2017 AP Calculus AB/BC 4a | AP Calculus AB solved exams | AP Calculus AB | Khan Academy

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 will 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 ( \frac{dH}{dt} ). The derivative of our internal temperature with respect to time is equal to negative ( \frac{1}{4} ) times ( R ), the difference between our internal temperature and the ambient room temperature, where ( H(t) ) is measured in degrees Celsius and ( H(0) ) is equal to 91.

So before I even read Part A, let's just make sense of what this differential equation is telling us. So let's see if it's consistent with our intuition. Let me draw some axes here. So this is my y-axis, and this right over here is my ( t ) axis. Now, if the ambient room temperature is 27 degrees Celsius, I'll just draw there; that's what the room temperature is doing.

We know at ( T = 0 ), our potato is at 91 degrees. So 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. 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. You would expect it to look something like this and then asymptote towards a temperature of 27 degrees Celsius. This is what you would expect to see, and this differential equation is consistent with that.

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 ( \frac{1}{4} ). You're constantly going to have a negative rate of change, which makes sense: the potato is cooling down.

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 = 0 ). Use this equation to approximate the internal temperature of the potato at time ( T = 3 ).

So what are we going to do? Well, we're going to think about what's going on at time ( T = 0 ), right over here. We want the equation of the tangent line, which might look something like this at ( T = 0 ). So this thing would be of the form ( Y = ) the slope of the equation of the tangent line. Well, it would be the derivative of our function at that point, so ( \frac{dH}{dt} ) times ( T ) plus our y-intercept.

Where does it intersect the y-axis here? Well, when ( T ) is equal to 0, the value of this equation is going to be 91 because it intersects our graph right at that point ( (0, 91) ). So, what is our derivative of ( H ) with respect to ( T ) at time ( T = 0 ), right at this point right over here? Well, we just have to look at this. You could also write this as ( H' (T) ) right over here.

So if we want to think about ( H' (0) ), that's going to be equal to negative one-fourth times ( H(0) - 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 - 27 = 64 ).

( 64 ) times negative ( \frac{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 = 0 ). I'll write it one more time: it is, a mini drumroll here, ( Y = -16T + 91 ).

That's the equation of that tangent line right over there. Then they say we want to use this equation to approximate the internal temperature of the potato at time ( T = 3 ). So let's say that this is time ( T = 3 ) right over here. We want to approximate the temperature that this model describes, right over here, but we're going to do it using the line.

So we're going to evaluate the line at ( T = 3 ). So then we would get, let's see, negative ( 16 ) times ( 3 ) 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 = 0 ), 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 = 3 ).