2026 AP Calculus AB FRQ #3 Answers

Added: 2026-05-14

[00:00:00]Welcome back to the channel. In the last two videos, I explained the first two Calc AB FRQs from 2026, and in this video, I'll explain FRQ 3. Here, we are told that a pie is taken from a hot oven and put on a table, and the internal temperature of the pie at t minutes is modeled by the function h, and then we're given h. It says then that h of 0 is 75, and that for t of 0, it is known that 20 is less than h of t is less than 75. And we're asked, which of the following could not be uh or explain why the following could not be a slope field for the differential equation. Well, I'm thinking here, usually, it's helpful to look at the like values that are like extreme. So, for example, if I have the value of, let's say, um we could use, let's say we could use um 20, right? If I have 20, so keep in mind that this is uh my value of h. Um Actually, I guess I couldn't really use uh the h there, cuz then it wouldn't be 20. Let's say we use the h value of, let's say, 30, right? If I have h is 30, then I'm going to have that dhdt should be -1/15 * 30 - 20, which would be a negative number, right? -1/15 * +10, right? And so, it would be negative. But, we see here that the slopes here are positive.

[00:01:26]And so, we could say something along the lines of, when dh when when h is greater than 20, this will be positive, and therefore, this whole thing will be negative, right? But, we see here that these slopes are positive. So, we could say something along the lines of, when I'll type this out, so um we could say, when h of t is greater than 20, dhdt is less than 0, but the given slope field shows that shows positive values of d h d t for h values for h is greater than 20. Something like that. And that would suffice.

[00:02:13]On to the next part, we want to find the slope of the line tangent to the graph of h at time t equals zero. Well, for that, literally all we need to do is get d h d t evaluated at t equals zero. And so what we would do is we would just plug in. Well, we know that h of zero is 75, so we would do -1/15 * 75 is h here, right? h of zero is 75, minus 20. So that would be -1/15 * 55. I believe that at this point you would get a point on the FRQ, but if we wanted to go out fully and solve it, we could say this, which would be -11/3, right? And we don't need any units.

[00:02:53]The next part says it can be shown that this is true. The line tangent to the graph at of h at time t equals zero is used to approximate h of five, the internal temperature of the pie at time t equals five. Is this approximation an overestimate or an underestimate for the actual value of h of five? Well, okay.

[00:03:10]At d squared d t or or d squared of h d t squared. Let's look. At time t equals zero, is this is this going to be concave up or concave down? Well, this is going to be concave up, right? This second derivative tells us if our graph is concave up, meaning looking like that, or concave down, looking like that.

[00:03:31]And this will be positive, d squared h of d t squared is positive uh for any value that h is greater than 24, right? And it's negative for uh it looks like any value where h is less than 20. However, we're always going to be greater than 20, right? h of t will always be greater than 20. And so because of that, d squared of h over dt squared is greater than zero, meaning that it is concave up. So, if I were to then So, let's draw a very obviously concave up graph. Something like this, right? If I were to take a point, let's say here, and use a tangent line to approximate it, if we make a straight line here, you'll see that that tangent line will give us an underestimate, right? You could say something along the lines of I mean, you could just say since d z squared h over dt squared is greater than zero for any value h is greater than 20 and h is 20 for any value t is greater than zero. So, basically you're saying that d squared h dt squared is greater than zero because h is greater than 20.

[00:04:43]The graph is concave up and therefore the estimate is an underestimate.

[00:04:51]Something like that.

[00:04:52]Okay, then the last part wants us to use separation of variables to find h of t.

[00:04:56]This is pretty common. So, all we have to do is get the dh's on one side, the dt's on a different side. So, we'll have I could write that dh * 1 / h - 20 is equal to -1/15 dt. So, if I take the integral here both sides, I'll get ln of h - 20 is equal to -t/15 + c. I could e both sides then and say that h - 20 = e to the -t/15 + c, which equals ce to the -t/15.

[00:05:32]At that point, I could say that h = ce to the -t/15 + 20, right? And so, by then, all I need to do is plug in the value of 0,75, say that 75 = ce to the 0 over 15 plus 20.

[00:05:50]Which gives us 75 equals CE to the 0 is just 1. So CE plus 20. So CE equals 55.

[00:05:58]And then it ends up just being if you plug this back into here, we get H is 55 E to the negative T over 15 plus 20. And that is our answer.