1. ## Chain Rule

Let s = cos (theta). Evaluate ds/dt when theta= $\displaystyle \frac {3pi}{2}$ and dtheta/dt= 5.

I have no clue what they want me to do.

2. Originally Posted by Truthbetold
Let s = cos (theta). Evaluate ds/dt when theta= $\displaystyle \frac {3pi}{2}$ and dtheta/dt= 5.
Hello,

you have:

$\displaystyle s=\cos(\theta)$ . Now derivate wrt t. Use chain rule!:

$\displaystyle \frac{ds}{dt}=-\sin(\theta) \cdot \frac{d \theta}{dt}$

Plug in the values you know:

$\displaystyle \frac{ds}{dt}=-\sin\left(\frac {3\pi}{2} \right) \cdot 5$

Since $\displaystyle -\sin\left(\frac {3\pi}{2} \right) = 1$ the result is 5.

3. Originally Posted by earboth
Hello,

you have:

$\displaystyle s=\cos(\theta)$ . Now derivate wrt t. Use chain rule!:

$\displaystyle \frac{ds}{dt}=-\sin(\theta) \cdot \frac{d \theta}{dt}$

Plug in the values you know:

$\displaystyle \frac{ds}{dt}=-\sin\left(\frac {3\pi}{2} \right) \cdot 5$

Since $\displaystyle -\sin\left(\frac {3\pi}{2} \right) = 1$ the result is 5.
Ooooooh! I see.
It threw me off that the derivative of 3pi/2 equaled 5.
Makes sense now.