Q. 47 - Calculus

Question

Dad’s casserole surprise is hot out of the oven, and its temperature after t minutes is given by the function T(t), measured in degrees. The casserole cools by changing at a rate of $\mathrm{T}\text{'}\left(\mathrm{t}\right)=-15 {\mathrm{e}}^{-0.5\mathrm{t}}.$degrees per minute.

Rate of change of casserole temperature $\mathrm{T}\text{'}\left(\mathrm{t}\right)=-15 {\mathrm{e}}^{-0.5\mathrm{t}} .$

(a) Discuss what the graph of T'(t) says about the behavior of the casserole’s temperature after the casserole is taken out of the oven. 

(b) Just as we can approximate distance travelled by approximating the area under the corresponding velocity curve, we can approximate the change in temperature T of the casserole by approximating the area under the graph of T' . Why? 

(c) Estimate the change in temperature of the casserole over the first 5 minutes it is out of the oven, using any Riemann sum you like, with n = 10. 

Step-by-Step Solution

Answer

$\mathrm{The} \mathrm{change} \mathrm{is} \boxed{\approx -24.2385}$

Step 1. Given information is .

$$\begin{gathered} \mathrm{Rate} \mathrm{of} \mathrm{change} \mathrm{of} \mathrm{casserole} \mathrm{temperature} \\ \mathrm{T}\text{'}\left(\mathrm{t}\right)=-15 {\mathrm{e}}^{-0.5\mathrm{t}}. \end{gathered}$$

Step 2. Part (a)

$$\begin{gathered} \mathrm{The} \mathrm{objective} \mathrm{is} \mathrm{to} \mathrm{discuss} \mathrm{about} \mathrm{the} \mathrm{behavior} \mathrm{of} \mathrm{the} \mathrm{graph} \mathrm{of} \mathrm{T}\text{'}\left(\mathrm{t}\right) \mathrm{after} \mathrm{the} \mathrm{casserole} \mathrm{is} \mathrm{taken} \mathrm{out} \mathrm{of} \mathrm{the} \mathrm{oven}. \\ \mathrm{The} \mathrm{graph} \mathrm{shows} \mathrm{an} \mathrm{exponential} \mathrm{increasing} \mathrm{function} \mathrm{and} \mathrm{reaches} \mathrm{almost} 0 \mathrm{with} \mathrm{increasing} \mathrm{time}. \end{gathered}$$

Step 3. Part (b)

$$\begin{gathered} \mathrm{The} \mathrm{objective} \mathrm{is} \mathrm{to} \mathrm{determine} \mathrm{why} \mathrm{the} \mathrm{rate} \mathrm{of} \mathrm{change} \mathrm{in} \mathrm{temperature} \mathrm{T} \mathrm{can} \mathrm{be} \mathrm{approximated} \mathrm{by} \mathrm{the} \mathrm{area} \mathrm{under} \mathrm{the} \mathrm{graph} \mathrm{of} \mathrm{T}\text{'}. \\ \mathrm{The} \mathrm{rate} \mathrm{of} \mathrm{change} \mathrm{in} \mathrm{T} \mathrm{is} \mathrm{given} \mathrm{by} \mathrm{T}\text{'}. \\ \mathrm{So}, \mathrm{by} \mathrm{finding} \mathrm{the} \mathrm{area} \mathrm{under} \mathrm{the} \mathrm{graph} \mathrm{ofT} \mathrm{the} \mathrm{T}\text{'} \mathrm{can} \mathrm{be} \mathrm{found} \mathrm{out}. \end{gathered}$$

Step 4. Part (c)

$$\begin{gathered} \mathrm{The} \mathrm{objective} \mathrm{is} \mathrm{to} \mathrm{approximate} \mathrm{the} \mathrm{change} \mathrm{in} \mathrm{temperature} \mathrm{of} \mathrm{the} \mathrm{casserole} \mathrm{over} \mathrm{the} \mathrm{first} 5 \\ \mathrm{minutes} \mathrm{if} \mathrm{it} \mathrm{is} \mathrm{out} \mathrm{of} \mathrm{the} \mathrm{oven} \mathrm{with} \mathrm{n} = 10 . \\ \mathrm{The} \mathrm{interval} \mathrm{is} [0,5]. \mathrm{The} \mathrm{right} \mathrm{sum} \mathrm{defined} \mathrm{for} \mathrm{n} \mathrm{rectangles} \mathrm{on} [\mathrm{a},\mathrm{b}] \mathrm{is} \sum _{\mathrm{k}=1 }^{\mathrm{n}}\mathrm{f}\left({\mathrm{x}}_{\mathrm{k}}\right)∆\mathrm{x}. \\ \mathrm{Where}, ∆\mathrm{x}= \frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}},{\mathrm{x}}_{\mathrm{k}}=\mathrm{a}+\mathrm{k}∆\mathrm{x}. \\ \mathrm{Now}, ∆\mathrm{x}= \frac{5-3}{4} =\frac{1}{2}. \\ \mathrm{So}, {\mathrm{x}}_{\mathrm{k}}= 0+\mathrm{k}\frac{1}{2}= \frac{\mathrm{k}}{2}. \\ \mathrm{The} \mathrm{right} \mathrm{sum} \mathrm{is}, \\ \sum _{\mathrm{k}=1}^{10}\left(-15 {\mathrm{e}}^{-0.5\frac{\mathrm{k}}{2}}\right)\frac{1}{2} \\ = \frac{1}{2}\left(-15 {\mathrm{e}}^{-0.5\frac{1}{2}}-15 {\mathrm{e}}^{-0.5\frac{2}{2}}-15 {\mathrm{e}}^{-0.5\frac{3}{2}}-15 {\mathrm{e}}^{-0.5\frac{4}{2}}-15 {\mathrm{e}}^{-0.5\frac{5}{2}}-15 {\mathrm{e}}^{-0.5\frac{6}{2}}-15 {\mathrm{e}}^{-0.5\frac{7}{2}}-15 {\mathrm{e}}^{-0.5\frac{8}{2}}-15 {\mathrm{e}}^{-0.5\frac{9}{2}}-15 {\mathrm{e}}^{-0.5\frac{10}{2}}\right) \\ \approx -24.2385 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{change} \mathrm{is} \boxed{\approx -24.2385}. \end{gathered}$$