# A conical tank has a circular base with radius 5 ft and height 12 ft. If water is flowing out of the tank at a rate of 3 ft^3/min, how fast is the height of the water changing when the height is 7 ft?

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To find the rate at which the height of the water is changing when the height is 7 ft, we can use related rates and the formula for the volume of a cone.

[ V = \frac{1}{3} \pi r^2 h ]

[ \frac{dV}{dt} = \frac{1}{3} \pi (2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}) ]

Given: [ r = 5 \text{ ft} ] [ h = 7 \text{ ft} ] [ \frac{dV}{dt} = -3 \text{ ft}^3/\text{min} ] [ \frac{dr}{dt} = 0 \text{ ft/min} ]

Solve for ( \frac{dh}{dt} ):

[ -3 = \frac{1}{3} \pi (2(5)(7)(0) + 5^2 \frac{dh}{dt}) ]

[ -3 = \frac{1}{3} \pi (25 \frac{dh}{dt}) ]

[ -9 = 25 \frac{dh}{dt} ]

[ \frac{dh}{dt} = -\frac{9}{25} ]

So, the height of the water is changing at a rate of ( -\frac{9}{25} ) ft/min when the height is 7 ft.

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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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