calculus question- relative rates

[rocadelpunk]

ugh my brain is tired, and I could use some help...

Water is leaking out of a hemisphere tank of radius 8 feet at a rate of 2 cubic ft per hour. At what rate is the depth of the water changing when it is 3 feet deep at the deepest point?

Value of a region of height, h units, of a hemisphere of radius 8 units is given by the forumula V = pi H^2 [8 - (h/3)]

I tried taking this a couple ways...

taking the derivative with respect to volume or with respect to time, my gut feeling is that taking the derivative with time is what I need to do.

so I think this is what I have...

know: dv/dt = 2ft^3/hr

want: dh/dt

when: h = 3ft

and the relationship is the formula...I guess I'm having problems with the derivative atm, like i said brain is frazzled : P.

so I divided by pi and multiplied through by 8 and i get

V/pi = 8h^2 - h^3/3

if someone could help me take the derivative of that ^, I'd be most appreciate, or any other hints.

thanks

 

[Shooters]

Don't divide by pi; leave it in the following form:

V = 8*pi*h^2 - (pi/3)*h^3

Now take the derivative of V with respect to t remembering to use the chain rule:

dV/dt = 16*pi*h*dh/dt - pi*h^2*dh/dt

Setting the above expression equal to your given of dV/dt = 2 gives

16*pi*h*dh/dt - pi*h^2*dh/dt = 2

Now just plug in 3 for h and solve for dh/dt.

 

[rocadelpunk]

thank you very much, that was really helpful

one question though, once you work it all out, my final answer becomes something like 2ft^3/hr over 39pi ft...final answer being ft^2/hr...which doesn't seem to make sense?!?

or should i just ignore the units 😀

 

[Shooters]

The units work out. Remember that in your original equation for V, the radius, 8, is in units of feet, and h is also in units of feet. When you take the time derivative of V, remember to carry the units through when you multiply. So, you have

V = pi*h^2(8ft - h/3)

Expanding it out gives

V = 8ft*pi*h^2 - (pi/3)*h^3

Taking the derivative and setting it equal to your given dV/dt gives

dV/dt = 16ft*pi*h*dh/dt - pi*h^2*dh/dt = 2ft^3/hr

If you plug in 3ft for h then you get

16ft*pi*3ft*dh/dt - pi*9ft^2*dh/dt = 2ft^3/hr

Now if you solve for dh/dt you should get an answer that has units of ft/hr, which is correct.