Question

hi guys
can you help me please?
Water flows out of a hemispherical tank at a rate which is 4 times the square root of its depth. If the radius of the tank is 8 ft, how fast is the water level falling when the water is 4 ft deep? Hint: Use the formula for the volume of a spherical segment, i.e. V=1/3 π h^2 (3r-h).

wher the answer is 16 pi

Answer 1

\[h= 4 ft, r=8ft \frac{ dh }{ dt }= ?\]

Answer 2

Hmm I think there is another piece of information you need before you do any differentiation :)

`Water flows out of a hemispherical tank at a rate`
v'

which is `4 times` the `square root of its depth`.
4*sqrt(h)

So we have: \(\large\rm v'=-4\sqrt{h}\)
ya? :o

Answer 3

@zepdrix continue please ...i seem to be going wrong somewhere

Answer 4

well... i tried this: dV= pi*r^2 dh (where r is the radius that corresponds to height h)

Answer 5

substitute for dV in V'
\[\pi r^2 \frac{ dh }{ dt }= -4\sqrt{h}\]

Answer 6

They gave us a nice formula to use,
\(\large\rm v=\frac{\pi}{3}h^2(3r-h)\)
This is the volume of the filled portion of the hemisphere.
I guess we want to differentiate this formula with respect to time.

Before we can do that though,
we would like to write the r in terms of h.
So I think we need to use another formula, let's call it v tilde,
\(\large\rm \tilde{v}=\frac{2\pi}{3}r^3\)
Oh no no no, scratch that... it's a lot simpler than that.
r is held constant, it is r=8.
So we don't even need to worry about any r' showing up.
AHHH my bad.

Answer 7

I dunno what you did baru :c sorry kinda confused by your formula thing

Answer 8

do you get 16pi by ur method?

Answer 9

\[\large\rm v=\frac{\pi}{3}h^2(3r-h)\]\[\large\rm v=\frac{\pi}{3}h^2(24-h)\]And then differentiate with respect to time, ya?\[\large\rm v'=\frac{\pi}{3}\left(h^2\right)'(24-h)+\frac{\pi}{3}h^2(24-h)'\]Product rule? :o

Answer 10

Oh she provided the answer, nice.
Hope so! Let's find out :d

Answer 11

\[\large\rm v'=\frac{2\pi}{3}2h(24-h)h'-\frac{\pi}{3}h^2h'\]\[\large\rm v'=\left(\frac{2\pi}{3}2h(24-h)-\frac{\pi}{3}h^2\right)h'\]

Answer 12

At \(\large\rm h=4\) and \(\large\rm v'=-4\sqrt{h}=-8\) we have,\[\large\rm -8=\left(\frac{2\pi}{3}2\cdot4(24-4)-\frac{\pi}{3}16\right)h'\]\[\large\rm -\frac{1}{2}=\left(\frac{\pi}{3}(20)-\frac{\pi}{3}\right)h'\]\[\large\rm -\frac{1}{2}=\left(\frac{19\pi}{3}\right)h'\]Hmm darn, no that's not leading to the answer which was provided :(\[\large\rm h'=-\frac{3}{38\pi}\]

Answer 13

Hmmm

Answer 14

i get -1/6pi .... in either case... it looks like 'pi' has to be in the denominator

Answer 15

Well I guess she ran off to bed or something XD
I should do the same heh

Answer 16

XD

Answer 17

hi @baru and @zepdrix
i got -1/6pi

is it okay if the answer given by the book is wrong?

Answer 18

oh and thanks for your help

Answer 19

it's possible that the text book has got it wrong...check with your teachers/classmates and let us know :)