Oil is leaking from an oil well at the bottom of a 300-foot deep lake and forms an oil slick that takes the form of a right circular cone. The radius on the lake's surface is increasing at 10ft/min at the moment the oil slick is 100 feet wide.
a. Find the rate at which the area of slick is increasing at this time. Indicate units.
b. Find the rate at which the volume of the slick is changing and express your answer in the proper units. (The volume V of a cone with radius r and height h is given by V=1/3(pi)(r^2)(h)
Thanks in advance for any help!
a. Find the rate at which the area of slick is increasing at this time. Indicate units.
b. Find the rate at which the volume of the slick is changing and express your answer in the proper units. (The volume V of a cone with radius r and height h is given by V=1/3(pi)(r^2)(h)
Thanks in advance for any help!
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h=300ft r=50ft (100ft wide means dimeter=100) dr/dt=10ft/min.
a) dA/dt= dA/dr x dr/dt A=(pi)xr^2 so dA/dr=2x(pi)xr =100x(pi) as r=50
dA/dt = 100x(pi)x10=1000pi=3141.6 sq.ft/min
b) v=(1/3)(pi)(r^2)(300) = 100(pi)r^2 so that dv/dr=200 x (pi) x r
dv/dt=dv/dr x dr/dt = 200x(pi) x 50x10 =100000pi=314159.3 cubic ft /min.
a) dA/dt= dA/dr x dr/dt A=(pi)xr^2 so dA/dr=2x(pi)xr =100x(pi) as r=50
dA/dt = 100x(pi)x10=1000pi=3141.6 sq.ft/min
b) v=(1/3)(pi)(r^2)(300) = 100(pi)r^2 so that dv/dr=200 x (pi) x r
dv/dt=dv/dr x dr/dt = 200x(pi) x 50x10 =100000pi=314159.3 cubic ft /min.
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a) pi*2500(square feet)
b) 525 square feet per minute
b) 525 square feet per minute