30ft^3 of chemicals were spilled into the river and caused a circular slick whose area is expanding while its thickness is decreasing to form. If the radius of the slick expands at the rate of 1 foot per hour, how fast is the thickness of the slick decreasing when the area is 100ft^3?
i let dh/dt be the rate of change of the thickness and dr/dt be the rate of change of the radius.
I have V = πr^2h and A = πr^2,
i did 100 = πr^2 so r = 10/sqrt(π) when area = 100ft^3. i also did 30 = π(10/sqrt(π)^2 (h) so h = 3/10 when area = 100ft^3.
i took dv/dt = π(r^2 h' + h(2r)r'). but i don't know what dv/dt is. i used dA/dt = π(2r)r' to find dA/dt = 20sqrt(π) and i did V = Ah so therefore dV/dt = Ah' + hA' and when i solved for dv/dt i get 100(dh/dt) + 6sqrt(π). when i plug this back in to dv/dt = π(r^2 h' + h(2r)r') i just got dh/dt = dh/dt which got me nowhere. how do i solve this problem?