Chapter 14, Problem 1RE

a.

To determine

Whether the statement “If $r\left(t\right)=\langle \cos t,e^t,t\rangle +C$ and $r\left(0\right)=\langle 0,0,0\rangle$, then $C=\langle 0,0,0\rangle$” is true or not.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

The given vector valued function is $r\left(t\right)=\langle \cos t,e^t,t\rangle +C$.

Substitute $t=0$ in the vector as follows.

$\begin{array}{l}r\left(0\right)=\langle \cos \left(0\right),e^{\left(0\right)},\left(0\right)\rangle +C\\ \langle 0,0,0\rangle =\langle 1,1,0\rangle +C\\ C=\langle 0,0,0\rangle -\langle 1,1,0\rangle \\ =\langle -1,-1,0\rangle \end{array}$

Thus, the constant vector is $C=\langle -1,-1,0\rangle$.

Hence the statement if $r\left(t\right)=\langle \cos t,e^t,t\rangle +C$ and $r\left(0\right)=\langle 0,0,0\rangle$, then $C=\langle 0,0,0\rangle$ is false.

b.

To determine

Whether the statement “The curvature of a circle of radius 5 is $\kappa =\frac{1}{5}$” is true or not.

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Definition used:

“The curvature of a curve at a point can be visualized in terms of a circle of curvature, which is a circle of radius R that is tangent to the curve at that point.

The curvature at the point is $\kappa =\frac{1}{R}$”

Description:

The given circle has the radius of 5.

By the above definition used, the curvature of a circle is reciprocal of the radius of the circle.

Thus, the curvature of the circle is $\kappa =\frac{1}{5}$.

Hence, the statement “the curvature of a circle of radius 5 is $\kappa =\frac{1}{5}$” is true.

c.

To determine

Whether the statement “The graph of the curve $r\left(t\right)=\langle 3\cos t,0,6\sin t\rangle$ is an ellipse in the xz-plane” is or not.

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Definition used:

One point to a curve corresponds to a single vector $r=\langle x,y,z\rangle$ where $x=x_0+at,y=y_0+bt\text{ and }z=z_0+ct$.

The entire curve can be represented by a vector-valued function $r\left(t\right)=\langle x\left(t\right),y\left(t\right),z\left(t\right)\rangle$.

Calculation:

The given graph is $r\left(t\right)=\langle 3\cos t,0,6\sin t\rangle$.

The values are $\langle x\left(t\right),y\left(t\right),z\left(t\right)\rangle =\langle 3\cos t,0,6\sin t\rangle$.

Here, $\langle 3\cos t,6\sin t\rangle$ can represent an ellipse in x and z axis.

That is $\frac{x^2}{3^2}+\frac{z^2}{6^2}=1$.

Therefore, the statement “the graph of the curve $r\left(t\right)=\langle 3\cos t,0,6\sin t\rangle$ is an ellipse in the xz-plane” is true.

d.

To determine

Whether the given statement “If $r^{\prime }\left(t\right)=0,\text{ then }r\left(t\right)=\langle a,b,c\rangle$, where a, b, and c are real numbers.” is true or not.

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Consider $r^{\prime }\left(t\right)=0$ and integrate with respect to t.

$\begin{array}{c}r\left(t\right)={\displaystyle \int r^{\prime }\left(t\right)dt}\\ ={\displaystyle \int \langle 0,0,0\rangle dt}\\ =\langle 0,0,0\rangle +C\\ =\langle 0,0,0\rangle +\langle a,b,c\rangle \left(\because C=\langle a,b,c\rangle \right)\\ =\langle a,b,c\rangle \end{array}$

Thus, the vector $r\left(t\right)$ is $\langle a,b,c\rangle$ where a, b, and c are real numbers.

Therefore, the given statement is true.

e.

To determine

Whether the statement “The parameterized curve $r\left(t\right)=\langle 5\cos t,12\cos t,13\sin t\rangle$ has arc length as a parameter.” is true or not.

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Suppose $r\left(t\right)$ is a smooth curve. If $\left|v\left(t\right)\right|=1$, then the curve uses arc length as a parameter.

Differentiate $r\left(t\right)$ to compute $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=\langle {\left(5\cos t\right)}^{\prime },{\left(12\cos t\right)}^{\prime },{\left(13\sin t\right)}^{\prime }\rangle \\ =\langle -5\sin t,-12\sin t,-13\cos t\rangle \end{array}$

Compute $\left|v\left(t\right)\right|$.

$\begin{array}{c}\left|v\left(t\right)\right|=\sqrt{{\left(-5\sin t\right)}^2+{\left(-12\sin t\right)}^2+{\left(-13\cos t\right)}^2}\\ =\sqrt{25{\sin }^{2}t+144{\sin }^{2}t+169{\cos }^{2}t}\\ =\sqrt{169\left({\sin }^{2}t+{\cos }^{2}t\right)}\\ =\sqrt{169}\\ =13\end{array}$

Since $\left|v\left(t\right)\right|\ne 1$, the curve does not use arc length as a parameter.

