# Higher Derivatives of trigonometric functions

The position of a particle is given by $s = 5 \cos (2t+ (\pi/4))$ at time $t$ . What are the maximum values of the displacement,the velocity,and the acceleration? The answers are displacement: $5$ velocity: $10$ acceleration: $20$

I tried to do it but I am not getting the right answer. Here is my solution:

$s(t) = 5\cos(2t+ \pi/4)$

$s'(t) = -10 \sin(2t + \pi/4)$

$0= -10[\sin2t\cos\pi/4 + \cos2t\sin\pi/4]$

$-\sin2t(\sqrt{2}/2) = \cos2t(\sqrt{2}/2)$

$-1=\tan2t$

$\tan2t = 3\pi/8, 7\pi/8$

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The displacement is $5\cos\left(2t+\frac{\pi}{4}\right)$. Use what you know about the geometry of the cosine curve.

The biggest a cosine ever gets is $1$. That will happen in our case, for example, when $2t+\frac{\pi}{4}=2\pi$. So the biggest displacement gets is $5$.

Instead, you differentiated and set the result equal to $0$. That's OK, we will need the derivative in order to talk about velocity.

The derivative turns out to be $-10\sin\left(2t+\frac{\pi}{4}\right)$. You wanted to find out where this is $0$, and used the formula for the sine of a sum. It is much easier to note that the derivative is $0$ when $2t+\frac{\pi}{4}$ is a multiple of $\pi$. But examining the derivative to find the maximum is not the best procedure here.

To finish the problems, calculate velocity (you already did) and acceleration. Using the simple method I used to find the max displacement, find the max velocity, and the max acceleration.

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Ok.. got that.can you show me the differentiation of this equation, I'm not sure if my formula is right. I got v'(t)= cos 2t(sqrt.2/2)-sin 2t(sqrt.2/2) –  DataLg2 Aug 1 '13 at 2:08
It is right, but it is better to keep things in terms of $2t+\frac{\pi}{4}$. –  André Nicolas Aug 1 '13 at 2:15
I got v'(t)=-20cos(2t+4) –  DataLg2 Aug 1 '13 at 2:17
Well, you mean $-20\cos(2t+\pi/4)$. And the smallest $\cos$ gets is $-1$, so biggest acceleration gets is $20$. –  André Nicolas Aug 1 '13 at 2:22
thank you very much! –  DataLg2 Aug 1 '13 at 2:46