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Calculus 2 : Derivatives of Vectors

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Example Questions

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

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

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

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

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

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

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

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

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

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

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Calculus 2 : Derivatives of Vectors

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

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

All Calculus 2 Resources

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