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How would I find the formula that relates the height of water to the volume of water in a sphere when the water is added at 0.5cm^3/s and the sphere's radius is 1cm (or if easier, any radius, if not, stick to 1cm radius)?

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How would I find the formula that relates the height of water in a sphere to the time taken when the water is added at 0.5cm^3/s and the sphere's radius is 1cm (again if easier, any radius, if not, stick to 1cm radius)?

Any one of those will do, but if you're feeling extra intrepid both would be extremely helpful, though again, I will still give best answer for just one.

Comment: I know (or at least think) it should use the chain formula thing from differentiation because it needs to relate time into the formula

V=(4/3) * pi * (h/2)^3

(I've used the sphere volume formula and replaced the r with h/2 because I want the formula to be V and h) I've put it into the formula and have

dv/dt = (dv/dh) * (dh/dt)

0.5cm^3/s = (pi * h^2)/2 * (dh/dt)

dh/dt = 1/(pi * h^2)

However I don't know what to do from there. I don't understand what that value is supposed to mean, rate at which h changes? Like, what? Any help would be appreciated, even if it doesn't go the way I just went using the chain rule.

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  • $\begingroup$ While pouring water in sphere its height will also be variable so by differentiate V and put the value. And then since the height is variable differentiate it using chain rule. $\endgroup$ – Adesh Tamrakar Oct 4 '15 at 13:22
  • $\begingroup$ You just need the formula for the volume of a spherical cap: en.wikipedia.org/wiki/Spherical_cap $\endgroup$ – Aretino Oct 4 '15 at 13:41
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If the sphere has a radius of $r$, and the height of the water is $h$, then the volume of a spherical cap (as mentioned by @Aretino) is given by $$\frac{\pi h}{6}\left(3a^2 + h^2\right)$$ but $a = \sqrt{r^2 - (r - h)^2}$, so $$V = \frac{\pi h^3}{6}\left(3(2r - h)^2 + 1\right)$$

There we have your equation relating the height $h$ of the water and the volume $V$. With some algebraic manipulation, we can write $h$ in terms of $V$ (I think), and then we know $V = 0.5t$ where $t$ is the time in seconds, $V$ is the volume in $\text{cm}^3$, and $h$ is the height of the water in $\text{cm}$.

We know that $\frac{dV}{dt} = 0.5\ \text{cm}^3\text{s}^{-1}$, so to find the rate of change in height with respect to time:

$$ \begin{align} V &= \frac{\pi h^3}{6}\left(3(2r - h)^2 + 1\right) \\ \implies \frac{dV}{dt} &= \frac{d}{dt}\left(\frac{\pi h^3}{6}\left(3(2r - h)^2 + 1\right)\right) \\ \implies 0.5 &= \frac{dh}{dt} \cdot \frac{d}{dh}\left(\frac{\pi h^3}{6}\left(3(2r - h)^2 + 1\right)\right) \\ &= \frac{dh}{dt}\left( \frac{d\left(\frac{\pi h^3}{6}\right)}{dh}\left(3(2r - h)^2 + 1\right) + \left(\frac{\pi h^3}{6}\right)\frac{d\left(3(2r - h)^2 + 1\right)}{dh}\right) \\ &= \frac{dh}{dt}\left(\frac{\pi h^2}{2}\left(3(2r - h)^2 + 1\right) + \frac{\pi h}{6}\left(6h - 12r\right)\right) \\ &= \frac{dh}{dt} \cdot \frac{\pi h}{6}\left(3h\left(3(2r - h)^2 + 1\right) + 6h - 12r\right) \\ \implies \frac{dh}{dt} &= 0.5 \left(\frac{\pi h}{6}\left(3h\left(3(2r - h)^2 + 1\right) + 6h - 12r\right)\right)^{-1} \end{align} $$

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