MCWG Worksheet 18-1b Prerequisites

Section 2 Prerequisites

Finding a vector equation for a curve with a specified orientation including bounds.
Finding the derivative of a vector equation of a curve.
Composing multivariable functions.
Taking the dot product of vectors with functions as components.
Calculating a definite integral of a single variable function.

Exercises Exercises

1.

Find a vector equation for the line in the direction of \(\vec{v}={9}\vec{i} +{9}\vec{j} -{5}\vec{k}\) passing through the point \(({-5},{5},{2})\text{.}\)

\(\vec{r}(t)=\)

Hint

Recall that a vector equation for a line passing through point \(P\) at \(t=0\) in the direction of the vector \(\vec{v}\) is \(\vec{r}(t)=\vec{P} +t\vec{v}\) where \(\vec{P}\) is the position vector with head at point \(P\text{.}\)

Solution

A vector equation for the line in the direction of \(\vec{v}={9}\vec{i} +{9}\vec{j} -{5}\vec{k}\) passing through the point \(P=({-5},{5},{2})\) at \(t=0\) is \[\vec{r}(t)=\vec{P}+t\vec{v},\] where \(\vec{P}\) is the position vector with head at point \(P\text{.}\) Thus we get

\begin{equation*} \newcommand{\amp}{\amp }\begin{aligned} \vec{r}(t)\amp ={-5}\vec{i}+{5}\vec{j}+{2}\vec{k}+t({9}\vec{i} +{9}\vec{j} -{5}\vec{k}) \\ \amp =({-5}+{9}t)\vec{i} +({5}+{9}t)\vec{j} +({2}-{5}t)\vec{k}. \end{aligned} \end{equation*}

2.

Find a vector equation for the line segment from \(({3},{5},{6})\) to \(({7},{5},{-3})\text{.}\)

\(\vec{r}(t) =\) for \(\leq t \leq\)

Hint

Recall that a vector equation for a line segment from point \(P\) to point \(Q\) is \(\vec{r}(t)=\vec{P} +\vec{PQ}t\) for \(0\leq t\leq 1\) where \(\vec{P}\) is the position vector with head at point \(P\) and \(\vec{PQ}\) is the vector pointing from \(P\) to \(Q\text{.}\)

Solution

The vector pointing from \(({3},{5},{6})\) to \(({7},{5},{-3})\) is \(\vec{v}={4}\vec{i} -{9}\vec{k}\text{.}\) Therefore, a vector equation for this line segment is

\begin{equation*} \vec{r}(t)={3}\vec{i} +{5}\vec{j} +{6}\vec{k} +({4}\vec{i} -{9}\vec{k})t=({3+4t})\vec{i} +{5}\vec{j} +({6}-{9}t)\vec{k} \end{equation*}

with \(0\leq t\leq 1\text{.}\)

3.

Find a vector equation for a quarter circle centered at \((0,0)\) of radius \({4}\) starting at the point \(({4},0)\) and oriented counter-clockwise.

\(x(t) =\)

\(y(t) =\)

for \(\leq t \leq\)

Hint

Recall that \(\vec{r}(t) =\cos t \vec{i} +\sin t \vec{j}\text{,}\) \(0\leq t\leq 2\pi\) gives a full circle oriented counter-clockwise starting (when \(t=0\)) at \((1,0)\text{.}\)

Solution

The equation \(\vec{r}(t) =\cos t \vec{i} +\sin t \vec{j}\text{,}\) \(0\leq t\leq 2\pi\) gives a full circle of radius \(1\) oriented counter-clockwise starting at \((1,0)\) when \(t=0\text{.}\) To get a circle of radius \({4}\text{,}\) we scale each component by \({4}\text{.}\) Thus \(\vec{r}(t) ={4}\cos t \vec{i} +{4}\sin t \vec{j}\) for \(0\leq t\leq 2\pi\) is a circle of radius \({4}\) centered at \((0,0)\) oriented counterclockwise starting at \(({4},0)\) when \(t=0\text{.}\) To get the quarter circle, we restrict \(0\leq t\leq \pi/2\text{.}\)

4.

