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## AP®︎/College Calculus AB

### Unit 9: Lesson 1

AP Calculus AB 2017 free response# 2017 AP Calculus AB/BC 4c

Potato problem from 2017 AP exam (Question 4, part c). Finding a particular solution for a separable differential equation.

## Want to join the conversation?

- where is the C for the integral with (G-27)^(-2/3)?(1 vote)
- The constants of integration are entirely arbitrary, so we don't need one on each side.

Let's, for argument's sake, have one on both sides ie

3(G - 27)^(1/3) + C₁ = -t + C₂

We could take C₁ from both sides:

3(G - 27)^(1/3) = -t + C₂ - C₁

But since they are arbitrary we can let C = C₂ - C₁ and so we have

3(G - 27)^(1/3) = -t + C(9 votes)

- What if the negative sign on (G-27) was not transferred to t? Wouldn't the constant be negative 12 instead? How did Sal know to move the negative over with the t?(3 votes)
- It doesn't matter. He could have moved it before, during, or after integration; G(t) would still be the same in all cases.(3 votes)

- i don't understand it when sai says"we were not only able to solve for the general solution. We were able to find the particular solution using this initial condition right here the G of zero is equal to 91."at6:27(1 vote)
- Sal finishes solving for the general solution at4:17; this is the equation for G that solves the given differential equation. With the general solution, C could equal any constant, and the equation would still hold true. However, Sal is also given the initial condition of G(0) = 91, which limits C to one single value, making G a particular solution. Hope that I helped.(4 votes)

- At2:40, how did Sal turn
`dG/((G-27)^2/3)`

into`(G-27)^2/3 dG`

?(2 votes)- Sal turned dG/((G-27)^2/3) into (G-27)^-2/3 dG , not (G-27)^2/3. A negative exponent turns it into a denominator of a fraction(1 vote)

- Why doesn't Sal just substitute G(0)=91 in the dG/dt equation?(2 votes)
- The question is to evaluate at t=3, so it's G(3)(1 vote)

- Did you solve this from material learned from differential equations? I'm studying for an upcoming integrals test and this seemed confusing. So hopefully I shouldn't have to know this right now.(2 votes)
- Why doesn't net change work here?(2 votes)

## Video transcript

- [Instructor] Let's now tackle part C which tells us, for T is less than 10, an alternate model for
the internal temperature of the potato at time T minutes is the function G that satisfies
the differential equation. The derivative of G with respect to T is equal to the negative of G minus 27 to the two third power. Where G of T is measured
in degrees Celsius and G of zero is equal to 91. Find an expression for G of T. Based on this model, what
is the internal temperature of the potato at time T is equal to three? So they gave us a differential equation. They want us to find an
expression for G of T, so essentially find a solution to this differential equation. And then they want us to
use that solution to find the internal temperature
time T equals three. So, the first thing to appreciate, this is in an AP calculus class. And so they're asking us
to solve a differential equation, or find a solution
to a differential equation it is unlikely to be a really
strange differential equation. It's likely to be a separable
differential equation. And then once we find that solution we just have to evaluate
it at time T equals three. So let's rewrite the
differential equation. And then, let's try to evaluate. Let's see if we can find the solution and then we'll evaluate. The derivative of G with respect to T is equal to the negative of G minus 27 to the two thirds power. So, if this is going to be a
separable differential equation I want to separate the DG and the DT. So I'm gonna treat my
differential like numbers so, or variables. So I'm gonna multiply both sides times the T differential. And so then I'm going to have the capital G differential DG is equal to negative times G minus 27 to the two thirds power DT. And the whole notion
here, I'm trying to get all of the G things on the side with the DG. And then all of the things that involve T in the side with the DT. So we don't see any Ts here so really, we just have to get the Gs over on this side while
leaving the DT there. And then we can integrate both sides. So let's see. If we were to divide both sides by G minus 27 to the two thirds. G minus 27 to the two thirds. G minus 27 to the two thirds, what do we have? Well, we could rewrite the left side as G minus 27 to the negative two thirds negative two thirds DG is equal to I'm left with just this negative DT. Is equal to negative DT. Now let's see. Now, we can, or actually just a simplify or just to make it a
little bit more obvious I could write this as negative one DT And now we just have to
integrate both sides. So, we could rewrite this as let me write it on both sides. I'm going to integrate I'm going to integrate both sides here. And so what does this give me? Well, on the left side here you could try to do some
U substitutions saying, U is equal to G minus 27. And then DU would be DG. Or, you might recognize that look, the derivative of G minus 27 is just DG. The derivative of G minus 27 is just going to be equal to one. So you could even say, look, I have my I have the derivative there. So I could really integrate
with respect to G minus 27. And so really, I would just
use the reverse power rule. I would take my G minus 27. I would increment this exponent by one. So let's see, negative two thirds plus one is positive one third. So positive one third power. And then I would divide
by this new exponent. So dividing by one third
is the same thing as multiplying by three. So that's the left hand side. This is going to be equal to the right hand side. This is just going to be negative T. And for good measure I'm going to have A plus C. So, how could we solve for C? Well they give us some
information, they say G of zero is equal to 91. So lets write that down. So, when T is zero G is 91. So we can write let me write it over here. So three times so when T is zero G is 91. Ninety one minus 27 to the one third power
is equal to negative T. Well now we're saying T is zero. So you can write negative zero there. Or we could just not write it. And then plus C. So that's what C is going to be equal to. Let's see, 91 minus 27 is 64. Sixty four to the one third power is positive four. And so we have C is equal to 12. C is equal to 12. And let's see, we want to
write an expression for G of T. So now let's just manipulate
this and solve for G. So, if we let me just take this and go right over here. So if we divide both sides by three we are going to get G minus 27 to the one third power is equal to negative T over three plus four. I just divided both sides by three. And I can take the cube of both sides and I would get G minus 27 is equal to negative T over three plus four to the third power. And I would just have
to add 27 to both sides. And I'll make it clear,
G is a function of T is equal to negative T over three plus four. All of that to the third power. Plus 27. So I did the first part. This is, what do we got cross this right over here is an expression for G of T. It was indeed a separable
differential equation. It took a little bit of algebraic manipulation to get us there. But we were able to do it and we were not only able to solve for the general solution. We were able to find
the particular solution using this initial condition right here the G of zero is equal to 91. Now, we'll do the easy part. Based on this model, what
is the internal temperature of the potato at time T equals three? So G of three is equal to negative three over three plus, I'll just write it out. Negative three over three plus four to the third power plus 27. Well this is negative one. So negative one plus four is three. So this is three to the
third power which is 27. Plus 27 is equal to 54 degrees Celsius. And, we are done.