Chapter 9, Problem 1RE

To determine

To calculate: The sine function that model the curve.

Answer to Problem 1RE

Solution:

The sine function of the curve is $\underset{\_}{f\left(x\right)=2\sin x+1}$.

Explanation of Solution

Given Information:

The provided curve is:

Formula Used:

The value of Amplitude in the sine function is the height of peak above the baseline and named it as A.

Then the vertical offset is the height of the baseline and it is named as C.

The angular frequency is the number of cycles in every interval of length $2\pi$ denoted by $\omega$.

The period or wavelength is the length of each cycle which is given by $P=\frac{2\pi }{\omega }$.

Then $\alpha$ which is the phase shift.

The sine function is given by $f\left(x\right)=A\sin \left[\omega \left(x-\alpha \right)\right]+C$.

Calculation:

Consider the provided curve.

Determine the value of Amplitude which is the height of peak above the baseline.

$A=2$

Then determine the vertical offset which is the height of the baseline.

$C=h-A$

Where h is the total height of the peak and A is Amplitude.

$\begin{array}{l}C=3-2\\ C=1\end{array}$

Then determine the period or wavelength which is the length of each cycle.

$P=2\pi$

Then determine the angular frequency which is the number of cycles in every interval of length $2\pi$.

$\begin{array}{l}\omega =\frac{2\pi }{P}\\ \omega =\frac{2\pi }{2\pi }\\ \omega =1\end{array}$

Then determine the phase shift which is the distance to which the function is shifted horizontally.

$\alpha =0$

Then substitute the values in the sine function given by $f\left(x\right)=A\sin \left[\omega \left(x-\alpha \right)\right]+C$.

$\begin{array}{l}f\left(x\right)=A\sin \left[\omega \left(x-\alpha \right)\right]+C\\ f\left(x\right)=2\sin \left[1\left(x-0\right)\right]+1\\ f\left(x\right)=2\sin x+1\end{array}$

Hence, the sine function of the curve is $f\left(x\right)=2\sin x+1$.