A conical paper cup is 10 cm tall with a radius of 30 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 9 cm?

Question

Ezekiel Alexander · Accepted Answer

Rate #= 1458pi \ cm^3s^-1 #
# \ \ \ \ \ \ \ = 4580.44 \ cm^3s^-1 \ \ (2dp)#

Let us set up the following variables:

# {(R, "Radius of conical cup (cm)",=30 \ cm), (H, "Height of the conical cup (cm)",=10 \ cm), (t, "time", "(s)"), (h, "Height of water in cup at time t","(cm)"), (r, "Radius of water at time t","(cm)" ), (V, "Volume of water at time t", "(cm"^3")") :} #

Our aim is to find #(dV)/dt# when #h=9# and we are given #(dh)/dt=2#

By similar triangles we have

# r:h = R:H #
# :. r/h=30/10 => r = 3h #

The volume of a cone is #1/3pir^2h#, So:

# V = 1/3 pi r^2 h #
# \ \ \ = 1/3 pi (3h)^2 h #
# \ \ \ = 1/3 pi 9h^2 h #
# \ \ \ = 3pi h^3 #

Differentiating wrt #h# we get:

# (dV)/(dh) = 9pih^2 #

And Applying the chain rule we have:

# (dV)/(dt) = (dV)/(dh) * (dh)/(dt) #
# \ \ \ \ \ \ \= 9pih^2 * 2 #
# \ \ \ \ \ \ \= 18pih^2 #

And so when h=9 we have:

# [ (dV)/(dt) ]_(h=9) = 18pi(9^2) = 1458pi \ cm^3s^-1 #

Samantha Adkison · Answer

undefined To find the rate at which water is being poured into the cup when the water level is 9 cm, we can use related rates.

Let $ r $ be the radius of the water surface and $ h $ be the height of the water in the cup. We are given $ \frac{{dh}}{{dt}} = 2 \, \text{cm/sec} $ and we want to find $ \frac{{dV}}{{dt}} $ when $ h = 9 $ cm.

Using similar triangles, $ \frac{r}{h} = \frac{30}{10} $ and $ r = \frac{3}{h} \cdot 9 $.

The volume of water in the cup at height $ h $ is $ V = \frac{1}{3} \pi r^2 h $. Substituting the expression for $ r $, $ V = \frac{1}{3} \pi \left(\frac{3}{h} \cdot 9\right)^2 h $.

Differentiating $ V $ with respect to $ t $ gives $ \frac{dV}{dt} = \pi \left( 2 \cdot 9 - \frac{3}{h^2} \cdot 9^2 \cdot \frac{dh}{dt} \right) $.

When $ h = 9 $ cm, $ r = 3 $ cm, and $ \frac{dV}{dt} = \pi \left( 2 \cdot 9 - \frac{3}{9^2} \cdot 9^2 \cdot 2 \right) $. Simplifying this gives $ \frac{dV}{dt} = 36 \pi $ cubic cm per second.