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)
\(\frac{dH}{dt}_{H=91} = -\frac{1}{4}(91-27)=\left ( -\frac{1}{4} \right )(64)=-16\)
A(t) = -16 (t-0) + 91
A(t) = -16t + 91
A(3) = – 16(3) + 91
= -48 + 91
430C
at t = 3 minutes
(b)
\(\frac{dH}{dt} = -\frac{1}{4}(H-27)\) H > 27 for all t > 0
\(\frac{d^{2}H}{dt^{2}} = -\frac{1}{4}\left ( \frac{dH}{dt} \right )\) \(\frac{d^{2}H}{dt^{2}}\) is always positive,
\(-\frac{1}{4}\left ( -\frac{1}{4} \right )(H-24)\) So part (a) is an underestimate.
\(\frac{d^{2}H}{dt^{2}} = \frac{1}{16}(H-24)\)
(c)
\(\frac{dG}{dt} = -(G-27)^{2/3}\)
\(\int \frac{dG}{(G-27)^{2/3}}=\int -1dt\)
\(\int (G-27)^{-2/3}dt = -t+c\)
3 (G-27) 1/3 = -t + c1
\((G-27)^{-2/3} = -\frac{t}{3}+c_{2}\)
\(G-27 = \left ( -\frac{t}{3}+c_{2} \right )^{3}\)
\(G = \left ( -\frac{t}{3}+c_{2} \right )^{3}+27\)
at (0, 91)
\(91= (0+c_{2})^{3}+27\)
91 = (c2)3 + 27
64 = (c2)3 →c2 = 4
\(G(t)= \left ( -\frac{t}{3}+4 \right )^{3}+27\)
\(G(3)= \left ( \frac{-3}{3}+4 \right )^{3}+27\)
= (3)3 + 27
= 27 + 27
G(3) = 540C