# A conical cup of radius 5 cm and height 15 cm is leaking water at the rate of 2 cm^3/min. What rate is the level of water decreasing when the water is 3 cm deep?

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To find the rate at which the water level is decreasing when the water is 3 cm deep, we'll use related rates involving similar triangles. Given that the volume of a cone is ( V = \frac{1}{3}\pi r^2h ), where ( r ) is the radius of the base and ( h ) is the height, and that ( \frac{dV}{dt} = -2 ) cm(^3)/min (since the water is leaking), we'll differentiate the volume formula with respect to time, ( t ), to find ( \frac{dh}{dt} ) when ( h = 3 ) cm. After differentiating and substituting the given values, we find ( \frac{dh}{dt} = -\frac{1}{9\pi} ) cm/min. Thus, the water level is decreasing at a rate of ( \frac{1}{9\pi} ) cm/min when the water is 3 cm deep.

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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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