# Question #84622

Mar 24, 2017

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

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

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

$\text{d} T = - 6$ $\text{d} x$

Then integrating:

$\int \text{d"T=-6int"d} x$

Which, adding a constant of integration, gives:

$T \left(x\right) = - 6 x + C$

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

$290 = - 6 \left(30\right) + C$

$C = 470$

$T \left(x\right) = - 6 x + 470$

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

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

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

$\left(\text{d"T)/("d} x\right) = k x$

Which can be solved by again separating variables:

$\text{d} T = k x$ $\text{d} x$

$\int \text{d} T = k \int x$ $\text{d} x$

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

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

Using $T \left(0\right) = 380$ shows that

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

$380 = C$

So:

$T \left(x\right) = \frac{k {x}^{2}}{2} + 380$

Then using $T \left(60\right) = 20$:

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

$- 360 = 1800 k$

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

So:

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