Related Rates: Finding the Area Change of a Snowball with Changing Radius

hunter55
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Homework Statement


a snowball is rolling donw a hill, its radius is changing at a rate of 2 cm/min. what is the rate of change of the area, when the radius is 8 cm?


Homework Equations



da/dt= 2pir2dr/dt
dat/dt=2pi(8)(2)
32pi ?
is that right


The Attempt at a Solution



32pi
 
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What formula are you using for A? If you are thinking of the snowball as a disk and A=pi*r^2, then yes, dA/dt=pi*2*r*dr/dt. You might want to put units on your answer.
 
As Dick suggested, you are using the wrong formula (because a snowball is NOT a disk). The "area'' referred to here is the SURFACE area of a sphere.
Surface area of a sphere is 4\pi r^2.

(And, of course, be sure to put units on your answer.)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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