Posted by Joel on October 30, 2002 at 01:15:50:
In Reply to: Related Rates Problems posted by C. Slayden on October 29, 2002 at 18:06:30:
: 4.The formula for a volume of a cone is v=(1/3)(pi)r^2(h). Find the rate of change of the volume if dr/dt is 2 inches per minute and h=3r when r=6 inches.
: My work. Replaced h for 3r in the formula and differentiated. Recieved dv/dt=3(pi)dr/dt(r^2+2r) plugged the numbers in again and recieved 288(pi)
Looks like you started out right, but I don't see where you got (r^2+2r).
v = (1/3)pi*r^2*h and you're given that h=3r so:
v = (1/3)pi*r^2*3r
v = pi*r^3
dv/dr = 3pi*r^2
dv/dt = 3pi*r^2*(dr/dt)
so with r given as 6 in and dr/dt 2 in/min
dv/dt = 3pi*(36 sq in)* (2 in/min) = 216 cu in/min
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