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Calculus 2 : Parametric, Polar, and Vector

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Example Questions

Example Question #2 : 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:

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

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

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

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

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

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 .

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Example Question #7 : 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}+x_{2}+\cdots nx_{n})\quad \cdots\quad ncos(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}+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})\]$

$[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})\]$

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Example Question #3 : 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:

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

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

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

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

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

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:

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

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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:

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

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

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

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

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

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.

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Example Question #2 : 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:

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

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

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

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:

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

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Example Question #3 : 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:

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

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

(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),cos(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 :

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

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

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

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Example Question #4 : 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,m,n)

(n!, m!, s!)

(n!,0,s!)

(n,0,0)

(n!,0,0)

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)

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Example Question #991 : Calculus Ii

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 $(-\infty,0,1)$

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

Never

$(0,\infty, 1)$

Yes at (0, 1 ,1)

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.

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Example Question #1 : 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:

$(1,\infty)$

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

$(0,\infty)$

(0, 1)

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

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Example Question #11 : Derivatives Of Vectors

We define the vector $\vec{u}=(t^{n}, e^{nt}, 1 , 1 , 1 , 1 , 1 ,1 )$ .

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

Possible Answers:

$\vec{u}=(n!, n^ne^{t}, 0, 0 , 0 , 0 , 0 ,0 )$

$\vec{u}=(n, n^ne^{nt}, 0, 0 , 0 , 0 , 0 ,0 )$

$\vec{u}=(n!, ne^{nt}, 0, 0 , 0 , 0 , 0 ,0 )$

$\vec{u}=(n!, n^ne^{nt}, 0, 0 , 0 , 0 , 0 ,0 )$

$\vec{u}=(n, n^ne^{n}, 0, 0 , 0 , 0 , 0 ,0 )$

Correct answer:

$\vec{u}=(n!, n^ne^{nt}, 0, 0 , 0 , 0 , 0 ,0 )$

Explanation:

To obtain the derivative we need to compute the derivative of each component. Since we are looking for the derivative of order n we have to differentiate each component n times to obtain the required result.

We know that :

$\{t^{n}\}^{[n]}=n!$

$\{e^{nt}\}^{[n]}=n\times n\cdots (e^{nt}) (ntimes)=n^{n}e^{nt}$

The nth derivatives of the constant is 0.

Thereore we obtain:

$\vec{u}^{[n]}=(n!, n^ne^{nt}, 0 , 0 , 0, 0 , 0 ,0 )$

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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #7 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #3 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #1 : Derivatives Of Vectors

Example Question #2 : Derivatives Of Vectors

Example Question #3 : Derivatives Of Vectors

Example Question #4 : Derivatives Of Vectors

Example Question #991 : Calculus Ii

Example Question #1 : Derivatives Of Vectors

Example Question #11 : Derivatives Of Vectors

All Calculus 2 Resources

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