Calculus, $i = 2 \cos (50πt - 0.4)$ Determine the minimum value and time when this minimum occurs.

Question

I have this question about minima and maxima. I have found the derivative of $i = 2 \cos (50πt - 0.4)$

$\frac{di}{dt}$ $=-100π \sin(50πt - 0.4)$

From here I attempt to find the turning points by making the derivative equal zero and solving for it. I have the final equation and steps below.

$0 = -100π \sin(50πt - 0.4)$

$100π = \sin(50πt - 0.4)$

$\sin^{-1}$ $(100π) = 50πt - 0.4$

$\sin^{-1}$ $(2) = t - 0.4$

$\sin^{-1}$ $(2) +0.4 = t$

This can't be right as you are unable to get an inverse sine value over $1$, my calculator simply says error.

Where have I gone wrong in trying to find the turning point of the derivative?

Answer 1

$0 = -100π \sin(50πt - 0.4)\implies \sin(50πt - 0.4)=0 \implies 50πt - 0.4=n\pi$, for $n \in \mathbb{Z}$

Now $ 50πt - 0.4=n\pi\implies t=\frac{n\pi+0.4}{50\pi}$, for $n \in \mathbb{Z}$

Answer 2

You don’t need any calculus at all. The minimum values of a cosine are all $-1$ and they come when the argument is an odd multiple of $\pi$.

So your minimum value is $-2$.

It happens in infinitely many places. One of them is given by $$50\pi t-0.4=\pi$$