Q. 70 - Calculus

Question

 Dad’s casserole surprise is for dinner again, and from previous experiments we know that after the casserole is taken out of the oven, the temperature of the casserole will cool by changing at a rate of T (t) =$-15e^{-0.5t}$ degrees Fahrenheit per minute.

(a) Calculate T (t) dt. Then, given that the casserole was T₀= 350 degrees when it was removed from the oven, find a formula for the temperature T(t) of the casserole after t minutes.

 (b) Calculate T(5) − T(0). Compare your answer with the answer to the last part of Exercise 47 from Section 4.2. Write a few sentences about the connection that this comparison suggests between Riemann sums and antidifferentiation

Step-by-Step Solution

Answer

Part(a) $7.5e^{-0.5t}$

Part(b) $T\left(5\right) -T\left(0\right)=-6.88$

Step 1. Given

The temperature of the casserole will cool by changing at a rate of T (t) = $-15e^{-0.5t}$degrees Fahrenheit per minute. 

Part(a) Step 2. Calculation T(t) and find the formula for the temperature.

$$\begin{gathered} \int T\left(t\right)dt \\ =\int -15e^{-0.5t}dt \\ =-15\int e^{-0.5t}dt \\ =-15e^{-0.5t}\times -0.5 \\ =7.5e^{-0.5t} \\ \mathrm{The} \mathrm{formula} \mathrm{for} \mathrm{temperature} \mathrm{is} 7.5e^{-0.5t} \end{gathered}$$

Part(b) Step 3. Calculation t(5)-t(0)

$$\begin{gathered} T\left(5\right) =7.5e^{-0.5*5} \\ =0.615 \\ T\left(0\right)=7.5e^{-0.5*0} \\ =7.5 \\ T\left(5\right) -T\left(0\right)=0.615-7.5 \\ -6.88 \end{gathered}$$