Related rate???? i hate them!?

Question

Container has shape of an open right circular cone. Height of container is 10 cm and diameter of top opening is 10 cm. water is evaporating so its depth, h, is changing at constant rate of -3/10 cm/hr.
a. find the volume V of the container when h=5 cm. ((Volume of cone = 1/3pie*r^2*h))
b. Find rate of change in the volume of the water with respect to time when h = 5 cm.
c. show that rate of change of volume of water due to evaporation is directly proportional to the exposed surface area of the water. what is the constnat of proportionality??
i did the first one, but i am not sure if it is right. i got (31.25*pie)/3
is this right?
also, part b, i took the derivative of...the volume equation?
and then i put in -3/10 cm for dh/dt, and 2.25 cm for r, but then i dont have dr/dt?
im confused. What is the equation i should be using for this part?
i am using dV/dt = 1/3*pie*2r*dr/dt*dh/dt
is this right...i dont think it is?
ANY help is greatly appreciated!! THANKS!

the DIAMETER of the base is 10 cm.

Answer 1

V = (1/3)π(r^2)h, but by similar triangles r/h = R/H
r = h(R/H)
V = (1/3)π(R/H)^2(h^3)
dV/dt = (dV/dh)(dh/dt) = π(R/H)^2h^2dh/dt

a)
V(5) = (1/3)π(5/10)^2(5^3) = 125π/12
V(5) = 32.725 cc

b) h = 5 cm
dV/dt = π(5/10)^2((-3/10) cm/hr)(5^2 cm^2)
dV/dt = -π(3 cm/hr)(25 cm^2)/40
dV/dt = -5.890 cc/hr

b) A = πr^2 = π(hR/H)^2
dV/dt = π(hR/H)^2dh/dt
π(hR/H)^2 = A = (dV/dt)/(dh/dt)
dV/dt = A(dh/dt)
dh/dt == (-3/10)
dV/dt = (-3/10)A

Answer 2

You're missing the size of the base...That's a key piece of data.
=============================================
Ooh...It's an inverted cone...I thought it was a cone with the top cut off
The other thing you know is that r = 0.5 h.
so the volume of the cone is (1/12)pi*h^3.
dV/dt = 1/4 pi h^2 dh/dt = -1.875 pi cm^3/hr
so since r=0.5h or 2r = h, dV/dt = pi r^2 dh/dt
pi r^2 = area of the the top of the cone.

Answer 3

I'm in AP Calc BC with a 102% so I'm qualified to answer this.

First, list what you know.

dh/dt = -3/10 cm/hr

h = 10 r= 5 so r = (1/2)h

V = (1/3)(pi)rrh

A] V=1/3(pi)(h/2)(h/2)h = (pi/12)(5*5*5) = 32.725 cubic cm

B] dv/dt = (pi/12)(3)hh*dh/dt --- dv/dt = (pi/4)(25)(-3/10)

dv/dt = -5.89 cubic cm per hour

C] A = (pi)rr V = 1/3(pi)rr(h) or because h = 2r:

V = 2/3(pi)rrr so dv/dt = 2*(pi)rr* dr/dt do you see in the ** the area equation? that's proof of the direct relationship!

Anything else? Add details to your question.