Chapter 7.2, Problem 68E

(a)

To determine

To graph: The function $y=f\left(t\right)$, where $f\left(t\right)=f_1\left(t\right)+f_2\left(t\right)$ given by $f_1\left(t\right)=5\sin t$ and $f_2\left(t\right)=5\cos t$ and observe graph has the form of sine curve $y=k\sin \left(t+\varphi \right)$.

Explanation of Solution

Given:

The functions are $f_1\left(t\right)=5\sin t$, $f_2\left(t\right)=5\cos t$ and $f\left(t\right)=f_1\left(t\right)+f_2\left(t\right)$.

Formula used:

Pythagorean identity: ${\cos }^{2}x+{\sin }^{2}x=1$.

The addition formula of cosine function is $\cos \left(a+b\right)=\cos a\cos b-\sin a\sin b$.

Calculation:

Substitute the value $f_1\left(t\right)=5\sin t$ and $f_2\left(t\right)=5\cos t$ in $f\left(t\right)=f_1\left(t\right)+f_2\left(t\right)$.

$\begin{array}{c}f\left(t\right)=f_1\left(t\right)+f_2\left(t\right)\\ =5\sin t+5\cos t\\ =5\left(\sin t+\cos t\right)\end{array}$

Use online graphing calculator to graph the function $f\left(t\right)=5\left(\sin t+\cos t\right)$ as shown in the Figure 1.

From Figure 1, it is observed that $f\left(t\right)=5\left(\sin t+\cos t\right)$ has the form of a sine curve $y=k\sin \left(t+\varphi \right)$.

(b)

To determine

To find: The values for $k$ and $\varphi$ for which the given function $y=f\left(t\right)$ satisfies the form of sine curve $y=k\sin \left(t+\varphi \right)$.

Answer to Problem 68E

The given function $y=f\left(t\right)$ satisfies the form of sine curve $y=k\sin \left(t+\varphi \right)$ for the values $k=5\sqrt{2}$ and $\varphi =\frac{\pi }{4}$.

Explanation of Solution

Formula used:

The addition formula of sine function is $\sin \left(a+b\right)=\sin a\cos b+\cos a\sin b$.

Calculation:

Set $y=f\left(t\right)=k\sin \left(t+\varphi \right)$.

Thus, $5\left(\sin t+\cos t\right)=k\sin \left(t+\varphi \right)$ (1)

Use the addition formula for sine function and expand $\sin \left(t+\varphi \right)$ as $\sin \left(t+\varphi \right)=\sin t\cos \varphi +\cos t\sin \varphi$.

Substitute $\sin \left(t+\varphi \right)=\sin t\cos \varphi +\cos t\sin \varphi$ in equation (1) and obtain that $k\sin \left(t+\varphi \right)=k\sin t\cos \varphi +k\cos t\sin \varphi$.

Thus, equation (1) can be rewritten as $5\sin t+5\cos t=k\sin t\cos \varphi +k\cos t\sin \varphi$.

Obtain the value of $k\cos \varphi$ as follows.

$\begin{array}{l}5\sin t=k\sin t\cos \varphi \\ k\cos \varphi =5\end{array}$

Similarly obtain the value of $k\sin \varphi$ as follows.

$\begin{array}{l}5\cos t=k\cos t\sin \varphi \\ k\sin \varphi =5\end{array}$

Use the Pythagorean identity and obtain ${\cos }^2\varphi +{\sin }^2\varphi =1$.

Multiply both side by $k^2$ and obtain the following.

$\begin{array}{l}{k}^2{\cos }^2\varphi +k^2{\sin }^2\varphi =k^2\\ {\left(k\cos \varphi \right)}^2+{\left(k\sin \varphi \right)}^2=k^2\end{array}$

Substitute the value $k\cos \varphi =5$ and $k\sin \varphi =5$ in ${\left(k\cos \varphi \right)}^2+{\left(k\sin \varphi \right)}^2=k^2$.

$\begin{array}{l}{\left(k\cos \varphi \right)}^2+{\left(k\sin \varphi \right)}^2=k^2\\ {\left(5\right)}^2+{\left(5\right)}^2=k^2\\ k^2=50\\ k=5\sqrt{2}\end{array}$

Substitute the value $k=5\sqrt{2}$ in $k\cos \varphi =5$ and $k\sin \varphi =5$.

Thus, $\cos \varphi =\frac{5}{5\sqrt{2}}=\frac{1}{\sqrt{2}}$ and $\sin \varphi =\frac{5}{5\sqrt{2}}=\frac{1}{\sqrt{2}}$.

Since, $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ and $\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$, the value of $\varphi$ is $\frac{\pi }{4}$

Hence, the given function $y=f\left(t\right)$ satisfies the form of sine curve $y=k\sin \left(t+\varphi \right)$ for the values $k=5\sqrt{2}$ and $\varphi =\frac{\pi }{4}$.