Find a vector equation for a half circle centered at \((1,{7},1)\) of radius \({5}\) in the plane \(y={7}\) starting at the point \(({6},{7},{1})\text{.}\)

\(x(t) =\)

\(y(t) =\)

\(z(t) =\)

for \(\leq t \leq\)

Hint

Solution

The equation \(\vec{r}(t) =\cos t \vec{i} +\sin t \vec{k}\text{,}\) \(0\leq t\leq 2\pi\) gives a full circle oriented counter-clockwise starting (when \(t=0\)) at \((1,0,0)\) in the \(xz\)-plane. To get a circle of radius \({5}\text{,}\) we scale each component by \({5}\text{,}\) and to move the center to \((1,{7},1)\) we add the vector \(\vec{i} +{7}\vec{j}+\vec{k}\) (which translates every point in the \(x\)-direction by \(1\text{,}\) the \(y\)-direction by \({7}\) and the \(z\)-direction by \(1\)). Thus

\begin{equation*} \vec{r}(t) =(1+{5}\cos t)\vec{i} +{7}\vec{j}+(1+ {5}\sin t)\vec{k} \end{equation*}

for \(0\leq t\leq 2\pi\) is a circle of radius \({5}\) centered at \((1,{7},1)\) in the plane \(y={7}\) starting at the point \(({6},{7},{1})\text{.}\) To get the half circle, we restrict \(0\leq t\leq \pi\text{.}\)

5.

Find a vector equation for a spiral starting at \(({6},0,0)\) rotating and moving up the \(z\)-axis with \(z=t\text{,}\) a rotation speed of one revolution per each vertical unit of displacement, and with radius \({6}\) oriented counterclockwise when view from above.

\(\vec{r}(t)=\) for \(t\geq\)

Hint

Recall that \(\vec{r}(t) =\cos t \vec{i} +\sin t \vec{j}\) parameterizes a circle of radius \(1\) in the \(xy\)-plane centered at \((0,0,0)\) oriented counter-clockwise when viewed from above starting (when \(t=0\)) at \((1,0,0)\text{.}\) How can you modify this equation to get the described spiral?

Solution

The equation \(\vec{r}(t) =\cos t \vec{i} +\sin t \vec{j}\) gives a circle of radius \(1\) in the \(xy\)-plane oriented counter-clockwise when viewed from above and centered at \((0,0,0)\) starting (when \(t=0\)) at \((1,0,0)\text{.}\) To get a circle of radius \({6}\) with starting point \(({6},0,0)\text{,}\) we scale each component by \({6}\text{,}\) and to make the curve spiral around the \(z\)-axis, we add the component \(t\vec{k}\text{.}\) Thus \(\vec{r}(t)={6}\cos(t)\vec{i} + {6}\sin(t)\vec{j}+t\vec{k}\) is a vector equation for the described curve.

6.

Find a vector equation for the segment of the graph of \(y=x^3\) starting at \(({4},{64})\) to the point \(({-5},{-125})\text{.}\)

\(x(t) =\)

\(y(t) =\)

for \(\leq t \leq\)

Hint

Recall that to parameterize a curve that follows the graph of a function \(y=f(x)\) oriented from right to left, we can choose \(x=-t\) and then \(y\) is completely determined by the function \(y=f(x)\text{.}\)

Solution

To parameterize a curve that follows the graph of a function \(y=f(x)\) oriented from right to left, we can choose \(x=-t\) and then we get \(y=f(-t)=-t^3\text{.}\) Thus, \(\vec{r}(t) =-t\vec{i} -t^3\vec{j}\) is a vector equation for \(y=x^3\) oriented from right to left. Since \(x=-t\) we choose \(-{4}\leq t\leq {5}\) to restrict to the appropriate domain.

7.

Find \(\vec{r}'(t)\) for \(\vec{r}(t)=t^2\vec{i}-\sin t\vec{j} +e^{{24} t}\vec{k}\text{.}\)

\(\vec{r}'(t)=\)

Hint

The components of \(\vec{r}'(t)\) are the derivatives of the components of \(\vec{r}(t)\text{,}\) so use standard single variable rules to differentiate each component.

