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Current time:0:00Total duration:6:08

Video transcript

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 at time T minutes can be modeled by the function H that satisfies the differential equation D H DT the derivative of our internal temperature with respect to time is equal to negative 1/4 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 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 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 if the ambient room temperature is 27 degrees Celsius I'll just draw there that's what the room temperature is doing and so we know at T equals 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 and when its temperature when there's a big difference between the potato in the room maybe its rate of change is steeper than when there's a little difference so it would you would expect the graph to look something like this so you would expect it to look something like this and then asymptotes 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 well 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 asked them to 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 0 use this equation to approximate the internal temperature of the potato at time T equals 3 so what are we going to do well we're going to think about what's going on at time T equals 0 right over here we want the equation of the tangent line which might look something like this at T equals 0 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 dhdt times T plus our y-intercept where does it intersect the y axis here well when T is equal to 0 the value the value of this equation is going to be 91 because it intersects our graph right at that point that point 0 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 one-fourth 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 0 I'll write it one more time it is a mini drumrolls 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 3 so let's say that this is time T equals 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 equals 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 is equal to 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 is equal to 3
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