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Question 4

Page history last edited by PBworks 16 years, 9 months ago

Question 4

 

You MAY NOT use a calculator to solve this question.

 

Water is draining at the rate of 48π ft3/minute from the vertex at the bottom of a conical tank whose diameter at its base is 40 feet and whose height is 60 feet.

 

(a) Find an expression for the volume of water in the tank in terms of its radius at the surface of the water.

 

(b) At what rate is the radius of the water in the tank shrinking when the radius is 16 feet?

 

(c) How fast is the height of the water in the tank dropping at the instant that the radius is 16 feet?

 

Solution

 

 

 

 

a) The formula for the volume of a cone is V=(1/3)πr²h

To find an expression that's solely in terms of radius, we have to find an

expression for height in terms of radius. You can use similar triangles to do that.             

 

 r 

20    60

 

h  = 3r

 

now you can substitute that back into the volume formula

 

v(r)=(1/3)πr²(3r)

v(r)=πr³

 

b) Now differentiating the answer from part a will give us an equation we can use to answer part b. Use implicit differentiation.

 

v(r)  =  πr³

 

dV   =  π(3r²)( dr )

dt                  dt

 

dV   =  3πr²( dr )

dt               dt

 

Plugging in the information from the question (dV/dt = 48π , r=16)

 

48π  =  3π(16)²( dr )

                                  dt

 

dr     =  ft./min.

dt         16

 

At the moment when the radius of the water is 16 feet, the radius is decreasing at a rate of 1/16th of a foot per minute.

 

c) Now you'll have to differentiate the original volume for a cone (the one that includes height) V=(1/3)πr²h.

This will allow you to solve for dh/dt.

To do this you will have to once again differentiate implicitly and also use the product rule. (f'(x) = r'h+h'r)

 

V   = (1/3)πr²h

 

dV = (1/3)π[2r( dr )(h)+r²( dh )]

dt                   dt             dt

 

To solve for dh/dt we'll first need to find h. This can be done by first finding the volume when r=16ft, then finding the height using that volume and radius.

 

1. v(r) = πr³

 

    v(r) = π(16)³

 

    v(r) = 4096π ft³

 

2. v     = (1/3)πr²h

 

 4096π = (1/3)π(16)²h

 

    h      = 48ft

 

 

Plug this height along with the other information into the eqation above (dV/dt = (1/3)π[2r(dr/dt)(h)+r²(dh/dt)]) to solve for dh/dt.

 

dV   = (1/3)π[2r( dr )(h)+r²( dh )]

dt                     dt             dt

 

48π = (1/3)π[2(16)(1/16)(48)+(16)²( dh )]

                                                   dt

 

dh   =  ft/min

dt       16

 

At the moment when the radius of the water is 16 feet, the height is decreasing at a rate of 3/16ths of a foot per minute.

 

 

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