# Thread: I need integration help

1. ## I need integration help

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 and best wishes,

Robert

2. #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.

Group the variables so that dy is with the y-terms, and dx is with the x-terms:
dy/(y-1)^2 = (2+x)dx
[(y-1)^(-2)]dy = (2+x)dx
Integrate bote sides,
-1(y-1)^(-1) = 2x +(x^2)/2 +C
1/(1-y) = 2x +(x^2)/2 +C ----------(i)
at (-1,2), or when x = -1, y=2, so,
1/(1-2) = 2(-1) +((-1)^2)/2 +C
-1 = -3/2 +C
C = 1/2
So, into (i),
1/(1-y) = 2x +(x^2)/2 +1/2
Multiply both sides by 2,
2/(1-y) = 4x +x^2 +1
(1-y) = 2/(4x +x^2 +1)
y = 1 -2/(x^2 +4x +1)
y = [x^2 +4x +1 -2]/(x^2 +4x +1)
y = (x^2 +4x -1)/(x^2 +4x +1) -------the equation of the curve, answer.

Vertical asymptotes:
x^2 +4x +1 = 0
x = {-4 +,-sqrt[4^2 -4(1)(1)]} /(2*1)
x = -3.732 or -0.268 ------------------two vertical asymptotes.

Horizontal asymptotes:
Lim(x-->infinity)[(x^2 +4x -1)/(x^2 +4x +1)]
= Lim(x-->infinity)[(1 +4/x -1/(x^2)) / (1 +4/x +1/(x^2))]
= 1
So, y = 1 is a horizontal asymptote

Oblique asymptote: None, because the numerator (x^2 +4x -1) is not one degree higher than the denominator (x^2 +4x +1).

3. #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?

Given : dV/dt = k*V ---------(i)

dV/V = k*dt
ln(V) = k*t +C
V = e^(kt +C)
V = e^(kt) *e^C
When t = 0, V = Vo, so,
Vo = e^(k*0) *e^C
e^C = Vo
Hence, V(t)= (Vo)e^(kt) --------------(ii)
Since Vo = volume when full, then
Vo = pi(R^2)H.
Therefore,
V(t) = [pi(R^2)H]e^(kt) -------(iii)
where R and H are constants. R is radius and H is height of the cylindrical tank.

Given: when t=20min, V = (1/2)[pi(R^2)H].
So,
(1/2)[pi(R^2)H = [pi(R^2)H]e^(k*20)
1/2 = e^(20k)
ln(1/2) = 20k*ln(e)
k = ln(0.5) / 20
k = -0.034657

So,
V(t) = [pi(R^2)H]e^(-0.034657t) -----------finally, equation for volume of remaining water at any time t in minutes.

Hence, after 1 hour,
V(60) = [pi(R^2)H]e^(-0.034657*60)
V(60) = [pi(R^2)H](0.125)

By proportion,
V(60) / (Full tank) = [pi(R^2)H](0.125) / [pi(R^2)H] = 0.125 = 1/8

Therefore, after one hour, only 1/8 of the full tank volume is still remaining in the tank. ---------answer.

4. Thank you very much, it helped me a lot!