Question
At time t = 0, 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 (°C) at time t = 0, and the internal temperature of the potato is greater than 27 C° for all times t > 0. The internal temperature of the potato at time t minutes can be modeled by the function H that satisfies the differential equation \(\frac{dH}{dt}=-\frac{1}{4}(H-27),\) where H(t) is measured in degrees Celsius and H(0) = 91 .
(a) Write an equation for the line tangent to the graph of H at t = 0. Use this equation to approximate the internal temperature of the potato at time t = 3.
(b) Use \(\frac{d^{2}H}{dt^{2}}\) to determine whether your answer in part (a) is an underestimate or an overestimate of the
internal temperature of the potato at time t = 3.
(c) For t < 10, an alternate model for the internal temperature of the potato at time t minutes is the function G that satisfies the differential equation \(\frac{dG}{dt}= -(G-27)^{2/3},\) where G (t) is measured in degrees Celsius and G(0) = 91. Find an expression for G(t). Based on this model, what is the internal temperature of the potato at time t = 3 ?
Answer/Explanation
Ans:
(a) \(H'(0)=-\frac{1}{4}(91-27)=-16\)
H(0) = 91
An equation for the tangent line is y = 91 – 16 . The internal temperature of the potato at time t = 3 minutes is approximately 91 – 16 ⋅ 3 = 43 degrees Celsius.
(b) \(\frac{d^{2}H}{dt^{2}}=-\frac{1}{4}\frac{dH}{dt}=\left ( -\frac{1}{4} \right )\left ( -\frac{1}{4} \right )(H-27)=\frac{1}{16}(H-27)\)
\(H > 27 for t>0\Rightarrow \frac{d^{2}H}{dt^{2}}=\frac{1}{16}(H-27)>0 for t>0\)
Therefore, the graph of H is concave up for t > 0. Thus, the answer in part (a) is an underestimate.
(c) \(\frac{dG}{(G-27)^{2/3}}=-dt\)
\(\int \frac{dG}{(G-27)^{2/3}}=\int (-1)dt\)
3(G-27)1/3 = -t + C
3(91-27)1/3 = 0 + C ⇒ C = 12
3(G-27)1/3 = 12 – t
\(G(t)=27 +\left ( \frac{12-t}{3} \right )^{3}for 0\leq t\leq 10\)
The internal temperature of the potato at time t = 3 minutes is
\(27 +\left ( \frac{12-t}{3} \right )^{3}=54\) degrees Celsius.