- #1

- 4

- 0

T = x^2 -y^2 + 8y.

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Megatron16
- Start date

- #1

- 4

- 0

T = x^2 -y^2 + 8y.

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

- #2

mfb

Mentor

- 35,392

- 11,743

Pick at least five different values for T, draw solutions to the equation.Also how can I draw at least five isotherms of the above temperature field?

That's the way to get all isotherms. Consider T as constant, find a relation between x and y. Your equation

I don't see a differential equation here.

- #3

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

8/2= 4 and [itex]4^2= 16[/tex] so [tex]y^2- 8y= y^2- 8y+ 16- 16= (y- 4)^2- 16[/tex]

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[/itex] so that [itex]x^2- y^2+ 8y= C[/itex] becomes [itex]x^2- (y- 4)^2- 16= C[/itex] or [itex]x^2- (y- 4)^2= C+ 16[/itex]. To get that to "standard form", divide through by C+ 16:

[tex]\frac{x^2}{C+ 16}- \frac{(y- 4)^2}{C+ 16}= 1[/tex]

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

- #4

mfb

Mentor

- 35,392

- 11,743

It is easier to find a function x=f(y).Use thedefinitionof "isotherm". For a given temperature, C, you are points where the temperature is C: [itex]x^2- y^2+ 8y= C[/itex]. 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.

- #5

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

Not when there is NO such function, as here.

- #6

mfb

Mentor

- 35,392

- 11,743

... or two functions.

Share: