Q. 91 - Precalculus Enhanced with Graphing Utilities

Question

Alternating Current (ac) Generators The voltage V produced by an ac generator is sinusoidal. As a function of time, the voltage V is

      $V\left(t\right)=V_0\sin \left(2\mathrm{\pi ft}\right)$

where f  is the frequency, the number of complete oscillations (cycles) per second. [In the United States and Canada, f is 60 hertz $\left(Hz\right)$.] The power P delivered to a resistance R at any time t is defined as

      $P\left(t\right)=\frac{{\left[V\right(t\left)\right]}^2}{R}$

(a) Show that $P\left(t\right)=\frac{{V_0}^2}{R}{\sin }^2\left(2\mathrm{\pi ft}\right)$.
(b) The graph of P is shown in the figure. Express P as a sinusoidal function.

(c) Deduce that  ${\sin }^2\left(2\mathrm{\pi ft}\right)=\frac{1}{2}[1-\cos (4\mathrm{\pi ft}\left)\right]$

Step-by-Step Solution

Answer

(a) It is proved $P\left(t\right)=\frac{{V_0}^2}{R}{\sin }^2\left(2\mathrm{\pi ft}\right)$

(b) The sine function whose graph is given is: 

     $P\left(t\right)=\frac{{V_0}^2}{2R}\sin \left(2\mathrm{\pi ft}\right)+\frac{{{\mathrm{V}}_0}^2}{2\mathrm{R}}$

(c)  It is proved that ${\sin }^2\left(2\mathrm{\pi ft}\right)=\frac{1}{2}[1-\cos (4\mathrm{\pi ft}\left)\right]$

Step 1.Given information

Given that $V\left(t\right)=V_0\sin \left(2\mathrm{\pi ft}\right)$ and $P\left(t\right)=\frac{{\left[V\right(t\left)\right]}^2}{R}$ 

Step 2.(a) Show that P ( t ) = V 0 2 R sin 2 ( 2 πft )

$$\begin{gathered} P\left(t\right)=\frac{{\left[V\right(t\left)\right]}^2}{R} \\ =\frac{\left[V_0\sin {\left(2\mathrm{\pi ft}\right)}^2\right]}{R} \\ =\frac{{V_0}^2}{R}{\sin }^2\left(2\mathrm{\pi ft}\right) \end{gathered}$$

Hence,it is proved that  $P\left(t\right)=\frac{{V_0}^2}{R}{\sin }^2\left(2\mathrm{\pi ft}\right)$

Step 3.(b) The graph of P is shown in the figure. Express P as a sinusoidal function.

In the given graph, there is a vertical shift up by $\frac{{V_0}^2}{2R}$ units. If we shift the graph down by $\frac{{V_0}^2}{2R}$ unit, the given graph has characteristics of a sine function as the graph will pass through origin. So, we view the equation as a sine function $y=A\sin \left(\omega x\right)$ with $\left|A\right|=\frac{{V_0}^2}{2R}$ as the curve lies between $-\frac{{V_0}^2}{2R}$ and $\frac{{V_0}^2}{2R}$ on y-axis, and period $T=\frac{1}{t}$ as one cycle in the graph begins at $t=0$ and ends at $t=\frac{1}{f}$. The given sine function will not be negative as power in an ac generator is always positive.
Now, 

             $$\begin{gathered} \omega =\frac{2\mathrm{\pi }}{T} \\ =\frac{2\mathrm{\pi }}{{\displaystyle \frac{1}{f}}} \\ =2\mathrm{\pi f} \end{gathered}$$

After adding the vertical shift value, equation will be of the form $y=A\sin \left(\omega x\right)+c$ where  is the vertical shift. Hence, the sine function whose graph is given is:  $P\left(t\right)=\frac{{V_0}^2}{2R}\sin \left(2\mathrm{\pi ft}\right)+\frac{{{\mathrm{V}}_0}^2}{2\mathrm{R}}$

Step 4.Deduce that sin 2 ( 2 πft ) = 1 2 [ 1 - cos ( 4 πft ) ]

To prove the given expression, we use the Pythagorean identity ${\cos }^{2}x+{\sin }^{2}x=1$ and double angle formula $\cos 2x=1-2{\cos }^{2}x$

$$\begin{gathered} {\sin }^2\left(2\mathrm{\pi ft}\right)=1-{\cos }^2\left(2\mathrm{\pi ft}\right) \\ =\frac{1}{2}(2-2{\cos }^2(2\mathrm{\pi ft}) \\ =\frac{1}{2}[1+(1-2{\cos }^2(2\mathrm{\pi ft}\left)\right)] \\ =\frac{1}{2}[1-\cos \left(4\mathrm{\pi ft}\right)] \end{gathered}$$

Hence,the given expression is proved