Related Rates in a Triangle

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Hi, This is the 5th problem i cant do, maybe im missing somthing. Could some one please help me?
The question is:

The included angle of the two sides of a constant equal length "s" of an isosceles triangle is "θ"
If θ is increasing at a rate of 1/2 radian per minute, find the rate of change of the area when θ=pi/6 The equation is Area=(1/2)s^2sinθ

I cant get the dirivetive, the answer is dA/dt=(1/2)s^2cos(θ)(dθ/dt)

Can some one please explain how to do this?
I got as far at dA/dt
 
You know the derivative of sin is cos. That's all it is.

\(\displaystyle \L\\\frac{d}{d{\theta}}\left[\frac{1}{2}s^{2}sin({\theta})\right]\)

=\(\displaystyle \L\\\frac{1}{2}s^{2}cos({\theta})\frac{d{\theta}}{dt}\)

You have \(\displaystyle {\theta}=\frac{\pi}{6} \;\ and \;\ \frac{d{\theta}}{dt}=\frac{1}{2}\)

Sub them in a solve.
 
Re:

Thank you,

but i got that, what i dont get is why dont i touch the (1/2)s^2? its just left there... dont i have to do something to it too? or becuase its multipled i just leave it?

or am i over thinking it?
 
You're overthinking it. It's a constant. You're letting that s throw you.

Suppose you had the derivative of \(\displaystyle 2sin({\theta})\). That's \(\displaystyle 2cos({\theta})\), is it not?. Same thing here.
 
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