Integration help needed

[Robert G]

I have two integration problems where I need help. Please help me with these problems, that would be very nice. It would be also very nice if you could give me explanations of what to do, how and why.


#1 dy/dx = (y-1)²(2+x) goes through (-1,2) I need to find the equation of the curve and the equations of the asymptotes to it.

#2 A cylindrical tank is full and it leaks water at a rate which is proportional to the volume of the remaining water. After 20 minutes the tank is half full. What fraction of water is remaining after one hour?

Please don't only tell me what to do, but also how exactly, because I don't have very much confidence in my calculus skills. Thank you very very much for your help.

Regards,

Robert

 

[galactus]

The best way to build confidence is by doing.

For the first one, separate variables, then integrate.

\(\displaystyle \frac{dy}{dx}=(y-1)^{2}(2+x)\)

Separate variables:

\(\displaystyle \frac{dy}{(y-1)^{2}}=(2+x)dx\)

Integrate:

\(\displaystyle \int\frac{1}{(y-1)^{2}}=\int{(2+x)}dx\)

Now, finish the integration. When you do, you will have a constant C on the right side along with your indefinite integral. Solve for y. Then, sub in your knowns and solve for C and you're done. Let me know what you get.
The integrations are not tough ones. A simple u substitution on the left one will suffice.

For the tank problem, is there a formula you're to use?.

A formula for a draining tank is commonly given by \(\displaystyle A(h)\frac{dh}{dt}=-k\sqrt{h}\).

Where the depth of the liquid, h(t), and the area of its surface, A(h), are related. This is Torricelli's law.

But your problem says proportional to height remaining. So, instead of \(\displaystyle \sqrt{h}\), we use h.

I am going out on a limb here.

\(\displaystyle \frac{dV}{dt}=-kh\)

\(\displaystyle V(t)=Ah(t)\)

\(\displaystyle \frac{dV}{dt}=A\frac{dh}{dt}\)

\(\displaystyle \frac{dh}{dt}=\frac{-k}{A}h\)

\(\displaystyle \int{A\frac{dh}{h}}=-k\int{dt}\)

\(\displaystyle Aln(h)=-kt+C\)

Now, use the IC, half full at 20 minutes, to find C.

 

[Dr. Flim-Flam]

Second Problem: dV/dt=kV, V(0)=1,V(1/3)=1/2 Solving this you'll get V(1)=1/8.