4.A paper cup, which is in the shape of a right circular cone, is 16 cm deep and has a radius of 4 cm. Water is poured into the cup at a constant rate of . 3 2cm / sec

Question

(a) At the instant the depth is 5 cm, what is the rate of change of the height?
(b) At the instant the radius is 3 cm, what is the rate of change of the radius?

Answer 1

Let the radius of the water level be r cm
let the height of the water be h cm
by ratio:
r/h = 4/16 = 1/4
h = 4r or r = h/4

V = (1/3)π r^2 h
= (1/3)π (h^2/16)(h) = (1/48)π r^3
dV/dt = (1/16)π r^2 dr/dt

a) for the given data:
.3 = (1/16)π(25) dr/dt
dr/dt = 16(.3)/(25π) cm/s
= ....

b) in the origininal V = ... equation, replace h to have only V and r
differentiate and sub in your values

Answer 2

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