# 3.2 Calculus of vector-valued functions  (Page 5/11)

 Page 5 / 11

## Key concepts

• To calculate the derivative of a vector-valued function, calculate the derivatives of the component functions, then put them back into a new vector-valued function.
• Many of the properties of differentiation from the Introduction to Derivatives also apply to vector-valued functions.
• The derivative of a vector-valued function $\text{r}\left(t\right)$ is also a tangent vector to the curve. The unit tangent vector $\text{T}\left(t\right)$ is calculated by dividing the derivative of a vector-valued function by its magnitude.
• The antiderivative of a vector-valued function is found by finding the antiderivatives of the component functions, then putting them back together in a vector-valued function.
• The definite integral of a vector-valued function is found by finding the definite integrals of the component functions, then putting them back together in a vector-valued function.

## Key equations

• Derivative of a vector-valued function
${r}^{\prime }\left(t\right)=\underset{\text{Δ}t\to 0}{\text{lim}}\frac{\text{r}\left(t+\text{Δ}t\right)-\text{r}\left(t\right)}{\text{Δ}t}$
• Principal unit tangent vector
$\text{T}\left(t\right)=\frac{{r}^{\prime }\left(t\right)}{\text{‖}\phantom{\rule{0.1em}{0ex}}{r}^{\prime }\left(t\right)\text{‖}}$
• Indefinite integral of a vector-valued function
$\int \left[f\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+g\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+h\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{k}\right]dt=\left[\phantom{\rule{0.1em}{0ex}}\int f\left(t\right)dt\right]\phantom{\rule{0.1em}{0ex}}\text{i}+\left[\phantom{\rule{0.1em}{0ex}}\int g\left(t\right)dt\right]\phantom{\rule{0.1em}{0ex}}\text{j}+\left[\phantom{\rule{0.1em}{0ex}}\int h\left(t\right)dt\right]\phantom{\rule{0.1em}{0ex}}\text{k}$
• Definite integral of a vector-valued function
${\int }_{a}^{b}\left[f\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+g\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+h\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{k}\right]dt=\left[\phantom{\rule{0.1em}{0ex}}{\int }_{a}^{b}f\left(t\right)dt\right]\phantom{\rule{0.1em}{0ex}}\text{i}+\left[\phantom{\rule{0.1em}{0ex}}{\int }_{a}^{b}g\left(t\right)dt\right]\phantom{\rule{0.1em}{0ex}}\text{j}+\left[\phantom{\rule{0.1em}{0ex}}{\int }_{a}^{b}h\left(t\right)dt\right]\phantom{\rule{0.1em}{0ex}}\text{k}$

Compute the derivatives of the vector-valued functions.

$\text{r}\left(t\right)={t}^{3}\phantom{\rule{0.1em}{0ex}}\text{i}+3{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{j}+\frac{{t}^{3}}{6}\phantom{\rule{0.1em}{0ex}}\text{k}$

$⟨3{t}^{2},6t,\frac{1}{2}{t}^{2}⟩$

$\text{r}\left(t\right)=\text{sin}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\text{cos}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+{e}^{t}\phantom{\rule{0.1em}{0ex}}\text{k}$

$\text{r}\left(t\right)={e}^{\text{−}t}\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\left(3t\right)\text{​}\phantom{\rule{0.1em}{0ex}}\text{j}+10\sqrt{t}\phantom{\rule{0.1em}{0ex}}\text{k}.$ A sketch of the graph is shown here. Notice the varying periodic nature of the graph.

