calc help plz :)?

Question

A conical icicle metls at a rate of 1.2 cm^3/h. At 10:00 A.M, the icicle is 25 cm long and 4 cm in diameter at its widest point. THe icicle keeps the same proportions as it melts. Determine the rate at which its length is decreasing at 10:00 A.M.

Answer 1

V=(1/3)Pi*r^2*h [Volume of a cone]
dV/dt = -1.2 [Given]
Find dh/dt when r = 2 and h = 25. [what you are asked to find]
By similar triangles: r/h = 2/25 so r = (2/25)h
Substitute into formula for V:
V = (1/3)Pi*(2/25)^2*h^3
Differentiate with respect to time, t, using the chain rule:
dV/dt = Pi*(2/25)^2h^2*(dh/dt)
Solve for dh/dt:
dh/dt = (dV/dt)(25/2)^2/(Pi*h^2)
Plug in dV/dt = -1.2 and h = 25:
dh/dt = (-1.2)*(625/4)/(Pi*625) = -0.3*Pi cm/hr

Answer 2

Since the volume of the original icicle contained 1/2 x Pi x r**2 x h, or 157 cm3, the ratio of the height to radius is 25/2, or 12.5:1. Since the rate of the change in volume is 1.2 cm3/hr, and the only variables are height and radius, while maintaining same proportions, the following will hold true:
1.2 cm3 = 1/2 Pi x d(r**2 x h), where h:r = 12.5 (25:2x2).

Reducing, we get (1.2 x 2)/Pi = .76397419, which will equal the change in r**2 x h. Since h:r = 12.5, it can be rewritten: r = h/12.5.

Substituting we get (h/12.5)**2 x h = .76397419, which becomes (h**3)/156.25 = .76397419
after further reduction becomes: h**3 = 119.37095, from which we get h = 4.9237. and r=4.9237/12.5 or r=.3939

Checking the answer, we get
1/2 x Pi x .3939**2 cm2 x .4.9237 cm = 1.2 cm3
so the change in height = 4.9237, or approx 5 cm/hr.