Science:Math Exam Resources/Courses/MATH102/December 2016/Question A 05

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MATH102 December 2016

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Question A 05

An object is placed on the kitchen table at time t =0(t in minutes). Let A be the average rate of change of the temperature of the object over the interval $5\leq t\leq 12$. Let $T(t)$ denote the temperature of the object at time t. Which of the following statements is true? (i) If the object is a pizza that was taken out of the oven, then $T'(5)>A$. (ii) If the object is a bottle of milk that was taken out of the fridge, then $|T'(5)|>|A|$. (iii) If the object is a pizza that was taken out of the oven, then $|T'(12)|>|A|$. (iv) If the object is a bottle of milk that was taken out of the fridge, then $T'(12)>A$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

The average rate of change $A={\frac {\Delta T}{\Delta t}}={\frac {T(12)-T(5)}{12-7}}$ is the slope of the secant line through the points $(5,T(5))$ and $(12,T(12))$ , whereas $T'(5)$ and $T'(12)$ represent the slope of the tangent line at the points $(5,T(5))$ and $(12,T(12))$ respectively. Recall that the (instantaneous) rate of change of the temperature of an object is given by $T'(t)=k(E-T)$ , where $E$ is the ambient temperature and k is a positive constant.

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Solution

Note that since $T'(t)=k(E-T)$ we have $T''(t)=k(E-T)'=k(-T')=-k^{2}(E-T)$. This will help us to determine the concavity and therefore, the comparison between slope of the tangent line and the slope of the secant line. 1st case: A pizza is out of the oven, so it is cooling down. i.e., the temperature is decreasing. This implies that the slope of the tangent line at $t=5$ and $t=12$, and the slope of the secant line through the two endpoints are negative. Since in this case $E<T$, $T''(t)>0$, if we sketch an exponential type graph which is concave up and decreasing, we see that the absolute value of the slope of the tangent line at $(5,T(5))$ is greater than the absolute value of the slope of the secant line i.e. $|T'(5)|>|A|$, however, the slope itself is more negative than the secant line i.e. $T'(5)<A$, similarly we have $|T'(12)|<|A|$, while $T'(12)>A$ 2nd case: A milk is out of the fridge, so it is heating up. i.e, the temperature is increasing. This implies that the slope of the tangent line at $t=5$ and $t=12$, and the slope of the secant line through the two endpoints are positive. Since in this case $E>T$, $T''(t)<0$, this means that an exponential type graph which is concave down and increasing, so the slope of the tangent line at $(5,T(5))$ is greater than the slope of the secant line i.e. $T'(5)>A$, similarly we have $T'(12)<A$. The only choice matches with these scenarios is $\color {blue}(ii)$ .

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