$⟨\text{−}{e}^{\text{−}t},3\phantom{\rule{0.1em}{0ex}}\text{cos}\left(3t\right),\frac{5}{\sqrt{t}}⟩$

$\text{r}\left(t\right)={e}^{t}\phantom{\rule{0.1em}{0ex}}\text{i}+2{e}^{t}\phantom{\rule{0.1em}{0ex}}\text{j}+\phantom{\rule{0.1em}{0ex}}\text{k}$

$\text{r}\left(t\right)=\text{i}+\phantom{\rule{0.1em}{0ex}}\text{j}+\phantom{\rule{0.1em}{0ex}}\text{k}$

$⟨0,0,0⟩$

$\text{r}\left(t\right)=t{e}^{t}\phantom{\rule{0.1em}{0ex}}\text{i}+t\phantom{\rule{0.1em}{0ex}}\text{ln}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\text{sin}\left(3t\right)\phantom{\rule{0.1em}{0ex}}\text{k}$

$\text{r}\left(t\right)=\frac{1}{t+1}\phantom{\rule{0.1em}{0ex}}\text{i}+\text{arctan}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\text{ln}\phantom{\rule{0.1em}{0ex}}{t}^{3}\phantom{\rule{0.1em}{0ex}}\text{k}$

$⟨\frac{-1}{{\left(t+1\right)}^{2}},\frac{1}{1+{t}^{2}},\frac{3}{t}⟩$

$\text{r}\left(t\right)=\text{tan}\left(2t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sec}\left(2t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+{\text{sin}}^{2}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{k}$

$\text{r}\left(t\right)=3\phantom{\rule{0.1em}{0ex}}\text{i}+4\phantom{\rule{0.1em}{0ex}}\text{sin}\left(3t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+t\phantom{\rule{0.1em}{0ex}}\text{cos}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{k}$

$⟨0,12\phantom{\rule{0.1em}{0ex}}\text{cos}\left(3t\right),\text{cos}\phantom{\rule{0.1em}{0ex}}t-t\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t⟩$

$\text{r}\left(t\right)={t}^{2}\phantom{\rule{0.1em}{0ex}}\text{i}+t{e}^{-2t}\phantom{\rule{0.1em}{0ex}}\text{j}-5{e}^{-4t}\phantom{\rule{0.1em}{0ex}}\text{k}$

For the following problems, find a tangent vector at the indicated value of t .

$\text{r}\left(t\right)=t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\left(2t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\text{cos}\left(3t\right)\phantom{\rule{0.1em}{0ex}}\text{k};t=\frac{\pi }{3}$

$\frac{1}{\sqrt{2}}⟨1,-1,0⟩$

$\text{r}\left(t\right)=3{t}^{3}\phantom{\rule{0.1em}{0ex}}\text{i}+2{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{j}+\frac{1}{t}\phantom{\rule{0.1em}{0ex}}\text{k};t=1$

$\text{r}\left(t\right)=3{e}^{t}\phantom{\rule{0.1em}{0ex}}\text{i}+2{e}^{-3t}\phantom{\rule{0.1em}{0ex}}\text{j}+4{e}^{2t}\phantom{\rule{0.1em}{0ex}}\text{k};$ $t=\text{ln}\left(2\right)$

$\frac{1}{\sqrt{1060.5625}}⟨6,-\frac{3}{4},32⟩$

$\text{r}\left(t\right)=\text{cos}\left(2t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}+{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{k};t=\frac{\pi }{2}$

Find the unit tangent vector for the following parameterized curves.

$\text{r}\left(t\right)=6\phantom{\rule{0.1em}{0ex}}\text{i}+\text{cos}\left(3t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+3\phantom{\rule{0.1em}{0ex}}\text{sin}\left(4t\right)\phantom{\rule{0.1em}{0ex}}\text{k},$ $0\le t<2\pi$

$\frac{1}{\sqrt{9\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{2}\left(3t\right)+144\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{2}\left(4t\right)}}⟨0,-3\phantom{\rule{0.1em}{0ex}}\text{sin}\left(3t\right),12\phantom{\rule{0.1em}{0ex}}\text{cos}\left(4t\right)⟩$

$\text{r}\left(t\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j}+\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{k},$ $0\le t<2\pi .$ Two views of this curve are presented here:

