Now if 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 it's 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 4th, 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 can imagine, becomes less and less and less negative as we asymptote towards the ambient temperature.