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Section 2 Prerequisites

Subsection 2.1 Prerequisite Topics and Skills

  • Finding a vector equation for a given curve including bounds on the parameter.
  • Differentiating a vector-valued function.
  • Composing vector-valued functions.
  • Integrating a function of a single variable.

Exercises 2.2 Exercises

The following exercises test your knowledge of the prerequisites for this section.

1.

Find a vector equation for the line segment from the point \((1,2)\) to the point \((-1,3)\text{.}\)

\(\vec{r}(t) =\) 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 \((1,2)\) to \((-1,3)\) is \(\vec{v}=-2\vec{i} +\vec{j}\text{.}\) Therefore, a vector equation for this line segment is

\begin{equation*} \vec{r}(t)=\vec{i} +2\vec{j} +(-2\vec{i} +\vec{j})t=(1-2t)\vec{i} +(2+t)\vec{j} \end{equation*}

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

There are many other correct answers.

2.

Find a vector equation for the half circle \((x-1)^2+(y+2)^2=4\) with \(y\geq -2\) 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

Observe that \((x-1)^2+(y+2)^2=4\) is a circle of radius \(2\) centered at \((1,-2)\text{.}\) 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 \(2\text{,}\) we scale each component by \(2\text{,}\) and to move the center to \((1,-2)\) we add the vector \(\vec{i} -2\vec{j}\) (which translates every point to the right by \(1\) and down by \(2\)). Thus \(\vec{r}(t) =(2\cos t +1)\vec{i} +(2\sin t -2)\vec{j}\) for \(0\leq t\leq 2\pi\) is a circle of radius \(2\) centered at \((1,-2)\) oriented counterclockwise starting at \((3,-2)\) when \(t=0\text{.}\) To get the half circle, we restrict \(0\leq t\leq \pi\text{.}\)

3.

Find a vector equation for the portion of the parabola \(y=2x^2-1\) from \(x=-1\) to \(x=2\text{.}\)

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

Hint

Recall that to parameterize a curve that follows the graph of a function \(y=f(x)\text{,}\) we can choose \(x=t\text{,}\) 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 \(y=f(x)\text{,}\) we choose \(x=t\text{,}\) so we get \(y=f(t)=2t^2-1\text{.}\) Thus, \(\vec{r}(t) =t\vec{i} +(2t^2-1)\vec{j}\) is a vector equation for \(y=2x^2-1\text{.}\) Since \(x=t\text{,}\) we choose \(-1 \leq t\leq 2\) to restrict to the appropriate domain.

4.

Find a parametric equation for the curve \(x=\sin{y}\) for \(0\leq y\leq 2\pi\text{.}\)

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

Hint

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

Solution

To parameterize a curve that follows the graph of a function \(x=f(y)\text{,}\) we choose \(y=t\text{,}\) so we get \(x=f(t)=\sin t\text{.}\) Thus, \(\vec{r}(t) =\sin t\vec{i} +t\vec{j}\) is a vector equation for \(x=\sin y\text{.}\) Since \(y=t\text{,}\) we choose \(0\leq t\leq 2\pi\) to restrict to the appropriate domain.

5.

Find \(f(\vec{r}(t))\) for \(f(x,y)=2x-3y\) and \(\vec{r}(t)=2t\vec{i} +t^2\vec{j}\text{.}\)

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

Hint

What are the \(x\)- and \(y\)-coordinates of \(\vec{r}(t)\text{?}\)

Solution

The \(x\)-coordinate of \(\vec{r}(t)\) is \(2t\text{,}\) and the \(y\)-coordinate is \(t^2\text{,}\) so \(f(\vec{r}(t))=4t-3 t^2\text{.}\)

6.

Find \(f(\vec{r}(t))\) for \(f(x,y)=x^2+y^2\) and \(\vec{r}(t)=\cos t\vec{i} +\sin t\vec{j}\text{.}\)

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

Hint

What are the \(x\)- and \(y\)-coordinates of \(\vec{r}(t)\text{?}\)

Solution

The \(x\)-coordinate of \(\vec{r}(t)\) is \(\cos t\text{,}\) and the \(y\)-coordinate is \(\sin t\text{,}\) so

\begin{equation*} f(\vec{r}(t))=(\cos t)^2+(\sin t)^2=1. \end{equation*}
7.

Find \(\vec{r}\,'(t)\) for \(\vec{r}(t)=2\cos t\vec{i} +3\sin t\vec{j}\text{.}\)

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

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,

\begin{equation*} \vec{r}\,'(t)=(-2\sin t)\vec{i} + (3\cos t)\vec{j}\text{.} \end{equation*}
8.

Find \(\vec{r}\,'(t)\) for \(\vec{r}(t)=(2t^2+1)\vec{i} +te^t\sin t\vec{j}\text{.}\)

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

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,

\begin{equation*} \vec{r}\,'(t)=(4t)\vec{i}+ e^t( \sin t + t \sin t + t \cos t)\vec{j}\text{.} \end{equation*}
9.

Find the value of the integral.

\(\displaystyle \int_{0}^{1} 2x\sqrt{1-x^2} dx=\)

Hint

It might be handy to use integration by substitution.

Solution

We use the substitution \(u=1-x^2\text{.}\) When \(x=0\text{,}\) we have \(u=1\text{,}\) and when \(x=1\text{,}\) we have \(u=0\text{.}\) We also have \(du=-2xdx\text{,}\) so

\begin{equation*} \newcommand{\amp}{\amp }\begin{aligned} \int_{0}^{1} 2x\sqrt{1-x^2} dx \amp =\int_{1}^{0} -\sqrt{u}du\\ \amp =\int_{0}^{1} \sqrt{u}du =\frac{2}{3} u^{\frac{3}{2}}\bigg|_{0}^{1}\\ \amp =\frac{2}{3} \left(1\right)^{\frac{3}{2}}-\frac{2}{3} \left(0\right)^{\frac{3}{2}}\\ \amp =\frac{2}{3}. \end{aligned} \end{equation*}
10.

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

\(\displaystyle \int_{0}^{1} (e^x+x^2) \sqrt{3} ~ dx =\)

Hint

Recall 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{3} dx \amp = \sqrt{3} \int_{0}^{1} e^x+ \sqrt{3}\int_{0}^{1} x^2 dx\\ \amp = \sqrt{3} (e-1) + \sqrt{3}(1/3) \\ \amp = (3 e-2)/\sqrt{3} \end{aligned} \end{equation*}
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