Chapter 11, Problem 1RE

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

1. a. Given two vectors u and v, it is always true that 2u + v = v + 2u.
2. b. The vector in the direction of u with the length of v equals the vector in the direction of v with the length of u.
3. c. If u ≠ 0 and u + v = 0, then u and v are parallel.
4. d. If r′(t) = 0, then r(t) = 〈a, b, c〉, where a, b, and c are real numbers.
5. e. The parameterized curve r(t) = 〈5 cos t, 12 cos t, 13 sin t〉 has arc length as a parameter.
6. f. The position vector and the principal unit normal are always parallel on a smooth curve.

To determine

The given statement is true or not and give an example or counterexample.

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Given:

“Given two vectors u and v, it is always true that $2u+v=v+2u$.”

Formula used:

Suppose the vectors $u=\langle a,b\rangle \text{ and }v=\langle c,d\rangle$ and the scalar is c. Then,

Vector addition is $u+v=\langle a+c,b+d\rangle$.

Scalar multiplication is $cu=\langle ca,cb\rangle$.

Commutative property $u+v=v+u$.

Calculation:

Suppose $u=\langle a,b\rangle \text{ and }v=\langle c,d\rangle$.

Use vector addition and scalar multiplication to compute the value of $2u+v$.

$\begin{array}{c}2u+v=2\langle a,b\rangle +\langle c,d\rangle \\ =\langle 2a,2b\rangle +\langle c,d\rangle \\ =\langle 2a+c,2b+d\rangle \end{array}$

Thus, the component of the vector $2u+v$ is $\langle 2a+c,2b+d\rangle$ (1)

Use vector addition and scalar multiplication to compute the value of $v+2u$.

$\begin{array}{c}v+2u=\langle c,d\rangle +2\langle a,b\rangle \\ =\langle c,d\rangle +\langle 2a,2b\rangle \\ =\langle c+2a,d+2b\rangle \\ =\langle 2a+c,2b+d\rangle \left[\because \text{vector addition is commutative}\right]\end{array}$

Thus, the component of the vector $v+2u$ is $\langle 2a+c,2b+d\rangle$ (2)

From the equations (1) and (2), it is observed that $2u+v=v+2u$.

Therefore, the given statement is true.

To determine

The given statement is true or not and give an example or counterexample.

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Given:

“The vector in the direction of u with the length of v equals the vector in the direction of v with the length of u.”

Formula used:

Suppose the two vectors are u and v.

The unit vector in the direction of u with the length of v is $\left|v\right|\frac{u}{\left|u\right|}$.

Calculation:

Suppose $u=\langle a,b\rangle \text{ and }v=\langle c,d\rangle$.

Let x be the unit vector in the direction of u with the length of v.

Use the above mentioned formula to compute the vector x.

$\begin{array}{c}x=\left|\langle c,d\rangle \right|\frac{\langle a,b\rangle }{\left|\langle a,b\rangle \right|}\\ =\sqrt{c^2+d^2}\frac{\langle a,b\rangle }{\sqrt{a^2+b^2}}\end{array}$

Thus, the vector x is $\sqrt{c^2+d^2}\frac{\langle a,b\rangle }{\sqrt{a^2+b^2}}$ (1)

Let y be the unit vector in the direction of v with the length of u.

Use the above mentioned formula to compute the vector y.

$\begin{array}{c}y=\left|\langle a,b\rangle \right|\frac{\langle c,d\rangle }{\left|\langle c,d\rangle \right|}\\ =\sqrt{a^2+b^2}\frac{\langle c,d\rangle }{\sqrt{c^2+d^2}}\end{array}$

Thus, the vector y is $\sqrt{a^2+b^2}\frac{\langle c,d\rangle }{\sqrt{c^2+d^2}}$ (2)

From the equations (1) and (2), it is observed that both the vectors are not equal.

Therefore, the given statement is false.

To determine

The given statement is true or not and give an example or counterexample.

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Given:

“If $u\ne 0$ and $u+v=0$, then u and v are parallel.”

Result used:

The vectors u and v are said to be parallel vectors, if one vector is the scalar multiple of the other vector.

Calculation:

Consider $u+v=0$ and simplify.

$\begin{array}{l}u+v=0\\ u=-v\end{array}$

This implies that the vector u is −1 times the vector v. By the result of parallel vectors, the two vector u and v are parallel.

Therefore, the given statement is true.

To determine

The given statement is true or not and give an example or counterexample.

Answer to Problem 1RE

The given statement is true.

Explanation of Solution

Given:

“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.”

Calculation:

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

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

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.

To determine

The given statement is true or not and give an example or counterexample.

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Given:

“The parameterized curve $r\left(t\right)=\langle 5\cos t,12\cos t,13\cos t\rangle$ has arc length as a parameter.”

Formula used:

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.

Calculation:

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

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

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

$\begin{array}{c}\left|v\left(t\right)\right|=\sqrt{{\left(-5\sin t\right)}^2+{\left(-12\sin t\right)}^2+{\left(-13\sin t\right)}^2}\\ =\sqrt{25{\sin }^{2}t+144{\sin }^{2}t+169{\sin }^{2}t}\\ =\sqrt{338{\sin }^{2}t}\end{array}$

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.

To determine

The given statement is true or not and give an example or counterexample.

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Given:

“The position vector and the principal unit normal are always parallel on a smooth curve.”

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 $N\left(t\right)=\frac{T^{\prime }\left(t\right)}{\left|T^{\prime }\left(t\right)\right|}$.

Calculation:

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.