# How to determine the isotherms of given temperature field?

1. Oct 11, 2013

### Megatron16

How can I determine the isotherms(curves of constant temperature) of given Temperature field like
T = x^2 -y^2 + 8y.
Also how can I draw at least five isotherms of the above temperature field?

2. Oct 11, 2013

### Staff: Mentor

Pick at least five different values for T, draw solutions to the equation.
That's the way to get all isotherms. Consider T as constant, find a relation between x and y. Your equation is such a relation, but you can express it in a nicer way.

I don't see a differential equation here.

3. Oct 11, 2013

### HallsofIvy

Use the definition of "isotherm". For a given temperature, C, you are points where the temperature is C: $x^2- y^2+ 8y= C$. Now, if you are asking "how can I write that as a function I know how to graph quickly?", the answer is, as so often for things like this, complete the square.

8/2= 4 and $4^2= 16[/tex] so $$y^2- 8y= y^2- 8y+ 16- 16= (y- 4)^2- 16$$ We can write [itex]x^2- y^2+ 8y= x^2- (y^2+ 8y)= x^2- ((y- 4)^2- 16)= x^2- (y- 4)^2- 16$ so that $x^2- y^2+ 8y= C$ becomes $x^2- (y- 4)^2- 16= C$ or $x^2- (y- 4)^2= C+ 16$. To get that to "standard form", divide through by C+ 16:
$$\frac{x^2}{C+ 16}- \frac{(y- 4)^2}{C+ 16}= 1$$

That is an equilateral hyperbola with center at (0, 4), vertices at $$(\sqrt{C+ 16}, 4)$$ and [tex](-\sqrt{C+ 16}, 4), and asymptotes y= x+ 4 and y= -x+ 4.

4. Oct 11, 2013

### Staff: Mentor

It is easier to find a function x=f(y).

5. Oct 12, 2013

### HallsofIvy

Not when there is NO such function, as here.

6. Oct 13, 2013

### Staff: Mentor

... or two functions.