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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition)>Unit 10

Lesson 5: Exponential models

# Exponential models & differential equations (Part 1)

Assuming a quantity grows proportionally to its size results in the general equation dy/dx=ky. Solving it with separation of variables results in the general exponential function y=Ceᵏˣ.

## Want to join the conversation?

• Even though we can never have a negative population in practice, do we really need to assume that the population is positive in order to get rid of the absolute value sign, mathematically speaking? Since our new constant C could be either positive or negative, doesn't that simply account for all of the values our population could be in and of itself, making the absolute value sign redundant?
• Well, the mathematical reason would be this:

The sign of P only depends on the sign of constant C, so you can notice that your initial population, P(0), is equal to C (just make the substitution).

Now, the initial population is always positive, therefore P is always positive.
• I don't understand why C only appears on the right side of the equation. Can somebody please elaborate?
• As Hengru said, you can combine the C's to make another C. Since C1 and C2 are arbitrary constants anyways, adding/subtracting them shouldn't do anything since something arbitrary + something arbitrary is still something arbitrary.
• In the final form of the equation, P = Ce^kt, shouldn't the caveat C > 0 be included, since C is really just e^C, which is always positive and nonzero?
• Mathematically that is not really a restriction; if for example `C₁ = 𝒾π`, then `C = e^(𝒾π) = -1`. In the context of the problem we know that `C₁` is never going to take imaginary values, but as a general solution to a differential equation, the restriction is not valid.
• At why is it Kt and not K^2/2?
• It's because K is a constant and we're integrating *with respect to t" - the dt bit in
∫ K dt
If we were integrating with respect to (a variable) K, then we'd have
∫ K dK
and you'd be right, the integral would be K²/2
• What is K and why do we need to include it?
• I think there are some videos in Pre-Algebra that discuss K. Just search up the Constant of Proportionality, and maybe also look at the video on Direct Vs. Inverse Variation.
• In the first equation : dP/dt = kP. P is a function or a real number ?
dP/dt is a slope , kP is a number , why are they equal ?
• P is a real number. Multiplying by k allows you to go from a number to a slope.
• The mass of an isotope decreases at a rate that is proportional to the mass at that time.

The mass of the isotope was 40 grams initially, and it was 10 grams after 16 days.

What was the mass of the isotope after 20 days?

Should my model be look like this: dM/dt = - kM

Is my model correct?
• Yes that is correct: Directly proportional means that the equation takes the form y = kx, where y is said to be proportional to x. In this case, it says the rate that the mass decreases is proportional to the mass. The rate is the derivative which must be equal to -kM where k is just a constant. Note that it is -kM since the mass is decreasing.

Thus, the model becomes dM/dt = -kM
• When integrating 1/p, is it necessary to write ln|x|? or could you say the integral is ln(x) since population can't be negative?
(1 vote)
• Would another solution to the differential equation in the video be P=C/1-kt?
at I tried doing the following:
dP/dt = kP \\-kP
-kP +dP/dt = 0 \\Integrate both sides with respect to t
-kPt+P = C
P(1-kt) = C \\divide by 1-kt
P = C/(1-kt)
However if we differentiate the following equation with respect to t we get:
dP/dt = (0*(1-kt)-(-k)*C) / (1-kt)^2
dP/dt = kC/(1-kt)^2 \\ substitute for P = c \(1-kt)
dP/dt = kP/(1-kt) not equaling to dP/dt = kP which was my initial condition.

I`d be happy if someone can help me understand where I made a mistake :)
(1 vote)
• Integrate -kP with respect to t won't result in -kPt, because P is a function of t.