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Water is flowing into a conical tank at a rate of 3 ft3/min. The height of the tank is 10 feet and its diameter is 6 feet. Find the rate at which the height of the water level is changing at the instant its height is 5 feet.

Respuesta :

Answer:

Dh/dt  =  1.273   ft/min

Step-by-step explanation:

Volume of cone   V =  1/3 * π *r² * h

Then

DV/dt   =  1/3*  π *r² * Dh/dt      (1)

We have to find out values of r   when  h = 5

By symmetry  in a cone (we have proportion between h and r

when  h  =  10  ft       r = 3   ft    from problem statement

Then  h   =  5   ft        r = 1.5 ft

That is from proportion       3/10  =  X/5

Then by subtitution in (1)

DV/dt    =    3    ft³/min      r  =  1,5  ft

3   =  1/3*π*(1.5)²* Dh/dt

3   =  2.355  Dh/dt

Dh/dt  =  1.273   ft/min

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