1. ## Integration help needed

Hey guys, I need some help with some integration problems. Some I could solve already but there are two where I need help. If you can help me, however, please don't just say me: integrate this, but also show me how to do it, that would be really nice. Thanks a lot for everyone who helps me.

1: A large bottle with, 4m high and with a radius of 2m, loses water. Originally, the bottle was full and lost water at a rate proportional to the square root of the depth of the remaining water.

I'm given that the depth of the water is 1m after 2 hours. The question is after how many hours the tank is empty. Please help me with this question.

2. In a freezing machine the temperature is constantly 5 degree. An 100 degree hot object is put in it. One minute later the object is only 80 degree hot. When will it reach a temperature of 10 degree?

Thanks a lot for every advice. But please also give me explanations if you can. I very much appreciate any help. Thanks a lot.

2. Can nobody help me with these problems?

3. Hello, Instigator!

I have a long, looong solution . . . maybe someone can shorten it.

1) A large bottle, 4m high and with a radius of 2m, loses water.
Originally, the bottle was full and lost water at a rate proportional
to the square root of the depth of the remaining water.

The depth of the water is 1m after 2 hours.
After how many hours the tank is empty?

The volume of a cylinder is: .$\displaystyle V \:=\:\pi r^2h$

Since $\displaystyle r = 2$, the volume of the water is: .$\displaystyle V \:=\:\pi(2^2)h \:=\:4\pi h$
. . Then: .$\displaystyle \frac{dV}{dt} \:=\:4\pi\left(\frac{dh}{dt}\right)$

But we are told that: .$\displaystyle \frac{dV}{dt} \:=\:k\sqrt{h}$

So we have: .$\displaystyle 4\pi\left(\frac{dh}{dt}\right) \:=\:k\!\cdot\!h^{\frac{1}{2}}$ . . . a differential equation

. . Separate variables: .$\displaystyle 4\pi h^{-\frac{1}{2}}dh \:=\:k\,dt$

. . Integrate: .$\displaystyle 8\pi h^{\frac{1}{2}} \:=\:kt + C$

When $\displaystyle t = 0,\;h = 4$
. . We have: .$\displaystyle 8\pi\sqrt{4} \:=\:k(0) + C\quad\Rightarrow\quad C \,=\,16\pi$

The equation (so far) is: .$\displaystyle 8\pi\sqrt{h} \:=\:kt + 16\pi$

When $\displaystyle t=2,\:h=1$
. . $\displaystyle 8\pi\sqrt{1} \:=\:k\!\cdot\!2 + 16\pi\quad\Rightarrow\quad k \,=\,-4\pi$

The equation is: .$\displaystyle 8\pi\sqrt{h} \;=\;-4\pi t + 16\pi\quad\Rightarrow\quad\sqrt{h} \:=\:2 - \frac{t}{2}$

Hence, the height function is: .$\displaystyle h \;=\;\left(2 - \frac{t}{2}\right)^2$

If the tank is empty, then $\displaystyle h = 0.$

We have: .$\displaystyle 0 \:=\:\left(2 - \frac{t}{2}\right)^2\quad\Rightarrow\quad t \,=\,4$

Therefore, it takes 4 hours to empty the tank.

But check my reasoning and my work . . . please!
.