# 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?

Rate

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

By similar triangles we have

# r:h = R:H #

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

The volume of a cone is

# V = 1/3 pi r^2 h #

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

# \ \ \ = 1/3 pi 9h^2 h #

# \ \ \ = 3pi h^3 #

Differentiating wrt

# (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 #

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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