Question

4. An electrical voltage E is given by: E = ( 8 sin50πt + 14 cos50 πt ) volts. Where t is the time in seconds. Determine the maximum value of the voltage

Answer 1

$E = 8 \sin(50\pi t) + 14 \cos(50 \pi t ) .$

The maximum and minimum values correspond to points, where the derivative is equal to 0. Let us take the derivative with respect to t:

$E' = 8\cdot 50\pi \cdot \cos ( 50\pi t) - 14\cdot 50\pi \cdot \sin ( 50\pi t) = 0.$

$8\cdot \cos ( 50\pi t) - 14 \cdot \sin ( 50\pi t) = 0. \\ 4\cdot \cos ( 50\pi t)= 7 \cdot \sin ( 50\pi t).$

$\dfrac47 = \tan (50\pi t) \Rightarrow 50\pi t = \mathrm{atan}\,\dfrac47 + \pi k, k\in\mathbb{Z}$

We should take only those values of argument, where the derivative changes its sign from positive to negative. On an interval $\left[ \mathrm{atan}\,\dfrac47 + 2\pi k, \mathrm{atan}\,\dfrac47 +\pi+ 2\pi k\right]$ lie the points $2\pi k + \dfrac{\pi}{2}$ and $2\pi k + \pi.$

$E'\left(2\pi k + \dfrac{\pi}{2}\right) = 8\cdot 50\pi \cdot \cos \left( 2\pi k + \dfrac{\pi}{2}\right) - 14\cdot 50\pi \cdot \sin \left(2\pi k + \dfrac{\pi}{2}\right) =- 14\cdot 50\pi ,$

$E'\left(2\pi k + {\pi} \right) = 8\cdot 50\pi \cdot \cos \left( 2\pi k + {\pi} \right) - 14\cdot 50\pi \cdot \sin \left(2\pi k + \pi \right) =- 8\cdot 50\pi ,$

we see the derivative is negative here, so on$\left[ \mathrm{atan}\,\dfrac47 + 2\pi k, \mathrm{atan}\,\dfrac47 +\pi+ 2\pi k\right]$ the function decreases, so $50\pi t = \mathrm{atan}\,\dfrac47 + 2\pi k, k \in \mathbb{Z}$ are the points of maximum.

Therefore, $t = \dfrac{1}{50\pi}\mathrm{atan}\,\dfrac47 + \dfrac{1}{25} k, k \in \mathbb{Z}$ .

Now we should put this answer into the formula for E. We know, that $50\pi t = \mathrm{atan}\,\dfrac47 + 2\pi k, k \in \mathbb{Z},$

so we may take k = 0 as a representative point. So, $E = 8\sin\left(\mathrm{atan}\dfrac47\right) + 14\cos\left (\mathrm{atan}\dfrac47\right) \approx 16.12\,\mathrm{V}.$