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Calculus 2 : Derivatives of Vectors

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Example Questions

Example Question #1003 : Calculus Ii

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

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (-4sin(4t),cos(t),2cos(2t)+2t)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (-4sin(4t),sin(t),sin(2t)+t^2)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (cos(4t),sin(t),sin(2t)+2t)$

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

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (-4sin(4t),cos(t),2cos(2t)+2t)$

Explanation:

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

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

$\\ \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}cos(4t),\frac{\mathrm{d} }{\mathrm{d} t}sin(t),\frac{\mathrm{d} }{\mathrm{d} t}(sin(2t)+t^2)\right) \\ \\= (-4sin(4t),cos(t),2cos(2t)+2t)$

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Example Question #1004 : Calculus Ii

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

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (5cos(5t)+3,2tcos(t^2),-sin(t))$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (5cos(5t)+3,cos(t^2),-sin(t))$

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

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (5cos(5t)+3,2tsin(t^2),-sin(t))$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (5cos(5t)+3,2tcos(t^2),-sin(t))$

Explanation:

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

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

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Example Question #1005 : Calculus Ii

$\newline {\vec{r}(t)} = (t^3+t,t^2+5,6t^3+t^2+t+1)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3t^2+1,2t,18^2+2t+1)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3t^2,2t,18^2+2t+1)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3t^2,2t,18t^2+2t+1)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3t^2+1,2,18^2+2t+1)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3t^2+1,2t,18^2+2t+1)$

Explanation:

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

$\\ \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}(t^3+t),\frac{\mathrm{d} }{\mathrm{d} t}(t^2+5),\frac{\mathrm{d} }{\mathrm{d} t}(6t^3+t^2+t+1)\right)\\ \\ = (3t^2+1,2t,18t^2+2t+1)$

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Example Question #1006 : Calculus Ii

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

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,15^2+2t,5e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,15t^2,5e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,15^2+2t,e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,15t^2+t^2,5e^{5t})$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2t,15^2+2t,5e^{5t})$

Explanation:

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

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = \left(\frac{\mathrm{d} }{\mathrm{d} t}t^2,\frac{\mathrm{d} }{\mathrm{d} t}(5t^3+t^2),\frac{\mathrm{d} }{\mathrm{d} t}e^{5t}\right) = (2t,15t^2+2t,5e^{5t})$

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Example Question #1003 : Calculus Ii

$\newline {\vec{r}(t)} = (20t,6t,3t^2)\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

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

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

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

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (20,6,3)$

Correct answer:

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

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, ${\vec{r}(t)} = (20t,6t,3t^2)$

$\frac{\mathrm{d} }{\mathrm{d} t}20t=20$

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

$\frac{\mathrm{d} }{\mathrm{d} t}3t^2=6t$

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

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

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Example Question #21 : Derivatives Of Vectors

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

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2e^{2t},3e^{3t},22e^{2t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2e^{2t},3e^{3t},22e^{t})$

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

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2,3,22)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2e^{2t},3e^{3t},22e^{2t})$

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:

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

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

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

$\frac{\mathrm{d} }{\mathrm{d} t}11e^{2t}=22e^{2t}$

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

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (2e^{2t},3e^{3t},22e^{2t})$

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Example Question #1009 : Calculus Ii

$\newline {\vec{r}(t)} = (e^{11t},e^{8t},e^{5t})\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11e^{11t},8e^{8t},5e^{4t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11e^{11t},8e^{8t},5e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11e^{11t},e^{8t},5e^{t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11,8,5)$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11e^{11t},8e^{8t},5e^{5t})$

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, ${\vec{r}(t)} = (e^{11t},e^{8t},e^{5t})$

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

$\frac{\mathrm{d} }{\mathrm{d} t}e^{8t}=8e^{8t}$

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

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

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (11e^{11t},8e^{8t},5e^{5t})$

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Example Question #231 : Ap Calculus Bc

$\newline {\vec{r}(t)} = (5t^3,e^{11t},e^{5t})\newline \textnormal{Calculate } \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}$

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2,11,5)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15,11,5)$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2,e^{11t},5e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2.11e^{11t},5e^{5t})$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2.11e^{11t},5e^{5t})$

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, ${\vec{r}(t)} = (5t^3,e^{11t},e^{5t})$

$\frac{\mathrm{d} }{\mathrm{d} t}5t^3=15t^2$

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

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

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

$\frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (15t^2,11e^{11t},5e^{5t})$

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

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

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Example Question #21 : Derivatives Of Vectors

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

Possible Answers:

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

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

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

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

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

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

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Calculus 2 : Derivatives of Vectors

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