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## AP Calculus AB 2017 free response

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# 2017 AP Calculus AB/BC 4a

## Video transcript

- [Instructor] We are now
going to cover the famous or perhaps infamous potato problem from the 2017 AP Calculus exam. At time t equals zero, a boiled potato is taken
from a pot on a stove and left to cool in a kitchen. The internal temperature of the potato is 91 degrees Celsius
at time t equals zero, and the internal temperature
of the potato is greater than 27 degrees for all times
t greater than zero. I would guess that the
ambient room temperature is 27 degrees Celsius. And so that's why the
temperature would approach this, but it would always stay a
little bit greater than that, as t gets larger and larger. The internal temperature of the potato at time t minutes can be
modeled by the function H that satisfies the
differential equation dH/dt, the derivative of our internal temperature with respect to time, is equal
to negative 1/4 times our, the difference between
our internal temperature and the ambient room temperature, where H of t is measured
in degrees Celsius and H of zero is equal to 91. So before I even read part
(a), let's just make sense of what this differential
equation is telling us, and let's see if it's
consistent with our intuition. So let me draw some axes here. So this is my y-axis. So that is my y-axis. And this right over here is my t-axis. Now the ambient room temperature
is 27 degrees Celsius, so I'll just draw there. That's what the room temperature is doing. And so we know at t equals zero, our potato is at 91 degrees. So let's see, that's 27, 91
might be right over there. 91, this is all in degrees Celsius. And what you would expect
intuitively is that it would start to cool and when its temperature, when there's a big difference between the potato and the room, maybe its rate of change is steeper than when there's a little difference. So you would expect the graph
to look something like this. So you would expect it to look something like
this and then asymptote towards a temperature
of 27 degrees Celsius. So this is what you would expect to see, and this differential equation
is consistent with that, where the rate of change, notice this is for all
t greater than zero, this is going to be a negative value because our potato is greater than 27 for t greater than zero. So this part here is going to be positive, but then you multiply
positive times negative 1/4. You're constantly going to
have a negative rate of change, which makes sense. The potato is cooling down. And it also makes sense
that your rate of change is proportional to the difference
between the temperature of the potato and the
ambient room temperature. When there's a big difference, you expect a steeper rate of change. But then when there's
less of a difference, the rate of change, you could imagine, becomes less and less and less negative, as we asymptote towards
the ambient temperature. So with this out of the way,
now let's tackle part (a). Write an equation for the
line tangent to the graph of H at t equals zero. Use this equation to approximate
the internal temperature of the potato at time t equals three. So what are we going to do? Well, we're gonna think about what's going on at time t equal zero, right over here. We want the equation of the tangent line, which might look something like this, at t equals zero. So this thing would be
of the form y is equal to the slope of the equation
of the tangent line. Well, it would be the
derivative of our function at that point, so dH/dt, times t, plus our y-intercept. Where does it intersect the y-axis here? Well, when t is equal to zero, the value of this
equation is going to be 91 because it intersects our
graph right at that point, that point zero comma 91, plus 91. So what is our derivative
of H with respect to t at time t equals zero, right
at this point right over here? Well, we just have to look at this. You could also write this as
H prime of t right over here. So if we want to think about H prime of zero, that's going to be equal to negative 1/4 times H of zero minus 27. What is our initial temperature minus 27? This is, of course, 91 degrees. They tell us that multiple times. We've even drawn it a few times. 91 minus 27 is 64. 64 times negative 1/4
is equal to negative 16. So this is negative 16 right over here. So just like that, we have the equation for the line tangent to the graph of H at t is equal to zero. I'll write it one more time. It is, do a mini drum roll here, y is equal to negative 16 t plus 91. That's the equation of that
tangent line right over there. And then they say we
want to use this equation to approximate the internal
temperature of the potato at time t equals three. So let's say that this is time t equals three right over here. We want to approximate the temperature that this model describes, right over here, but we're
gonna do it using the line. So we're gonna evaluate
the line at t equals three. So then we would get, let's see, negative 16 times three plus 91 is equal to, this is negative 48 plus 91 is equal to, what is that? 43, so this is equal to 43 degrees Celsius. So this right over here is the equation for the line tangent to the graph of H at t is equal to zero, and this right over here
is our approximation using that equation of the tangent line of the internal temperature
of the potato at time t is equal to three.