Chapter 16, Problem 16.42P

(a)

To determine

The value of $A$ from scalar quantity $4\left(7+3\right)$.

Answer to Problem 16.42P

The value of $A$ from scalar quantity $4\left(7+3\right)$ is $40$.

Explanation of Solution

Given info: The scalar quantity $A$ is $4\left(7+3\right)$.

The formula to calculate scalar quantity $A$ is,

$A=a\left(b+c\right)$

Here,

$a$ is the scalar value.

$b$ is the scalar value.

$c$ is the scalar value.

Substitute $4$ for $a$, $7$ for $b$ and $3$ for $c$ in the above expression.

$\begin{array}{c}A=4\left(7+3\right)\\ =40\end{array}$

Conclusion:

Therefore, the value of $A$ from scalar quantity $4\left(7+3\right)$ is $40$.

(b)

To determine

To write: The value of $A,B$ and $C$ in the vector quantity $700\widehat{\text{i}}+\text{3}\text{.00}\widehat{\text{k}}=\text{A}\widehat{\text{i}}+\text{B}\widehat{\text{j}}+\text{C}\widehat{\text{k}}$.

Answer to Problem 16.42P

The value of $A$ is $7$, $B$ is $0$ and $C$ is $3$.

Explanation of Solution

Given info: The scalar quantity $A$ is $4\left(7+3\right)$ and the vector quantity is $700\widehat{\text{i}}+\text{3}\text{.00}\widehat{\text{k}}=\text{A}\widehat{\text{i}}+\text{B}\widehat{\text{j}}+\text{C}\widehat{\text{k}}$.

The given vectors are,

$700\widehat{\text{i}}+3.\text{00}\widehat{\text{k}}$ and $\text{A}\widehat{\text{i}}\text{7+B}\widehat{\text{j}}+\text{C}\widehat{\text{k}}$

The given two vectors are said to be equal if the magnitude of one vector equals the magnitude of other vector and direction in all the three axes are equal.

Compare the respective coefficient of $\widehat{\text{i}}$, $\widehat{\text{j}}$ and $\widehat{\text{k}}$,

$\begin{array}{c}A=700\\ B=0\\ C=3.00\end{array}$

Conclusion:

Therefore, the value of $A$ is $7$, $B$ is $0$ and $C$ is $3$.

(c)

To determine

The proper way to get the answer and convince the student.

Answer to Problem 16.42P

The comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

Explanation of Solution

Given info: The scalar quantity $A$ is $4\left(7+3\right)$ and the vector quantity is $700\widehat{\text{i}}+\text{3}\text{.00}\widehat{\text{k}}=\text{A}\widehat{\text{i}}+\text{B}\widehat{\text{j}}+\text{C}\widehat{\text{k}}$.

The dot product generates a scalar as output. The coefficients of same unit vector are multiplied and added to get the answer.

The unit vectors in three mutually perpendicular axes are $\widehat{\text{i}}$ in $x$ direction $\widehat{\text{j}}$ in $y$ direction and $\widehat{\text{k}}$ in $z$ direction.

The coefficients of $\widehat{\text{i}}$, $\widehat{\text{j}}$ and $\widehat{\text{k}}$ are compared to get the three unknowns.

The coefficient of $\widehat{\text{j}}$ is $0$ in $700\widehat{\text{i}}+\text{3}\text{.00}\widehat{\text{k}}$. So the value of $B$ is $0.$ The coefficient of $\widehat{\text{i}}$ is $700$ in $700\widehat{\text{i}}+3.\text{00}\widehat{\text{k}}$. So the value of $A$ is $700$ and the coefficient of $\widehat{\text{k}}$ is $3.00$ in $700\widehat{\text{i}}+3.\text{00}\widehat{\text{k}}$. So the value of $C$ is $3.00$.

Conclusion:

Therefore, the comparison of the unit vectors on both sides of equation the given equation is the proper way to convince the student how we arrive at the answer.

(d)

To determine

The values of $A$, $B$, $C$, $D$ and $E$ from the functional equality.

Answer to Problem 16.42P

The value of $A$ is $0$, $B$ is $7.00$, $C$ is $3\text{rad}/\text{m}$, $D$ is $4\text{rad}/\text{s}$ and $E$ is $2\text{rad}$.

Explanation of Solution

Given info: The scalar quantity $A$ is $4\left(7+3\right)$ and the vector quantity is $700\widehat{\text{i}}+\text{3}\text{.00}\widehat{\text{k}}=\text{A}\widehat{\text{i}}+\text{B}\widehat{\text{j}}+\text{C}\widehat{\text{k}}$. The functional equality is $A+B\cos \left(Cx+Dt+E\right)=7.00\cos \left(3x+4t+2\right)$.

Compare the given functional equality for the values of $A$, $B$, $C$, $D$ and $E$.

The given functional equality is,

$A+B\cos \left(Cx+Dt+E\right)=7.00\cos \left(3x+4t+2\right)$

Rewrite the above equation as,

$A+B\cos \left(Cx+Dt+E\right)=0+7.00\cos \left(3x+4t+2\right)$

Compare the above expression.

$\begin{array}{c}A=0\\ B=7.00\\ C=3\text{rad}/\text{m}\\ D=4\text{rad}/\text{s}\\ E=2\text{rad}\end{array}$

Conclusion:

Therefore, the value of $A$ is $0$, $B$ is $7.00$, $C$ is $3\text{rad}/\text{m}$ and $D$ is $4\text{rad}/\text{s}$.

(e)

To determine

The way to get the answer.

Answer to Problem 16.42P

The comparison of the given functional equality is helped to get the answer $A$ is $0$, $B$ is $7.00$, $C$ is $3\text{rad}/\text{m}$ and $D$ is $4\text{rad}/\text{s}$.

Explanation of Solution

Given info: The scalar quantity $A$ is $4\left(7+3\right)$ and the vector quantity is $700\widehat{\text{i}}+\text{3}\text{.00}\widehat{\text{k}}=\text{A}\widehat{\text{i}}+\text{B}\widehat{\text{j}}+\text{C}\widehat{\text{k}}$. The functional equality is $A+B\cos \left(Cx+Dt+E\right)=7.00\cos \left(3x+4t+2\right)$.

The given functional equality is,

$A+B\cos \left(Cx+Dt+E\right)=7.00\cos \left(3x+4t+2\right)$

Add $0$ on the $R.H.S$ of the above equation.

$A+B\cos \left(Cx+Dt+E\right)=0+7.00\cos \left(3x+4t+2\right)$

Compare the R.H.S and L.H.S of the above equality,

$\begin{array}{c}A=0\\ B=7.00\\ C=3\text{rad}/\text{m}\\ D=4\text{rad}/\text{s}\\ E=2\text{rad}\end{array}$

Conclusion:

Therefore, the comparison of the given functional equality is helped to get the answer $A$ is $0$, $B$ is $7.00$, $C$ is $3\text{rad}/\text{m}$ and $D$ is $4\text{rad}/\text{s}$.