+1}},\frac{2t}{\sqrt{t^2+1}},\cdots \frac{2t}{\sqrt{t^2+1}}\right)$

Correct answer:

$\left(\frac{t}{\sqrt{t^2+1}},\frac{2t}{\sqrt{t^2+1}},\cdots \frac{nt}{\sqrt{t^2+1}}\right)$

Explanation:

To find the derivative of this vector, all we need to do is to differentiate each component with respect to t.

Use the Power Rule and the Chain Rule when differentiating.

$(\sqrt{1+t^2})dt=(1+t^2)^\frac{1}{2}dt=\frac{1}{2}(1+t^2)^{-\frac{1}{2}}\cdot 2t=\frac{t}{\sqrt{t^2+1}}$ is the derivative of the first component.

$(2\sqrt{1+t^2})dt=2(1+t^2)^\frac{1}{2}dt=2\frac{1}{2}(1+t^2)^{-\frac{1}{2}}\cdot 2t=\frac{2t}{\sqrt{t^2+1}}$ of the second component.

$\vdots$

$(n\sqrt{1+t^2})dt=n(1+t^2)^\frac{1}{2}dt=n\frac{1}{2}(1+t^2)^{-\frac{1}{2}}\cdot 2t=\frac{nt}{\sqrt{t^2+1}}$ is the derivative of the last component . we obtain then:

$\left(\frac{t}{\sqrt{t^2+1}},\frac{2t}{\sqrt{t^2+1}},\cdots \frac{nt}{\sqrt{t^2+1}}\right)$

Report an Error

Example Question #1 : Derivatives Of Vectors

Let be given by:

$\vec{u}=(artan(t),artan(2t),\cdots,artan(n-1)(t))$

Find the derivative of $\vec{u}$ .

Possible Answers:

$\left(\frac{1}{1+t^2}, \frac{2}{1+4t^2},\frac{3}{1+9t^2}, \cdots\quad ,\frac{n-1}{1+(n-1)^2t^2}\right)$

$\left(\frac{1}{1+t^2}, \frac{2}{1+4t^2},\frac{3}{1+9t^2}, \cdots\quad ,\frac{n-1}{1+(n-1)^2t}\right)$

$\left(\frac{1}{1+t^2}, \frac{2}{1+4t^2},\frac{3}{1+3t^2}, \cdots\quad ,\frac{n-1}{1+(n-1)^2t^2}\right)$

$\left(\frac{1}{1+t^2}, \frac{2}{1+2t^2},\frac{3}{1+9t^2}, \cdots\quad ,\frac{n-1}{1+(n-1)^2t^2}\right)$

$\left(\frac{1}{1+t^2}, \frac{2}{1+4t^2},\frac{3}{1+9t^2}, \cdots\quad ,\frac{n-1}{1+(n-1)t^2}\right)$

Correct answer:

$\left(\frac{1}{1+t^2}, \frac{2}{1+4t^2},\frac{3}{1+9t^2}, \cdots\quad ,\frac{n-1}{1+(n-1)^2t^2}\right)$

Explanation:

To find the derivative of u, we will have to find the derivative of every component.

Using the Chain Rule, we note that :

$aratan(t)'=\frac{1}{1+t^2}$

$artan(2t)'=\frac{2}{1+(2t)^2}=\frac{2}{1+4t^2}$

$artan(3t)'=\frac{3}{1+(3t)^2}=\frac{3}{1+9t^2}$

Continuing in the same fashion we have :

$artan((n-1)t)'=\frac{(n-1)}{1+((n-1)t)^2}=\frac{n-1}{1+(n-1)^2t^2}$

Ordering these components pairwise , we have then:

$\vec{u}'=\left(\frac{1}{1+t^2}, \frac{2}{1+4t^2},\frac{3}{1+3t^2}, \cdots\quad ,\frac{n-1}{1+(n-1)^2t^2}\right)$

this is what we needed to show.

Report an Error

Example Question #1 : Derivatives Of Vectors

Let be given by:

$u(x_{1},x_{2},\cdots,x_{n})=e^{b_{1}^2x_{1}+{b_{2}^2x_{2}}+\cdots+{b_{n}^2x_{n}}$

with $b_{1}^2+b_{2}^2+\cdots b_{n}^2=1$ .

