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December 18, 2014

December 18, 2014

Posted by **andrew** on Saturday, April 2, 2011 at 4:09pm.

tank with height of 7m and radius of 4m. If V is the volme of water in the tank and h is the depth

of water,

find the relationship between dV/dt and dhdt.

How is the volume changing when h = 3m and

dh/dt = 0.2m/s?

- calculus -
**Damon**, Saturday, April 2, 2011 at 5:06pmdV/dt = surface area * dh/dt (geometry , no calculus really required, draw picture)

dV/dt = pi r^2 dh/dt

- calculus -
**Reiny**, Saturday, April 2, 2011 at 5:08pmconsider a cross-section of the paraboloid placed with vertex at the origin an opening upwards

then its equation would be

y = ax^2

when x=4, y = 7

y = (7/16)x^2

or

h = (7/16)r^2 , using r and h

r^2 = 16h/7

V = (1/2)πr^2h

= (1/2)π(16h/7)h

= (8π/7) h^2

dV/dt = 16π/7 h dh/dt ----> the relationship between dV/dt and dh/dt

Whe h=3 and dh/dt = .2

dV/dt = (16π/7)(3)(.2) = ......

I will let you do the button pushing.

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