Question 3
(a)-Topic-7.3 Sketching Slope Fields
(b)-Topic-7.4 Reasoning Using Slope Fields
(c)-Topic-7.8 Exponential Models with Differential Equations
(d)-Topic-7.6 Finding General Solutions Using Separation of Variables
3. A bottle of milk is taken out of a refrigerator and placed in a pan of hot water to be warmed. The increasing function M models the temperature of the milk at time t, where M(t) is measured in degrees Celsius (°C) and t is the number of minutes since the bottle was placed in the pan. M satisfies the differential equation \(\frac{dM}{dt}=\frac{1}{4}(40-M)\) At time t = 0, the temperature of the milk is 5°C. It can be shown that M(t) < 40 for all values of t.
(a) A slope field for the differential equation \(\frac{dM}{dt}=\frac{1}{4}(40-M)\) is shown. Sketch the solution curve through the point (0, 5).
(b) Use the line tangent to the graph of M at t = 0 to approximate M(2), the temperature of the milk at time t = 2 minutes.
(c) Write an expression for \(\frac{d^{2}M}{dt^{2}}\) in terms of M . Use \(\frac{d^{2}M}{dt^{2}}\) to determine whether the approximation from part (b) is an underestimate or an overestimate for the actual value of M(2). Give a reason for your answer.
(d) Use separation of variables to find an expression for M(t), the particular solution to the differential equation \(\frac{dM}{dt}=\frac{1}{4}(40-M)\) with initial condition M(0) = 5.
▶️Answer/Explanation
3(a) A slope field for the differential equation \(\frac{dM}{dt}=\frac{1}{4}(40-M)\) is shown. Sketch the solution curve through the point (0, 5) .
3(b) Use the line tangent to the graph of M at t = 0 to approximate M (2 ,) the temperature of the milk at time t = 2 minutes.
\(\frac{DM}{dt} |_{t=0}=\frac{1}{4}(40-5)=\frac{35}{4}\)
The tangent line equation is\(y=5+\frac{35}{4}(t-0)\).
\(M(2)\approx 5+\frac{35}{4}.2=22.5\)
The temperature of the milk at time t = 2 minutes is approximately \(22.5^{\circ}\) Celsius.
3(c) Write an expression for \(\frac{d^{2M}}{dt^{^{2}}}\) in terms of M. Use \(\frac{d^{2M}}{dt^{^{2}}}\) to determine whether the approximation from part (b) is an underestimate or an overestimate for the actual value of M (2 .) Give a reason for your answer.
\(\frac{d^{2M}}{dt^{^{2}}}=\frac{1}{4}\frac{DM}{dt}=\frac{1}{4}\left ( \frac{1}{4}(40-M) \right )=\frac{1}{16}(40-M)\)
Because \(M(t)< 40,\frac{d^{2}M}{dt^{2}}< 0\) , so the graph of M is concave down. Therefore, the tangent line approximation of M (2) is an overestimate.
3(d) Use separation of variables to find an expression for M (t ), the particular solution to the differential equation \(\frac{DM}{dt}=\frac{1}{4}(40-M)\) with initial condition M (0 )=5.
\(\frac{dm}{40 -M} = \frac{1}{4}dt\)
\(\int \frac{dm}{40 -M} = \int \frac{1}{4}dt\)
\(-In|40 – M| = \frac{1}{4}t +C\)
\(-In |40 – 5| = 0+C \Rightarrow C = -In 35\)
\(M(t) < 40 \Rightarrow 40 – M > 0 \Rightarrow |40 -M| = 40 -M\)
\(-In(40-M)=\frac{1}{4}t-In35\)
\(In(40-M)= -\frac{1}{4}t-In35\)
\(40-M= 35e^{-t/4}\)
\(M=40- 35e^{-t/4}\)