What is the value of $\frac{\partial u}{\partial x_{1}}+\frac{\partial u}{\partial x_{2}}+\cdots +\frac{\partial u}{\partial x_{n}}$ ?

Possible Answers:

$1-b_{2}^2+\cdots b_{n}^2$

$b_{1}^2+b_{2}^2+\cdots b_{n}^3$

$b_{1}^2+b_{2}^2+\cdots b_{n}^2$

Correct answer:

Explanation:

To obtain the desired result, we will have to differentiate the expression of u with respect to each variable and add the derivatives.

We have

$\frac{\partial u}{\partial x_{1}}=b_{1}^2e^{b_{1}^2x_{1}+b_{2}^2x_{2}+\cdots +b_{n}^2x_{n}}$

$\frac{\partial u}{\partial x_{2}}=b_{2}^2e^{b_{1}^2x_{1}+b_{2}^2x_{2}+\cdots +b_{n}^2x_{n}}$

$\frac{\partial u}{\partial x_{3}}=b_{3}^2e^{b_{1}^2x_{1}+b_{2}^2x_{2}+\cdots +b_{n}^2x_{n}}$

$\vdots$

$\frac{\partial u}{\partial x_{n}}=b_{n}^2e^{b_{1}^2x_{1}+b_{2}^2x_{2}+\cdots +b_{n}^2x_{n}}$

we can also write the above expression as :

$\frac{\partial u}{\partial x_{1}}=b_{1}^2u$

$\frac{\partial u}{\partial x_{2}}=b_{2}^2u$

$\vdots$

$\frac{\partial u}{\partial x_{n}}=b_{n}^2u$

Now summing up we have:

$\frac{\partial u}{\partial x_{1}}+\frac{\partial u}{\partial x_{2}}+\cdots +\frac{\partial u}{\partial x_{n}}=(b_{1}^2+b_{2}^2+\cdots b_{n}^2)u}}$

But we are given that :

$(b_{1}^2+b_{2}^2+\cdots b_{n}^2)=1$

This gives:

$\frac{\partial u}{\partial x_{1}}+\frac{\partial u}{\partial x_{2}}+\cdots +\frac{\partial u}{\partial x_{n}}=u$

Report an Error

Example Question #1 : Derivatives Of Vectors

Let $\vec{u}=(t,tsin(t))$ and $\vec{v}=(tcos(t),t)$

Let $\vec{w}=\vec{u}+\vec{v}$ . What is the derivative of $\vec{w}$ ?

Possible Answers:

((1-tsin(t)+cos(t),1+tcos(t)+sin(t))

(cos(t)-(t+1)sin(t),cos(t)+cos(t)(t+1))

(cos(t)-(t+1)sin(t),sin(t)+cos(t)(t-1))

(cos(t)+(t+1)sin(t),sin(t)+cos(t)(t+1))

(cos(t)-(t-1)sin(t),sin(t)+cos(t)(t+1))

Correct answer:

((1-tsin(t)+cos(t),1+tcos(t)+sin(t))

Explanation:

One way to obtain the solution is that we first add the two functions to obtain $\vec{w}$ and we differentiate with each component to get the result.

Adding the two functions we obtain:

$\vec{w}=(t+tcos(t), tsin(t)+t)$

Now we will differentialte each component to get the derivative of $\vec{w}$ .

We have :

$\{t+tcos(t)\}'=1-tsin(t)+cos(t)$

$\{t+tsin(t)\}'=1+tcos(t)+sin(t)$

$\vec{w}'=((1-tsin(t)+cos(t),1+tcos(t)+sin(t))$

Report an Error

Example Question #5 : Derivatives Of Vectors

We let $\vec{u}=(t^{n},t^{m},t^{s})$ with $n>\{m,s\}, m, n , s\in\mathbb{N}$ .

What is the derivative of order of $\vec{u}$ ?

Possible Answers:

(n!,0,s!)

(n,0,0)

(n!, m!, s!)

(n!,0,0)

(n,m,n)

Correct answer:

(n!,0,0)

Explanation:

To obtain the derivative we will need to compute the derivative of each component. Note that we have a polynomial in this case.

We are also given that n>m and n> s with m,n and s positive integers.

