Question #84622

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Question #84622

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Answer 1 · 2017-03-24T21:14:53

Define the function $T (x)$ $T(x)$ as the temperature $T$ $T$ in degrees Celsius at $x$ $x$ centimeters away from the heated end.

The prompt tells us that $\frac{d T}{d x} = - 6$ $("d"T)/("d"x)=-6$ and that halfway along the $60$ $60$ centimeter long rod, the temperature is $290$ $290$ . That is, $T (30) = 290$ $T(30)=290$ .

We can solve for $T$ $T$ from $\frac{d T}{d x} = - 6$ $("d"T)/("d"x)=-6$ by separating the variables. Treating $\frac{d T}{d x}$ $("d"T)/("d"x)$ like a quotient, we can say that:

$d T = - 6$ $"d"T=-6$ $d x$ $"d"x$

Then integrating:

$\int d T = - 6 \int d x$ $int"d"T=-6int"d"x$

Which, adding a constant of integration, gives:

$T (x) = - 6 x + C$ $T(x)=-6x+C$

Use the original condition $T (30) = 290$ $T(30)=290$ to solve for $C$ $C$ :

$290 = - 6 (30) + C$ $290=-6(30)+C$

$C = 470$ $C=470$

$T (x) = - 6 x + 470$ $T(x)=-6x+470$

Then, the temperature difference between the two ends of the rod is given by $T (0) - T (60) = 470 - (- 6 (60) + 470) = 360^{\circ} C$ $T(0)-T(60)=470-(-6(60)+470)=360^@"C"$ .

In fact, to find the difference between the two ends, doing the actual integration isn't necessary. The differential $\frac{d T}{d x} = - 6$ $("d"T)/("d"x)=-6$ means that for every centimeter $x$ $x$ traveled, the temperature $T$ $T$ decreases constantly by $6$ $6$ degrees. Thus, if the rod is $60$ $60$ centimeters long, it will decrease $6 \times 60 = 360^{\circ} C$ $6xx60=360^@"C"$ in total.

In part B, we see that the rate of change described is again $\frac{d T}{d x}$ $("d"T)/("d"x)$ . It's proportional to $x$ $x$ with some proportionality constant $k$ $k$ , which is written as:

$\frac{d T}{d x} = k x$ $("d"T)/("d"x)=kx$

Which can be solved by again separating variables:

$d T = k x$ $"d"T=kx$ $d x$ $"d"x$

$\int d T = k \int x$ $int"d"T=kintx$ $d x$ $"d"x$

$T (x) = k (\frac{1}{2} x^{2}) + C$ $T(x)=k(1/2x^2)+C$

With this model, we see that $T (0) = 380$ $T(0)=380$ and $T (60) = 20$ $T(60)=20$ , which we can use to solve for $k$ $k$ and $C$ $C$ :

Using $T (0) = 380$ $T(0)=380$ shows that

$380 = k (\frac{1}{2} (0^{2})) + C$ $380=k(1/2(0^2))+C$

$380 = C$ $380=C$

So:

$T (x) = \frac{k x^{2}}{2} + 380$ $T(x)=(kx^2)/2+380$

Then using $T (60) = 20$ $T(60)=20$ :

$20 = \frac{k (60^{2})}{2} + 380$ $20=(k(60^2))/2+380$

$- 360 = 1800 k$ $-360=1800k$

$k = - \frac{1}{5}$ $k=-1/5$

So:

$T (x) = - \frac{x^{2}}{10} + 380$ $T(x)=-x^2/10+380$

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Question #84622

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