Solution

The components of \(\vec{r}'(t)\) are the derivatives of the components of \(\vec{r}(t)\text{.}\) Thus, \(\vec{r}'(t)=2t\vec{i}-\cos t\vec{j} +{24} e^{{24} t}\vec{k}\text{.}\)

8.

Find \(\vec{r}'(t)\) for \(\vec{r}(t)=\cos t \vec{i}+\sin t\vec{j} +\ln({27} t)\vec{k}\text{.}\)

\(\vec{r}'(t)=\)

Hint

The components of \(\vec{r}'(t)\) are the derivatives of the components of \(\vec{r}(t)\text{,}\) so use standard single variable rules to differentiate each component.

Solution

The components of \(\vec{r}'(t)\) are the derivatives of the components of \(\vec{r}(t)\text{.}\) Thus, \(\vec{r}'(t)=\displaystyle -\sin t \vec{i}+\cos t\vec{j} +\frac{1}{t} \vec{k}\text{.}\)

9.

Find \(\vec{F} (\vec{r}(t))\) for \(\vec{F} (x,y)=-{30} y\vec{i} +{30} x\vec{j}\) and \(\vec{r}(t)=\cos t \vec{i}+\sin t\vec{j}\text{.}\)

\(\vec{F} (\vec{r}(t))=\)

Hint

Recall that in a vector equation \(\vec{r}(t)\) for a curve, the \(\vec{i}\) component is the \(x\)-coordinate, the \(\vec{j}\) component is the \(y\)-coordinate, and, where relevant, the \(\vec{k}\) component is the \(z\)-coordinate. Therefore, to compose we replace \(x\text{,}\) \(y\text{,}\) and \(z\) with these components.

Solution

The \(x\) coordinate of \(\vec{r}(t)\) is \(\cos t\text{,}\) and the \(y\)-coordinate is \(\sin t\text{,}\) so \(\vec{F} (\vec{r}(t)) =-{30} \sin t \vec{i} + {30} \cos t\vec{j}\text{.}\)

10.

Find \(\vec{F} (\vec{r}(t))\) for

\begin{equation*} \vec{F} (x,y)=-xy\vec{i} +\sin x\vec{j} -z^2x\vec{k} \end{equation*}

and

\begin{equation*} \vec{r}(t)={5} t \vec{i}+e^t\vec{j}-t^2\vec{k}\text{.} \end{equation*}

\(\vec{F} (\vec{r}(t))=\)

Hint

Solution

The \(x\) coordinate of \(\vec{r}(t)\) is \({5} t\text{,}\) the \(y\)-coordinate is \(e^t\) and the \(z\)-coordinate is \(-t^2\text{,}\) so

\begin{equation*} \newcommand{\amp}{\amp }\begin{aligned} \vec{F} (\vec{r}(t)) \amp =-{5} te^t\vec{i} +\sin ({5} t) \vec{j} -(t^2)^2\cdot {5} t\vec{k}\\ \amp =-{5} te^t\vec{i} +\sin ({5} t) \vec{j} -{5} t^5\vec{k}. \end{aligned} \end{equation*}

11.

Find \(\vec{r} (t) \cdot \vec{q} (t)\) for

\begin{equation*} \vec{r}(t)=\sin ({8} t) \vec{i} +\cos ({8} t)\vec{j} \end{equation*}

and

\begin{equation*} \vec{q}(t)=-\cos ({8} t) \vec{i} +\sin ({8} t)\vec{j}\text{.} \end{equation*}

\(\vec{r} (t) \cdot \vec{q} (t)=\)

Hint

Recall that the dot product of two vectors \(\vec{u} =a\vec{i} +b\vec{j}\) and \(\vec{v} =c\vec{i} +d\vec{j}\) is \(\vec{u} \cdot \vec{v}=ac+bd\text{.}\)

Solution

We have

\begin{equation*} \newcommand{\amp}{\amp }\begin{aligned} \vec{r} (t) \cdot \vec{q} (t)\amp = (\sin ({8} t) \vec{i} +\cos ({8} t)\vec{j} )\cdot (-\cos ({8} t) \vec{i} +\sin ({8} t)\vec{j}) \\ \amp = -\sin ({8} t) \cos ({8} t) +\cos ({8} t) \sin ({8} t) \\ \amp = 0. \end{aligned} \end{equation*}

12.

