A conical paper cup is 10 cm tall with a radius of 10 cm. The cup is being filled with water so that the water level rises at a rate of 2 cm/sec. At what rate is water being poured into the cup when the water level is 8 cm?

Question

Christopher Allers · Accepted Answer

# = 32 pi \ cm^3 "/ sec"#

the volume of a cone is #V=( pi r^2 h)/3#

in this case, we can relate #h# and #r# as we know that the slope = #h/r = 10/5 = 2#

we can make the volume a formula in a single variable

#V=( pi (h/2)^2 h)/3 = ( pi h^3)/12#

differentiating wrt time, it follows that

#dot V = (3 pi h^2 dot h)/12= ( pi h^2 dot h)/4#

so to address the question

#dot V = ( pi 8^2 * 2)/4 = 32 pi \ cm^3 "/ sec"#

Isaiah Allen · Answer

undefined We can use related rates to solve this problem. Let $ V $ represent the volume of water in the cup and $ h $ represent the height of the water in the cup. The volume $ V $ of water in a conical cup with radius $ r $ and height $ h $ is given by $ V = \frac{1}{3}\pi r^2 h $. Taking the derivative of both sides with respect to time gives us $ \frac{dV}{dt} = \frac{1}{3}\pi r^2 \frac{dh}{dt} $. We are given that $ \frac{dh}{dt} = 2 $ cm/sec when $ h = 8 $ cm. Plugging in the values $ r = 10 $ cm and $ h = 8 $ cm into the formula, we can solve for $ \frac{dV}{dt} $. Thus, $ \frac{dV}{dt} = \frac{1}{3}\pi (10)^2 (2) = \frac{200}{3}\pi $ cm$^3$/sec. Therefore, the rate at which water is being poured into the cup when the water level is 8 cm is $ \frac{200}{3}\pi $ cm$^3$/sec.