calculus rates problems!?

Question

A cone shaped cofee filter of radius 6cm and depth 10 cm contains water, which drips out through a hole at the bottom at a constant rate of 1.5cm cubic per secind.
a. if the filter starts out full how long does it take to empty?
b. find the volume of water in the filterwhen the depth of the water
is h cm.
c. how fast is the water level falling when the depth is 8 cm?
c.

Answer 1

a. the volume of the filter, in cm^3, is V = pi*r^2*h/3, where r=6 and h=10. So V = 120*pi cm^3. At a rate of 1.5 cm^3/s, it will take 120pi/1.5 = 80pi seconds to empty.

b. When the depth is h=10, the radius is r=6. In general, when the depth is h, the radius is r = h*6/10 = 3h/5. Therefore, the volume is
V =pi*r^2*h/3 = pi*(3h/5)^2*h/3 = pi*9h^2/25*h/3 = pi*3h^3/25 cm^3.

c. You want dh/dt, given that h=8 cm, and dV/dt = -1.5 cm^3/s. To find dh/dt, differentiate V = pi*3h^3/25 with respect to t. You get

dV/dt = pi*9h^2/25 dh/dt

Plug in dV/dt = -1.5 and h=8, and you get dh/dt = -0.0207 cm/s.

Answer 2

a. the volume of the filter, in cm^3, is V = pi*r^2*h/3, where r=6 and h=10. So V = 120*pi cm^3. At a rate of 1.5 cm^3/s, it will take 120pi/1.5 = 80pi seconds to empty.

b. When the depth is h=10, the radius is r=6. In general, when the depth is h, the radius is r = h*6/10 = 3h/5. Therefore, the volume is
V =pi*r^2*h/3 = pi*(3h/5)^2*h/3 = pi*9h^2/25*h/3 = pi*3h^3/25 cm^3.

c. You want dh/dt, given that h=8 cm, and dV/dt = -1.5 cm^3/s. To find dh/dt, differentiate V = pi*3h^3/25 with respect to t. You get

dV/dt = pi*9h^2/25 dh/dt

Plug in dV/dt = -1.5 and h=8, and you get dh/dt = -0.0207 cm/s.