The base of a conical machine part is being milled such that the height is decreasing at the rate of 0.050cm/min. If the part originally had a radius of 1.0 cm and a height of 3.0 cm, how fast is the volume changing when the height is 2.8 cm ?
Related rate calculus problem, please help!?
Step 1: Define radius in terms of height.
r = 1 cm
h = 3 cm
r = 1/3 h
Step 2: Define volume in terms of height, since this assumes it%26#039;s a right circular cone:
V = 1/3 锜簉^2h (volume of right circular cone)
V = 1/3 锜?1/3h)^2h
V = 1/3 锜?* 1/9 h^2 * h
V = 1/27 锜篽^3
Step 3: Take the derivative to find the rate of change of volume.
V = 1/27 锜篽^3
dV/dh = 1/27 * 3 * 锜?* h^2
dV/dh = 锜篽^2/9
Solve for dV given dh:
dV = 锜篽^2/9 dh
dV = 锜?2.8)^2/9 * .05
dV = 0.137 cm^3/min (solution!)
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