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 4th times r, 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.