$\text{r}\left(t\right)=3\phantom{\rule{0.1em}{0ex}}\text{cos}\left(4t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+3\phantom{\rule{0.1em}{0ex}}\text{sin}\left(4t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+5t\phantom{\rule{0.1em}{0ex}}\text{k},1\le t\le 2$

$\text{T}\left(t\right)=\frac{-12}{13}\text{sin}\left(4t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\frac{12}{13}\text{cos}\left(4t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+\frac{5}{13}\phantom{\rule{0.1em}{0ex}}\text{k}$

$\text{r}\left(t\right)=t\phantom{\rule{0.1em}{0ex}}\text{i}+3t\phantom{\rule{0.1em}{0ex}}\text{j}+{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{k}$

Let $\text{r}\left(t\right)=t\phantom{\rule{0.1em}{0ex}}\text{i}+{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{j}-{t}^{4}\phantom{\rule{0.1em}{0ex}}\text{k}$ and $s\left(t\right)=\text{sin}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+{e}^{t}\phantom{\rule{0.1em}{0ex}}\text{j}+\text{cos}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{k}.$ Here is the graph of the function:

Find the following.

$\frac{d}{dt}\left[r\left({t}^{2}\right)\right]$

$⟨2t,4{t}^{3},-8{t}^{7}⟩$

$\frac{d}{dt}\left[{t}^{2}·s\left(t\right)\right]$

$\frac{d}{dt}\left[r\left(t\right)·s\left(t\right)\right]$

$\text{sin}\left(t\right)+2t{e}^{t}-4{t}^{3}\text{cos}\left(t\right)+t\phantom{\rule{0.1em}{0ex}}\text{cos}\left(t\right)+{t}^{2}{e}^{t}+{t}^{4}\text{sin}\left(t\right)$

Compute the first, second, and third derivatives of $\text{r}\left(t\right)=3t\phantom{\rule{0.1em}{0ex}}\text{i}+6\phantom{\rule{0.1em}{0ex}}\text{ln}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+5{e}^{-3t}\phantom{\rule{0.1em}{0ex}}\text{k}.$

Find $\text{r}\prime \left(t\right)·\phantom{\rule{0.1em}{0ex}}\text{r}\text{″}\left(t\right)\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}\text{r}\left(t\right)=-3{t}^{5}\phantom{\rule{0.1em}{0ex}}\text{i}+5t\phantom{\rule{0.1em}{0ex}}\text{j}+2{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{k}.$

$900{t}^{7}+16t$

The acceleration function, initial velocity, and initial position of a particle are
$\text{a}\left(t\right)=-5\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{i}-5\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.1em}{0ex}}\text{j},\phantom{\rule{0.1em}{0ex}}\text{v}\left(0\right)=9\phantom{\rule{0.1em}{0ex}}\text{i}+2\phantom{\rule{0.1em}{0ex}}\text{j},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{r}\left(0\right)=5\phantom{\rule{0.1em}{0ex}}\text{i}.$
Find $\text{v}\left(t\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{r}\left(t\right).$

The position vector of a particle is $\text{r}\left(t\right)=5\phantom{\rule{0.1em}{0ex}}\text{sec}\left(2t\right)\phantom{\rule{0.1em}{0ex}}\text{i}-4\phantom{\rule{0.1em}{0ex}}\text{tan}\left(t\right)\phantom{\rule{0.1em}{0ex}}\text{j}+7{t}^{2}\phantom{\rule{0.1em}{0ex}}\text{k}.$

1. Graph the position function and display a view of the graph that illustrates the asymptotic behavior of the function.
2. Find the velocity as t approaches but is not equal to $\pi \text{/}4$ (if it exists).

1. Undefined or infinite

Find the velocity and the speed of a particle with the position function $\text{r}\left(t\right)=\left(\frac{2t-1}{2t+1}\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\text{ln}\left(1-4{t}^{2}\right)\phantom{\rule{0.1em}{0ex}}\text{j}.$ The speed of a particle is the magnitude of the velocity and is represented by $\text{‖}{r}^{\text{'}}\left(t\right)\text{‖}.$

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