What is the limiting value of P as t approaches infinity?

[nick.martinez]

What is the limiting value of P as t->infinity

dp/dt=0.4P(10-P)

My attempt at the solution was to serperate the function and get each side in terms of one variable

dp/(P(10-P)) = 0.4/dt

[-ln(|p-10|/|p|)]/10=0.4t

-ln(|p-10|/|p|)=4t
take e^ of both sides of equation

(p-10/p)=e^4t and we know that the equation is undefined on the left when the p=10 therefore there is some type of discontinuity. is this correct?

 

[slider142]

What is the limiting value of P as t->infinity

dp/dt=0.4P(10-P)

My attempt at the solution was to serperate the function and get each side in terms of one variable

dp/(P(10-P)) = 0.4/dt

[-ln(|p-10|/|p|)]/10=0.4t

Unlike differentiation, when you integrate two functions that are equal, their integrals are not necessarily equivalent: they can differ by a constant term. This line should be [-ln(|p-10|/|p|)]/10=0.4t + C where C is an undetermined constant. This constant affects your next step, where you take the exponential of both sides.

-ln(|p-10|/|p|)=4t+ C
take e^ of both sides of equation

In your next step, I'm not sure if you took into account the "-" sign in front of the logarithm, the effect of which is to flip the fraction over:
p/(p - 10)=e^(4t + C) = e^(4t)e^C

However, I guess that was just a typo. We would usually just write e^C as an undetermined constant A, so the right side of your equation would be Ae^(4t).

and we know that the equation is undefined on the left when the p=10 therefore there is some type of discontinuity. is this correct?

Yes. However, since P is not the independent variable, it is not necessarily a discontinuity in the domain, merely a value that the function P(t) does not have in its range. The equation also doesn't have a solution when P = 0. It could be an asymptote, or it may merely be a hole, or just a value completely out of the range of the function. In order to get further information, you should try to solve for P as an explicit function of t.