Course:MATH102/Question Challenge/2006 December Q10

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Question

Consider Newton's Law of cooling

$\frac{dT}{dt}=2-\frac{1}{5}T$

with initial temperature $T(0)=37$.

(a) Find values of the constants $a,b,k$ such that

$T(t)=a+b{e}^{-kt}$

is a solution to the initial value problem given above.

(b) Using the solution obtained in (a), find the time $\tau$ at which $T(\tau )=13$. Express your answer in terms of $m$, where $m=\ln (3)$.

(c) What is the steady state of this differential equation? Is it stable?

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Hint
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Solutions

Solution

$dT/dt=2-\frac{1}{5}T$ The solution to the differential equation is given by: $T(t)=\frac{2}{1/5}+(y_0-\frac{2}{1/5})e^{-t/5}$ $T(t)=10+27e^{t/5}$ To find the time at which $T(t)=0$ we will isolate for $t$: $13=10+27e^{-t/5}$ $3=27e^{-t/5}$ $\frac{3}{27}=e^{-t/5}$ $\frac{1}{9}=e^{-t/5}$ $9=e^{t/5}$ $\ln 9=t/5$ $2\ln 3=2m=t/5$ $t=10m$ To find the steady state of the differential equation we will set $dy/dx=0$: $0=2-\frac{1}{5}T$ $\frac{1}{5}T=2$ $T=10$ To find the stability of this steady state we can draw a phase diagram. It will look like a linear equation with a y-intercept of 2, an x-intercept of 10, and a slope of -1/5. The arrows on the x-axis will indicate that when to the left of the steady state the function is positive, and when to the right of the steady state the function is negative. Therefore any values less then or greater than the steady state will tend to go towards it over time. This would describe the steady state as being stable. Part A) $a=10$ , $b=27$ , $k=-1/5$ Part B) The time at which $T(t)=13$ is at time $10ln(3)$ , or $10m$ Part C) There is a steady state when $T(t)=10$. This steady state is stable.