# 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?

we can make the volume a formula in a single variable

differentiating wrt time, it follows that

so to address the question

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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.

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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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