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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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All Calculus 2 Resources

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

Example Questions

Example Question #21 : Derivatives Of Vectors

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

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2e^t+1,6e^{2t}+1,10t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2e^t+1,6e^t+1,10t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2e^t,6e^{2t},10t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2e^t,6e^{2t},t)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2e^t+1,6e^{2t}+1,10t)$

Explanation:

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$
When taking derivatives of sums, evaluate with the sum rule which states that the derivative of the sum is the same as the sum of the derivative of each term: $\frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}$
Special rule when differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

In this problem, ${\vec{r}(t)} = ((2e^t + t),3e^{2t} + t,5t^2)$

$\frac{\mathrm{d} }{\mathrm{d} t}(2e^t + t)= \frac{\mathrm{d} }{\mathrm{d} t}2e^t + \frac{\mathrm{d} }{\mathrm{d} t}t = 2e^t + 1$

$\frac{\mathrm{d} }{\mathrm{d} t}(3e^{2t} + t)=\frac{\mathrm{d} }{\mathrm{d} t}3e^{2t} + \frac{\mathrm{d} }{\mathrm{d} t}t = 6e^{2t} + 1$

$\frac{\mathrm{d} }{\mathrm{d} t}5t^2=10t$

Put it all together to get $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2e^t + 1,6e^{2t} + 1,10t)$

Report an Error

Example Question #1011 : Calculus Ii

$\newline {\vec{r}(t)} = (3e^t+t^3,e^{2t}+t,9e^t+e^{2t})\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

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

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

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3t^2,0,9e^t+2e^{2t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3e^t,2e^{2t},9e^t)$

Correct answer:

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

Explanation:

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$
When taking derivatives of sums, evaluate with the sum rule which states that the derivative of the sum is the same as the sum of the derivative of each term: $\frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}$
Special rule when differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

In this problem, ${\vec{r}(t)} = ((3e^t + t^3),(e^{2t} + t),(9e^t+e^{2t}))$

$\frac{\mathrm{d} }{\mathrm{d} t}(3e^t + t^3)= \frac{\mathrm{d} }{\mathrm{d} t}3e^t + \frac{\mathrm{d} }{\mathrm{d} t}t^3 = 3e^t + 3t^2$

$\frac{\mathrm{d} }{\mathrm{d} t}(e^{2t} + t)= \frac{\mathrm{d} }{\mathrm{d} t}e^{2t}+ \frac{\mathrm{d} }{\mathrm{d} t}t =2e^{2t} + 1$

$\frac{\mathrm{d} }{\mathrm{d} t}(9e^t+e^{2t})= \frac{\mathrm{d} }{\mathrm{d} t}9e^t + \frac{\mathrm{d} }{\mathrm{d} t}e^{2t} = 9e^t + 2e^{2t}$

Put it all together to get $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

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

Report an Error

Example Question #31 : Derivatives Of Vectors

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

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (121e^{11t},3e^t,11e^t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (121e^{11t},e^t,11e^t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (121e^{11t},3e^t,11e^t+1)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (121e^{10t},3e^t,11e^t+1)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (121e^{11t},3e^t,11e^t+1)$

Explanation:

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$
When taking derivatives of sums, evaluate with the sum rule which states that the derivative of the sum is the same as the sum of the derivative of each term: $\frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}$
Special rule when differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

In this problem, ${\vec{r}(t)} = (11e^{11t},3e^t,(11e^t+t))$

$\frac{\mathrm{d} }{\mathrm{d} t}11e^{11t}=121e^{11t}$

$\frac{\mathrm{d} }{\mathrm{d} t}3e^t=3e^t$

$\frac{\mathrm{d} }{\mathrm{d} t}(11e^t+t)= \frac{\mathrm{d} }{\mathrm{d} t}11e^t + \frac{\mathrm{d} }{\mathrm{d} t}t =11e^t+1$

Put it all together to get $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (121e^{11t},3e^t,11e^t+1)$

Report an Error

Example Question #501 : Parametric, Polar, And Vector

$\newline {\vec{r}(t)} = (11t,10t,e^{10t})\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,10,10e^{10t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,10,10)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11.10,e^{10t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,0,10e^{10t})$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,10,10e^{10t})$

Explanation:

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$
Special rule when differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

In this problem, ${\vec{r}(t)} = (11t,10t,e^{10t})$

$\frac{\mathrm{d} }{\mathrm{d} t}11t=11$

$\frac{\mathrm{d} }{\mathrm{d} t}10t=10$

$\frac{\mathrm{d} }{\mathrm{d} t}e^{10t}=10e^{10t}$

Put it all together to get $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,10,10e^{10t})$

Report an Error

Example Question #1011 : Calculus Ii

$\newline {\vec{r}(t)} = (20t^2,6t,5t)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (20t,6,5)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (20,6,5)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (40t,6,5)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (40t,0,0)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (40t,6,5)$

Explanation:

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$

In this problem, ${\vec{r}(t)} = (20t^2,6t,5t)$

$\frac{\mathrm{d} }{\mathrm{d} t}20t^2=40t$

$\frac{\mathrm{d} }{\mathrm{d} t}6t=6$

$\frac{\mathrm{d} }{\mathrm{d} t}5t=5$

Put it all together to get $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (40t,6,5)$

Report an Error

Example Question #34 : Derivatives Of Vectors

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Special rule when differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

In this problem, ${\vec{r}(t)} = (2e^t,3e^t,5e^t)$

$\frac{\mathrm{d} }{\mathrm{d} t}2e^t=2e^t$

$\frac{\mathrm{d} }{\mathrm{d} t}3e^t=3e^t$

$\frac{\mathrm{d} }{\mathrm{d} t}5e^t=5e^t$

Put it all together to get $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

