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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition) > Unit 12

Lesson 1: AP Calculus AB 2017 free response# 2017 AP Calculus AB/BC 4b

Potato problem from 2017 AP exam (Question 4, part b). Taking the second derivative using the chain rule to think about concavity.

## Want to join the conversation?

- is it an underestimate because you assumed the same dH/dt for all t>0?(6 votes)
- Since the problem stated that H is always greater than 27, then dH/dt will always be negative. Since the second derivative is (-1/4)(dH/dt), the second derivative will always be positive. A positive second derivative means the curve is concave up, so the curve will begin to curve up while the tangent line will continue its same downward path. Thus the tangent line will be below the true curve, meaning it will underestimate.(13 votes)

- Does anyone have an example of any AP solved exam? I'm just curious to know how we would flesh out our language.(10 votes)
- I believe on the College Board website you can download the answer key to the 2017 exam. You can see a solved exam there.(3 votes)

- When answering the question, how should I format my answer? Do I have to answer in paragraph form?(3 votes)
- You should check out some sample AP exam papers. There, they'll give you an idea on how to write out your answers.(3 votes)

- For question (B) can we also use the Taylor polynomial centered at t=0 and approx. the function H(t) with 3 terms to find out that the temperature at t=3 is an understatement in question (A)?(3 votes)
- You
*could*, though for that, you'll need to use dH/dt to find H(t) and then find multiple derivatives of it (not necessarily only three) to get a good approximation for the function. And then you substitute t = 3 to find the value of H.

This is a lengthier method than just using the information given, and I wouldn't recommend doing this on an exam. But yes, if you just want to experiment, this method would work!(2 votes)

## Video transcript

- [Instructor] We're now
going to tackle part (b) of the potato problem, and it says use the second derivative of H with respect to time to
determine whether your answer in part (a) is an underestimate
or an overestimate of the internal temperature of the potato at time t is equal to three. So in part (a), we found the
equation of the tangent line at time equals zero and we used that to estimate what our
internal temperature would be at time is equal to three. So how is the second derivative
going to help us think whether that was an overestimate
or an underestimate? Well, the second derivative can help us know about concavity. It'll let us know, well, is our slope increasing over this interval,
or is our slope decreasing? And then we can use that to estimate whether we
overestimated or not. So first, let's just find
the second derivative. So we have the first
derivative written up here. Let me just rewrite it, and
I'll distribute the negative 1/4 'cause it'll be a little bit
more straightforward then. So if I write the derivative
of H with respect to time is equal to negative 1/4 times our internal temperature, which it itself is a function of time, and then negative 1/4 times negative 27, that would be plus 27 over four. Let me scroll down a little bit. So now, let me leave that graph up there 'cause I think that might be useful. What is the second derivative going to be with respect to time? So I'll write it right over here. D, the second derivative of H with respect to time is
going to be equal to, well, the derivative of this
first term with respect to time is going to be the derivative
of this with respect to H times the derivative of
H with respect to time. So this is equal to negative 1/4. That's the derivative of
this term with respect to H, and then we want to multiply
that times the derivative of H with respect to time. This comes just straight
out of the chain rule. And then the derivative of a constant, how does that change with respect to time? It's not gonna change, that
is just going to be zero. So just like that, we were able to find the second derivative
H with respect to time. And now what does this tell us? Well, we talked about,
in the previous video, that over the interval that we care about, for, actually, I can
show you from this graph. Over the interval that we care about, for t greater than zero, it says that our internal temperature is always going to be greater than 27. And so when you look
at this expression here or when you look at this
expression here for dH/dt, we talked in the previous video how this is always going
to be negative here. Because H is always going
to be greater than 27, so that part's going to be positive. But then we're gonna multiply
it by a negative 1/4, so our slope dH/dt, our derivative of our
temperature with respect to time, is always going to be negative. So we could write that this or this, this is going to be negative, let me write it this way. Since H is greater than 27 for t is greater than zero, we know that dH/dt is negative, is negative. So we could say that this right over here, since dH/dt is less than zero for t is greater than zero, the second derivative
of H with respect to t is going to be greater than zero for t is greater than zero. And so what does that mean? It means that we are, if your second derivative is positive, that means you're concave upward, concave upward, which means slope is increasing, slope increasing. Or you could say slope
becoming less negative, slope becoming less negative. Now what does that mean? And you could see it intuitively. If the slope is becoming
less and less negative, then that means when we approximated what our temperature is at t equals three, we used a really negative slope. When, in reality, our
slope is getting less and less and less negative. So what we would've done is
we would've over-decreased our temperature from t equals
zero to t equals three. So that would mean that we
would have underestimated it. So let me write that down. And I'm running out of
a little bit of space, but let me write it right over here. So that implies, so this implies that underestimate, underestimate in part (a). And if I were taking this on the AP exam, I would flesh out my
language a little bit here to make it clear, but
hopefully that makes sense.