Question 1163278
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Use the product rule and the chain rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(t)\ =\ 10t\sqrt{4\,-\,t^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dV}{dt}\ =\ 5t\(4\,-\,t^2\)^{-\frac{1}{2}}\(-2t\)\ +\ 10\(4\,-\,t^2\)^{\frac{1}{2}}]


Evaluate *[tex \Large \frac{dV}{dt}\|_{t=1.5}]


If *[tex \Large \frac{dV}{dt}\|_{t=1.5}\ >\ 0] then rising


If *[tex \Large \frac{dV}{dt}\|_{t=1.5}\ <\ 0] then falling


And, of course, if it is equal to zero it is doing neither, it is either at a minimum or a maximum.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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