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

Report an Error

Example Question #35 : Derivatives Of Vectors

$\newline {\vec{r}(t)} = (3e^t,6e^{2t},11e^t)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3e^t,12e^{2t},11e^t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3e^t,12e^t,11e^t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3,12,11)$

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

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3e^t,12e^{2t},11e^t)$

Explanation:

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Special rule when differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

In this problem, ${\vec{r}(t)} = (3e^t,6e^{2t},11e^t)$

$\frac{\mathrm{d} }{\mathrm{d} t}3e^t=3e^t$

$\frac{\mathrm{d} }{\mathrm{d} t}6e^{2t}=12e^{2t}$

$\frac{\mathrm{d} }{\mathrm{d} t}11e^t=11e^t$

Put it all together to get $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3e^t,12e^{2t},11e^t)$

Report an Error

Example Question #36 : Derivatives Of Vectors

$\newline {\vec{r}(t)} = (e^t+t^2,e^t+2t^2,5t^4+t)\newline$

Calculate $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (e^t+2t,e^t+4t,20t^3+1)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (e^t,e^t,20t^3)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (e^t+2t,e^t,20t^3)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (e^t,e^t+4t,20t^3+1)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (e^t+2t,e^t+4t,20t^3+1)$

Explanation:

In general:

If $\newline \ {\vec{r}(t)} = (f(t),g(t),z(t))$ ,

then $\newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))$

Derivative rules that will be needed here:

Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$
When taking derivatives of sums, evaluate with the sum rule which states that the derivative of the sum is the same as the sum of the derivative of each term: $\frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}$
Special rule when differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

In this problem, ${\vec{r}(t)} = (e^t+t^2,e^t+2t^2,5t^4+t)$

Use the sum rule and the power rule on each of the components.

$\frac{\mathrm{d} }{\mathrm{d} t}(e^t+t^2)= \frac{\mathrm{d} }{\mathrm{d} t}e^t + \frac{\mathrm{d} }{\mathrm{d} t}t^2 = e^t+2t$

$\frac{\mathrm{d} }{\mathrm{d} t}(e^{t}+2t^2)= \frac{\mathrm{d} }{\mathrm{d} t}e^{t} + \frac{\mathrm{d} }{\mathrm{d} t}2t^2 =e^{t}+4t$

$\frac{\mathrm{d} }{\mathrm{d} t}(5t^4+t)=\frac{\mathrm{d} }{\mathrm{d} t}5t^4 + \frac{\mathrm{d} }{\mathrm{d} t}t = 20t^3+1$

Put it all together to get $\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (e^t+2t,e^t+4t,20t^3+1)$

Report an Error

Example Question #1 : Vector Calculations

Find the cross product of the following two vectors:

$\vec{a}=\left \langle -4,3,-1 \right \rangle$

$\vec{b}=\left \langle 7,-2,3 \right \rangle$

Possible Answers:

$\left \langle -6,8,11 \right \rangle$

$\left \langle 4,-2,17 \right \rangle$

$\left \langle 12,4,7 \right \rangle$

$\left \langle 7,5,-13 \right \rangle$

$\left \langle 5,-2,-1 \right \rangle$

Correct answer:

$\left \langle 7,5,-13 \right \rangle$

Explanation:

We obtain the cross product of the two vectors by setting up the following matrix:

$\begin{bmatrix} \vec{i} &\vec{j} &\vec{k} \\-4 &3 &-1 \\7 &-2 &3 \end{bmatrix}$

Where our first row represents the unit vectors, the second row represents vector a, and the third row represents vector b. The first component of our cross product is obtained by taking the determinant of the matrix left by crossing out the row and column in which $\vec{i}$ is located. Accordingly, our second and third components are found by taking the determinant of the matrix left by crossing out the rows and columns in which $\vec{j}$ and $\vec{k}$ are located, respectively. This process gives us the following simple equation for expressing the cross product of two vectors, into which we can plug in the components of our vectors to find the cross product:

$\vec{a}\times \vec{b}=\left \langle a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1 \right \rangle$

$=\left \langle (3)(3)-(-1)(-2),(-1)(7)-(-4)(3),(-4)(-2)-(3)(7) \right \rangle$

$=\left \langle 7,5,-13 \right \rangle$

Report an Error

Example Question #1 : Vector Calculations

Evaluate the dot product: $\left \langle 3,2\right \rangle\cdot\left \langle -6,9\right \rangle$

Possible Answers:

$\left \langle -18,18\right \rangle$

$\left \langle -3,11\right \rangle$

$\left \langle 0,0\right \rangle$

Correct answer:

Explanation:

Let vector $\vec{a}=\left \langle a1,a2\right \rangle$ and $\vec{b}=\left \langle b1,b2\right \rangle$ .

The dot product is equal to:

$\vec{a}\cdot\vec{b} = a1\cdot b1+ a2\cdot b2$

Following this rule for the current problem, simplify.

$\left \langle 3,2\right \rangle\cdot\left \langle -6,9\right \rangle = (3)(-6)+(2)(9)=0$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #21 : Derivatives Of Vectors

Example Question #1011 : Calculus Ii

Example Question #31 : Derivatives Of Vectors

Example Question #501 : Parametric, Polar, And Vector

Example Question #1011 : Calculus Ii

Example Question #34 : Derivatives Of Vectors

Example Question #35 : Derivatives Of Vectors

Example Question #36 : Derivatives Of Vectors

Example Question #1 : Vector Calculations

Example Question #1 : Vector Calculations

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