1. anonymous

Yes that would be the case if pi was in the numerator, but I think you said pi was in the denominator though?

2. anonymous

$f(t)=20[1-\cos^2(\frac{ t }{ 1000\pi })$ like that ^

3. anonymous

$\frac{ \sin(\frac{ x }{ 500\pi }) }{ 50\pi }$ is this your answer?

4. anonymous

$f(t) = 20 - 20\cos ^{2}(t/(1000\pi))$ $f'(t) = 0 - 20 * 2\cos (t/(1000\pi)) * -\sin (t/(1000\pi)) * (1/(1000\pi))$ $f`(t) = 20/1000\pi * 2\cos(t/1000\pi)*\sin(t/1000\pi)$ Use identity sin (2x) = 2sinxcosx $f'(t) = 1/50\pi * \sin (2*t/1000\pi)$

5. anonymous

that acually looks right to me. ^

6. anonymous

You can also check your answers by punching the equations into wolframalpha.com

7. anonymous

beautiful, thank you.

8. anonymous

No problem, good luck!