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Calculus 2 : Vector

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Example Questions

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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Example Question #12 : Derivatives Of Vectors

Let $\vec{u}=(sin(2t),cos(2t),t^3)$ .

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

Possible Answers:

(-8cos(2x),8sin(2x),t)

(-8cos(2x),8sin(2x),3)

(-8cos(x),8sin(x),3t)

(-8cos(2x),8sin(2x),6)

(-8cos(2x),8sin(2x),2t^2)

Correct answer:

(-8cos(2x),8sin(2x),6)

Explanation:

To obtain the derivative, we simply differentiate 3 times each of the components.

Using the Chain Rule and the Power Rule we have :

$f(x)=sin(2t)\rightarrow f'(x)=2cos(2t)\rightarrow f''(x)=-4sin(2t)$

$\rightarrow cos(2x)^{[3]}=8sin(2x)$

$g(x)=cos(2t)\rightarrow g'(x)=-2sin(2t)\rightarrow g''(x)=-4cos(2t)$

$\rightarrow sin(2x)^{[3]}=-8cos(2x)$

$h(x)=t^3 \rightarrow h'(x)=3t^2 \rightarrow h''(x)=6t$

$\rightarrow t^3^{[3]}=6$

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Example Question #13 : Derivatives Of Vectors

$\newline {\vec{r}(t)} = (t^{2},3t,5t)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,3,5)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2,3,5)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,3t,5t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,0,0)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,3,5)$

Explanation:

$\newline \textnormal{In general:}\newline \textnormal{If} \ {\vec{r}(t)} = (f(t),g(t),z(t))\newline \textnormal{Then} \ \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t)\right)$

$\newline \textnormal{Derivative of a constant is always 0. For example, } \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0\newline \textnormal{Power Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}\newline \textnomal{Sum Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}\newline \textnormal{Use the power rule to differentiate each component of the vector.}\newline \textnormal{For example} \ \frac{\mathrm{d} }{\mathrm{d} t} t^2 = 2 * t ^ {2-1} = 2*t$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}t^{2},\frac{\mathrm{d} }{\mathrm{d} t}3t,\frac{\mathrm{d} }{\mathrm{d} t}5t\right) = (2t,3,5)$

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Example Question #14 : Derivatives Of Vectors

$\newline {\vec{r}(t)} = (4t,8t,10t^{3})\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (4,8,30t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (0,8,10t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (4,8,10t^{2})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (4,8,30t^{2})$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (4,8,30t^{2})$

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}4t,\frac{\mathrm{d} }{\mathrm{d} t}8t,\frac{\mathrm{d} }{\mathrm{d} t}10t^{3}\right) = (4,8,30t^{2})$

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Example Question #15 : Derivatives Of Vectors

$\newline {\vec{r}(t)} = (4t^{4},9t^{2},4t)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (24,9t,4)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (16t^3,9t,4)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (12,9,4)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (16^{3},18t,4)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (16^{3},18t,4)$

Explanation:

$\newline \textnormal{Derivative of a constant is always 0. For example, } \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0\newline \textnormal{Power Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1} \newline \textnormal{Use the power rule to differentiate each component of the vector.}\newline \textnormal{For example} \ \frac{\mathrm{d} }{\mathrm{d} t}4t^4 = 4*4*t^{4-1} = 16*t^3$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}4t^{4},\frac{\mathrm{d} }{\mathrm{d} t}9t^{2},\frac{\mathrm{d} }{\mathrm{d} t}4t\right) = (16t^{3},18t,4)$

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Example Question #16 : Derivatives Of Vectors

$\newline {\vec{r}(t)} = (11t,6t,3t)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,0,0)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,6,0)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (0,0,0)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,6,3)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,6,3)$

Explanation:

$\newline \textnormal{Derivative of a constant is always 0. For example, } \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0\newline \textnormal{Power Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1} \newline \textnormal{Use the power rule to differentiate each component of the vector.}\newline \textnormal{For example} \ \frac{\mathrm{d} }{\mathrm{d} t}11t = 11*1*t^{1-1} = 11$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}11t,\frac{\mathrm{d} }{\mathrm{d} t}6t,\frac{\mathrm{d} }{\mathrm{d} t}3t\right) = (11,6,3)$

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Example Question #17 : Derivatives Of Vectors

$\newline {\vec{r}(t)} = (3t^3,10t+1,t)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (9t^2,10,1)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (9t^3,11,1)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (9t^2,0,0)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (9t,10,1)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (9t^2,10,1)$

Explanation:

$\newline \textnormal{Derivative of a constant is always 0. For example, } \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0\newline \textnormal{Power Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}\newline \textnomal{Sum Rule:} \ \frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}\newline \textnormal{Use the power rule to differentiate each component of the vector.}\newline \textnormal{For example} \ \ \frac{\mathrm{d} }{\mathrm{d} t}3t^3 = 3*3*t^{3-1} = 9*t^2$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}3t^3,\frac{\mathrm{d} }{\mathrm{d} t}(10t+1),\frac{\mathrm{d} }{\mathrm{d} t}t\right) = (9t^2,10,1)$

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Calculus 2 : Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #4 : Derivatives Of Vectors

Example Question #991 : Calculus Ii

Example Question #1 : Derivatives Of Vectors

Example Question #11 : Derivatives Of Vectors

Example Question #12 : Derivatives Of Vectors

Example Question #13 : Derivatives Of Vectors

Example Question #14 : Derivatives Of Vectors

Example Question #15 : Derivatives Of Vectors

Example Question #16 : Derivatives Of Vectors

Example Question #17 : Derivatives Of Vectors

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