Find \(\vec{r} (t) \cdot \vec{q} (t)\) for

\begin{equation*} \vec{r}(t)=t \vec{i} +t^2\vec{j}+e^t\vec{k} \end{equation*}

and

\begin{equation*} \vec{q}(t)=({11} +t) \vec{i} -t\vec{j}+({29} -t)\vec{k}\text{.} \end{equation*}

\(\vec{r} (t) \cdot \vec{q} (t)=\)

Hint

Recall that the dot product of two vectors \(\vec{u} =a\vec{i} +b\vec{j}\) and \(\vec{v} =c\vec{i} +d\vec{j}\) is \(\vec{u} \cdot \vec{v}=ac+bd\text{.}\)

Solution

We have

\begin{equation*} \newcommand{\amp}{\amp }\begin{aligned} \vec{r} (t) \cdot \vec{q} (t)\amp = (t \vec{i} +t^2\vec{j}+e^t\vec{k})\cdot (({11} +t) \vec{i} -t\vec{j}+({29} -t)\vec{k}) \\ \amp = t\cdot ({11} +t) +t^2\cdot (-t) +e^t\cdot ({29} -t) \\ \amp = {11} t + t^2 -t^3+ {29} e^t-te^t. \end{aligned} \end{equation*}

13.

Find the value of the integral. Give the exact answer or round to three decimal places.

\(\displaystyle \int_{0}^{1} (e^x+x^2) \sqrt{{11}} dx=\)

Hint

Break up the integral using the fact that the integral of a sum is the sum of the integrals.

Solution

\begin{equation*} \newcommand{\amp}{\amp }\begin{aligned} \int_{0}^{1} (e^x+x^2) \sqrt{{11}} dx \amp = \sqrt{{11}} \int_{0}^{1} e^x+ \sqrt{{11}}\int_{0}^{1} x^2 dx\\ \amp = \sqrt{{11}} \left( e^x\right)\bigg|_{0}^{1} + \sqrt{{11}}\left( \frac{x^3}{3} \right)\bigg|_{0}^1 \\ \amp = \sqrt{{11}} ( e-1) + \sqrt{{11}}\left(\frac{1}{3} -0\right) \\ \amp = \frac{\sqrt{{11}}}{3}(3e-2) \end{aligned} \end{equation*}

14.

Find the value of the integral. Give the exact answer or round to three decimal places.

\(\displaystyle \int_0^1 -{7} t+{5} dt=\)

Hint

Break up the integral using the fact that the integral of a sum is the sum of the integrals.

Solution

\begin{equation*} \newcommand{\amp}{\amp }\begin{aligned} \int_{0}^{1} -{7} t+{5} dt \amp = -{7} \int_{0}^{1} t dt + {5} \int_{0}^{1} dt\\ \amp = -{7} \left( \frac{t^2}{2} \right)\bigg|_{0}^{1} + {5} \left( t\right)\big|_{0}^{1} \\ \amp = -{7} \left( \frac{1}{2} -0\right) + {5} \left( 1-0\right) \\ \amp = {5}-\frac{{7}}{2} = \frac{{3}}{2}. \end{aligned} \end{equation*}

15.

Find the value of the integral. Give the exact answer or round to three decimal places.

\(\displaystyle \int_0^1 -{13} t ~ dt=\)

Hint

Remember you can always take constants that are multiplying the whole integrand out of the integral.

Solution

\begin{equation*} \int_0^1 -{13} t dt = -{13}\int_0^1 t dt = -{13}\left( \frac{t^2}{2} \right)\bigg|_{0}^{1} =-\frac{{13}}{2} \end{equation*}