Therefore, the given statement is false.

f.

To determine

Whether the statement “The position vector and the principal unit normal are always parallel on a smooth curve.” is true or not.

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Formula used:

Suppose r is a smooth parameterized curve and s is the arc length.

The unit tangent vector T is $\frac{r^{\prime }}{\left|r^{\prime }\right|}$.

The principal unit normal vector is $N\left(t\right)=\frac{T^{\prime }\left(t\right)}{\left|T^{\prime }\left(t\right)\right|}$.

Counter example:

Consider $r\left(t\right)=\langle 4\sin t,4\cos t,10t\rangle$.

Differentiate $r\left(t\right)$ to compute $r^{\prime }\left(t\right)$.

$\begin{array}{c}{r}^{\prime }\left(t\right)=\langle {\left(4\sin t\right)}^{\prime },{\left(4\cos t\right)}^{\prime },{\left(10t\right)}^{\prime }\rangle \\ =\langle 4\cos t,-4\sin t,10\rangle \end{array}$

Use magnitude formula to obtain the value of $\left|r^{\prime }\left(t\right)\right|$.

$\begin{array}{c}\left|r^{\prime }\left(t\right)\right|=\left|\langle 4\cos t,-4\sin t,10\rangle \right|\\ =\sqrt{{\left(4\cos t\right)}^2+{\left(-4\sin t\right)}^2+{10}^2}\\ =\sqrt{16{\cos }^{2}t+16{\sin }^{2}t+100}\\ =\sqrt{16\left({\cos }^{2}t+{\sin }^{2}t\right)+100}\end{array}$

On further simplification,

$\begin{array}{c}\left|r^{\prime }\left(t\right)\right|=\sqrt{16\left(1\right)+100}\\ =\sqrt{116}\\ =2\sqrt{29}\end{array}$

Use unit tangent formula to compute $T\left(t\right)$.

$\begin{array}{c}T\left(t\right)=\frac{\langle 4\cos t,-4\sin t,10\rangle }{2\sqrt{29}}\\ =\frac{1}{\sqrt{29}}\langle \frac{4\cos t}{2},\frac{-4\sin t}{2},\frac{10}{2}\rangle \\ =\frac{1}{\sqrt{29}}\langle 2\cos t,-2\sin t,5\rangle \end{array}$

Thus, the unit tangent vector $T\left(t\right)$ is $\frac{1}{\sqrt{29}}\langle 2\cos t,-2\sin t,5\rangle$.

Differentiate $T\left(t\right)$ to compute $T^{\prime }\left(t\right)$.

$\begin{array}{c}{T}^{\prime }\left(t\right)=\frac{1}{\sqrt{29}}\langle {\left(2\cos t\right)}^{\prime },{\left(-2\sin t\right)}^{\prime },5^{\prime }\rangle \\ =\frac{1}{\sqrt{29}}\langle -2\sin t,-2\cos t,0\rangle \end{array}$

Use magnitude formula to obtain the value of $\left|T^{\prime }\left(t\right)\right|$.

$\begin{array}{c}\left|T^{\prime }\left(t\right)\right|=\left|\frac{1}{\sqrt{29}}\langle -2\sin t,-2\cos t,0\rangle \right|\\ =\frac{1}{\sqrt{29}}\sqrt{{\left(-2\sin t\right)}^2+{\left(-2\cos t\right)}^2+0^2}\\ =\frac{1}{\sqrt{29}}\sqrt{4{\sin }^{2}t+4{\cos }^{2}t}\\ =\frac{1}{\sqrt{29}}\sqrt{4\left({\sin }^{2}t+{\cos }^{2}t\right)}\end{array}$

On further simplification,

$\begin{array}{c}\left|T^{\prime }\left(t\right)\right|=\frac{1}{\sqrt{29}}\sqrt{4\left(1\right)}\\ =\frac{2}{\sqrt{29}}\end{array}$

Use principal unit normal formula to compute the value of $N\left(t\right)$.

$\begin{array}{c}N\left(t\right)=\frac{\frac{1}{\sqrt{29}}\langle -2\sin t,-2\cos t,0\rangle }{\frac{2}{\sqrt{29}}}\\ =\frac{1}{\sqrt{29}}\left(\frac{\sqrt{29}}{2}\right)\langle -2\sin t,-2\cos t,0\rangle \\ =\frac{1}{2}\langle -2\sin t,-2\cos t,0\rangle \\ =\langle -\frac{\sin t}{2},-\frac{\cos t}{2},\frac{0}{2}\rangle \end{array}$

         $=\langle -\sin t,-\cos t,0\rangle$

Thus, the principal unit normal vector $N\left(t\right)$ for the curve $r\left(t\right)$ is $\langle -\sin t,-\cos t,0\rangle$.

It is observed that the position vector and the principal unit normal vector are not equal.

Therefore, the given statement is false.