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Calculus 2 : Vector

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Example Questions

Example Question #23 : Graphing Vectors

In which quadrant does the vector terminate from the origin?

8i-13j

Possible Answers:

III

Correct answer:

Explanation:

8i-13j=8*<1,0>-13*<0,1>

=<8,0>-<0,13>

=<8,-13>

The resulting vector terminates in

$Quadrant\,IV$

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Example Question #24 : Graphing Vectors

In which quadrant does the vector terminate from the origin?

-i+2j

Possible Answers:

III

Correct answer:

Explanation:

-i+2j=-1*<1,0>+2*<0,1>

=<-1,0>+<0,2>

=<-1,2>

The resulting vector terminates in

$Quadrant\,II$

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Example Question #25 : Graphing Vectors

In which quadrant does the vector terminate from the origin?

5i+7j

Possible Answers:

III

Correct answer:

Explanation:

5i+7j=5*<1,0>+7*<0,1>

=<5,0>+<0,7>

=<5,7>

The resulting vector terminates in

$Quadrant\,I$

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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:

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

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

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

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

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

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:

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

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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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Calculus 2 : Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #23 : Graphing Vectors

Example Question #24 : Graphing Vectors

Example Question #25 : Graphing Vectors

Example Question #1 : Derivatives Of Vectors

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

All Calculus 2 Resources

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