How to determine the isotherms of given temperature field?

[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?

 

[mfb]

Also how can I draw at least five isotherms of the above temperature field?

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.

 

[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 [itex]4^2= 16[/tex] so $$y^2- 8y= y^2- 8y+ 16- 16= (y- 4)^2- 16$$<br /> <br /> We can write $x^2- y^2+ 8y= x^2- (y^2+ 8y)= x^2- ((y- 4)^2- 16)= x^2- (y- 4)^2- 16$ 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:<br /> $$\frac{x^2}{C+ 16}- \frac{(y- 4)^2}{C+ 16}= 1$$<br /> <br /> That is an equilateral hyperbola with center at (0, 4), vertices at $$(\sqrt{C+ 16}, 4)$$ and $$(-\sqrt{C+ 16}, 4), and asymptotes y= x+ 4 and y= -x+ 4.$$[/itex]

 

[mfb]

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.

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

 

[HallsofIvy]

Not when there is NO such function, as here.

 

[mfb]

... or two functions.