Question

Question #84622

Answer 1

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

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

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

`"d"T=-6` `"d"x`

Then integrating:

`int"d"T=-6int"d"x`

Which, adding a constant of integration, gives:

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

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

`290=-6(30)+C`

`C=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^@"C"`.

In fact, to find the difference between the two ends, doing the actual integration isn't necessary. The differential `("d"T)/("d"x)=-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 `6xx60=360^@"C"` in total.

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

`("d"T)/("d"x)=kx`

Which can be solved by again separating variables:

`"d"T=kx` `"d"x`

`int"d"T=kintx` `"d"x`

`T(x)=k(1/2x^2)+C`

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

Using `T(0)=380` shows that

`380=k(1/2(0^2))+C`

`380=C`

So:

`T(x)=(kx^2)/2+380`

Then using `T(60)=20`:

`20=(k(60^2))/2+380`

`-360=1800k`

`k=-1/5`

So:

`T(x)=-x^2/10+380`