dT/dt = - k (T - Tm) where k is a positive constant and Tm is the constant tempe
ID: 2973719 • Letter: D
Question
dT/dt = - k (T - Tm) where k is a positive constant and Tm is the constant temperature of the surroundings. Assume Tm >0 . Is the DE autonomous? Justify your answer. Now you will sketch the phase diagram associated with the DE. Draw the vertical and horizontal axes and label them appropriately. Now sketch the phase diagram. What shape is the curve? What is the intercept on the horizontal axis? Now you will begin to sketch a solution curve in the plane with t along the horizontal axis and T for the vertical axis. First sketch and label any equilibrium solutions. Next pick some value of T, say T1, between 0 and Tm and label that point on both the phase diagram and the solution curve graph. For that value, T1, is dT/dt positive or negative? Draw a short tangent line segment with an approximate and relevant slope on the solution curve that would go through the point you labeled on the solution curve graph. If there are equilibrium solutions, are they stable or unstable? Justify your answer. Is the solution curve you drew for NewtonExplanation / Answer
Starting from dT/(T^4 - (T_m)^4) = k dt:
Let a = T_m for simplicity.
By partial fractions,
1/(T^4 - a^4) = A/(T - a) + B/(T + a) + (CT + D)/(T^2 + a^2).
Clearing denominators:
1 = A(T + a)(T^2 + a^2) + B(T - a)(T^2 + a^2) + (CT + D)(T^2 - a^2)
Letting T = a yields 1 = (2a^3)A ==> A = 1/(2a^3)
Letting T = -a yields 1 = (-2a^3)B ==> B = -1/(2a^3)
Letting T = ai yields 1 = (Cai + D)(-2a^2) ==> C = 0 and D = -1/(2a^2).
Hence, (1/(2a^3)) [1/(T - a) - 1/(T + a) - a/(T^2 + a^2)] = k dt
==> 1/(T - a) - 1/(T + a) - a/(T^2 + a^2) = 2ka^3 dt.
Integrate both sides:
ln(T - a) - ln(T + a) - arctan(T/a) = 2ka^3 t + C.
==> ln [(T - a)/(T + a)] - arctan(T/a) = 2ka^3 t + C.
Assuming that T(0) = T (initial temperature), we find that
ln(T - a)/(T + a)] - arctan(T/a) = C.
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For your second part:
Assume that T - T_m is negligible compared to T_m itself.
Working with the original DE,
dT/dt = k(T^4 - (T_m)^4)
.........= k[((T - T_m) + T_m)^4 - (T_m)^4], expanding in powers of T - T_m
.........= k[(T - T_m)^4 + 4T_m (T - T_m)^3 + 6(T_m)^2 (T - T_m)^2 + 4(T_m)^3 (T - T_m)]
Any power of T - T_M is especially negligible, so ignoring these terms yields
dT/dt = k[0 + 4(T_m)^3 (T - T_m)]
........= (4k(T_m)^3) * (T - T_m),
which is essentially Newton's Law of cooling with constant 4k(T_m)^3.
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