AP Calculus AB 2017 free response
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2017 AP Calculus AB/BC 4a
- [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.