@thesmartone

Question

@thesmartone

sleepyjess · Answer

Just need to know what equation to use on this one then I think I can finish it...

If the surface area of a sphere is shrinking at the constant rate of 0.1cm^2/h, find the rate of change of its radius when the radius is 20/$\pi$ cm.

thesmartone · Answer

SA of a sphere = 4$\pi r^2$

sleepyjess · Answer

Would dS/dt be the .1?

thesmartone · Answer

Yes and we're solving for dr/dt

sleepyjess · Answer

Derivative would be dS/dt = $4\pi (2r) dr/dt$. 


.1cm^2/hr = $4\pi 2(20/\pi)dr/dt$?

sleepyjess · Answer

Would the 2 * 20/pi turn in to 40/pi?

thesmartone · Answer

yeah, and then multiplying it by the 4 pi, you're left with 160 dr/dt

sleepyjess · Answer

Gotcha, cuz the pi cancels... so that would make it .1/160 cm^2/hr = dr/dt?

thesmartone · Answer

you forget that the radius has units of cm

and so when you divide by 160 cm on both sides you get

0.1/160 cm/hr

remember, the units need to make sense

can radius have units of cm?
if the radius is changing with respect to time, it has to have the radius units and a time units

sleepyjess · Answer

bleh....