We know that if given f=t^n , then $f^{[k]}=n(n-1)...(n-(k-1))t^{n-k}$

so for k=n, we have $[t^{n}]^{[n]}=n!$

and

${\t^{m}\}^{[k]}=0$ $t^{m}^{[n]}=0$ for m< n .

In the same manner we have :

$t^{s}^{[n]}=0$

This shows that the derivative of order n is given by:

(n!,0,0)

Report an Error

Example Question #1 : Derivatives Of Vectors

Assuming that $\vec{u}=(-e^{-t},artan(t),t-\frac{1}{t})$ is the vector position of a moving vehicle. Can the velocity be zero?

Possible Answers:

Yes at (0, 1 ,1)

Never

$(0,\infty, 1)$

Yes at $(-\infty,0,1)$

Yes at $(-\infty, \infty,\infty)$

Correct answer:

Never

Explanation:

We first have to determine the expression of the velocity and see what happens as we move toward infinity.

Let us first compute the derivative. To do that, we do it componentwize.

We have,

$(-e^{-t})'=e^{-t}$

$(artant)'=\frac{1}{1+t^2}$

$(t-\frac{1}{t})'=1+\frac{1}{t^2}$

Therefore the expression of the velocity .

Since we have $e^{-t}>0$ , $\frac{1}{1+t^2}>0$ , $1+\frac{1}{t^2}>0$ for all t.

The velocity can never be zero.

Report an Error

Example Question #2 : Derivatives Of Vectors

Let

$\vec{u}=(tln(t),arctan(t),\frac{1}{1-t})$ .

What are the values of for which $\vec{u}$ is defined?

Possible Answers:

$(0,\infty)$

$(0 , 1)\cup (1 , \infty)$

[0,1]

$(1,\infty)$

(0, 1)

Correct answer:

$(0 , 1)\cup (1 , \infty)$

Explanation:

To have $\vec{u}$ defined, we need to have all the components defined on the same interval, ln(t) for t>0

arctan(t) is defined for all t.

$\frac{1}{1-t}$ is defined if $t\neq 1$

This gives $(t>0)\cap(t\neq 1)\cap (-\infty, \infty )=(0,1)\cup (1,\infty)$

Report an Error

← Previous 1 2 3 4 Next →

View Calculus 2 Tutors

Samita
Certified Tutor

University of Houston, Bachelor of Science, Computer Science. University of Houston, Non Degree Bachelors, Biology, General.

View Calculus 2 Tutors

Eric
Certified Tutor

national autonomous university of Mexico , Doctorate (PhD), Mathematics .

View Calculus 2 Tutors

Chandana
Certified Tutor

University of Colombo, Bachelor of Science, Mathematics. University of Wyoming, Doctor of Philosophy, Mathematics.

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Algebra Tutors in Chicago, ACT Tutors in San Francisco-Bay Area, Reading Tutors in Denver, MCAT Tutors in New York City, SAT Tutors in Boston, SSAT Tutors in Atlanta, Chemistry Tutors in Washington DC, GRE Tutors in Los Angeles, ACT Tutors in Denver, SAT Tutors in Miami

Popular Courses & Classes

SSAT Courses & Classes in Seattle, SAT Courses & Classes in Houston, ACT Courses & Classes in Phoenix, SAT Courses & Classes in Miami, GRE Courses & Classes in Los Angeles, SAT Courses & Classes in New York City, LSAT Courses & Classes in Washington DC, LSAT Courses & Classes in Phoenix, GRE Courses & Classes in Boston, GRE Courses & Classes in Philadelphia

Popular Test Prep

SSAT Test Prep in Atlanta, SAT Test Prep in New York City, GRE Test Prep in Boston, ACT Test Prep in Boston, ACT Test Prep in Phoenix, ISEE Test Prep in Boston, ACT Test Prep in New York City, ISEE Test Prep in Phoenix, MCAT Test Prep in Philadelphia, LSAT Test Prep in Boston

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 2 : Derivatives of Vectors

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Derivatives Of Vectors

Example Question #5 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #1 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #1 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #1 : Derivatives Of Vectors

Example Question #1 : Derivatives Of Vectors

Example Question #1 : Derivatives Of Vectors

Example Question #5 : Derivatives Of Vectors

Example Question #1 : Derivatives Of Vectors

Example Question #2 : Derivatives Of Vectors

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: