Derivatives of Trigonometric Functions Help

Question

The voltage of a power supply is given by the equation V(t)=100sin(4πt)+40. State the maximum voltage and the times when it occurs.

(I'm fairly confident that I have the derivative correct: V'(t)=400πcos(4πt), and I know I need to set V'(t) to equal 0 and solve for t but I am not sure how exactly the process is done for that, I keep getting stuck, thanks).

Answer 1

You don't need Calculus to be able to solve this problem. See the below:

The voltage of the power supply is given by the equation:

V(t) = 100sin⁡(4πt) + 40

Finding the Maximum Voltage

The maximum value of the sine function, sin⁡(x), is 1. Therefore, the maximum value of 100sin⁡(4πt) is:

100 × 1 = 100

Thus, the maximum voltage V(t) is:

V_max= 100 + 40 = 140V

Finding the Times When Maximum Voltage Occurs

The sine function sin⁡(4πt) reaches its maximum value of 1 at:

4πt = (π/2) + 2kπ for integer k

Solving for t:

t = 1/8 ​+ k/2​

Thus, the times when the maximum voltage occurs are:

t = 1/8 ​+ k/2​ for integer k

If you do it the Calculus way, you'll set the derivative function (your derivative is correct) equal to zero to find critical points (solve for t), and you'll get:

t = 1/8 + k/4 for integer k

But remember, critical points are where a maximum or a minimum may occur, so you need to evaluate the original sine function at the critical points to check if a maximum or a minimum occurred. You'll find that maximums only occur at t = 1/8 + k/2 for integer k (when the sine function equals 1).

Hope this was helpful.

Answer 2

The sine function has a maximum value of 1 which it obtains when its argument is exactly π/2 (refer to the unit circle). This also is the case when the cosine function obtains one of its two roots. When sinθ = 1, V(t) = 140. To find the time at which this occurs, you need to solve the equation θ = 4πt = π/2, which gives t = 1/8.