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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?

Respuesta :

rodolforodriguezr

Answer:

2 times

Step-by-step explanation:

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

Take the derivative using the chain rule:

[tex]\bf S(x)=e^{sin(x)}\Rightarrow S'(x)=cos(x)e^{sin(x)}[/tex]

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]