There is a right cylindrical bucket filled with sand and a tiny leakage in it. The base radius of this cylinder is 15 cm and height is 25 cm.
Initially, the sand occupies the entirety of the bucket and, after that, gradually loses volume and height over time. The rate of change of volume is proportional to the depth of the sand(height) in the bucket. This rate of change of volume is initially -5 cm^3 per minute.
Use natural logarithms to express the time (t) in terms of the height (h):
I think the answer was something like, t = 1125*ln((pi)h/25) or something quite similar.
I have done pages of calculation doing all kinds of stuff for hours and I still don't get it.
$\frac{dV}{dt}=k\cdot h$ assuming k is negative.
$V=h\pi r^2$ so substitute $$\frac{\pi r^2dh}{dt}=kh$$ Solve this differential for h using only variable letters. [I bet that just putting the answer won't get you very far.] Let the initial condition be $h_0$ at $t=0$. When you finally get an expression either identical or similar to $$h=h_{0}e^{\left(\frac{k}{r^2\pi}\right)t}$$ then you can go back and substitute in for $h_0$, $k=-5$ and $r$. I say "similar to" because the way the question is worded, you may not want to completely solve for $h$ and instead leave the answer in log form.