Question

related rates... i dont know how to do them at all, help me?
water is being poured into a bowl with radius of 6 inches at a rate of 4in^3/sec.
a)given that the volume of the water in the spherical segment is V=pih^2(R-(h/3)), R=the radius of the sphere. find the rate that the water level is rising when the water is 2 inches deep.
take deriv&plug in values?
b) find an expression for r, the radius of the surface of the spherical segment of water in terms of h. (idk what that means)
c) how fast is the circular area of the segment of water growing (in in^2/sec) when the water is 2inches

Answer 1

(dh/DV) * (dV/dt) = dh/dt

Answer 2

huh?

Answer 3

so far for the derivative i have: 2pi*h(dh/dt)*(R-(h/3))+((dR/dt)...... and stuck