Question

Please help...super confused on this for calc.

A particle moves along the x-axis with position function s(t) = e^sin(x). How many times in the interval [0, 2π] is the velocity equal to 0?

Answer 1

2 times

Explanation:

If S(x) is the position, the derivative S'(x) is the velocity.

Take the derivative using the chain rule:

`\bf S(x)=e^(sin(x))\Rightarrow S'(x)=cos(x)e^(sin(x))`

Now, if the velocity = 0 then S'(x) = 0, and cos(x) must be 0 because the exponential is always different to 0.

In the interval [0, 2π] cos(x) = 0 when x= π/2 (90°) or x = 3π/2 (270°)

So the velocity equals 0 only 2 times in the interval [0, 2π]