Sketch the graphs

Altair · #1 $Old$ November 23rd, 2008, 01:47 AM

Actually this may not be calculus but I am posting it here because its in the Calculus book and is in the curves and arc length etc which sure is calculus...

The question says...

Determine the isotherms of the temperature fields in the plane given by the following scalar functions.

$T = \frac{y}{x^2 + y^2}$
This one looks like a close relative of circle but I can't figure out mathematically...That y is kinda disturbing.

$T = x^2 - Y^2 + 8y$
lost

and lastly

how is $T= xy$ a hyperbola ?

HallsofIvy · #2 $Old$ November 23rd, 2008, 06:08 AM

Quote:

Originally Posted by Altair $View Post$

Actually this may not be calculus but I am posting it here because its in the Calculus book and is in the curves and arc length etc which sure is calculus...

The question says...

Determine the isotherms of the temperature fields in the plane given by the following scalar functions.

$T = \frac{y}{x^2 + y^2}$
This one looks like a close relative of circle but I can't figure out mathematically...That y is kinda disturbing.

$T = x^2 - Y^2 + 8y$
lost

and lastly

how is $T= xy$ a hyperbola ?

"iso" means "same" so "isotherms" are lines of constant temperature.
To take your second question first, it should be obvious that xy= constant is a hyperbola: it has the x and y axes as asymptotes.

If you learned only that a hyperbola must be of the form $\frac{x^2}{a^2}- \frac{y^2}{b^2}= 1$ then you should also have learned that this formula is valid only for hyperbolas with foci on the x-axis: "standard position". Quadratic formulas with "xy" terms are conic sections with axes rotated away from the x and y axes. In this case, if we take new axes at 45 degree angles to the x and y axes, that is, take x'= x- y and y'= x+ y, so that, adding the two equations, 2x= x'+ y' so x= (1/2)x'+ (1/2)y', and, subtracting, 2y= y'- x' so y= (1/2)y'- (1/2)y', then xy= (1/4)(x'+ y')(x'- y') we get $xy= \frac{x'^2}{4}- \frac{y'^2}{4}= T$ or $\frac{x'^2}{4T}- \frac{y'^2}{4T}= 1$ which is clearly a hyperbola.

For your first problem, again T is a constant so you have $T= \frac{y}{x^2+ y^2}$ . Multiply both sides by $x^2+ y^2$ and divide by T: $x^2+ y^2= \frac{1}{T}y$ or $x^2+ y^2- \frac{1}{T}y= 0$ . Complete the square in the y terms: $\frac{1}{T}$ divided by 2 is $\frac{1}{2T}$ and the square is [math\frac{1}{4T^2}[/math]:
$x^2 y^2+ \frac{1}{T}y= y^2+ \frac{1}{T}y+ \frac{1}{4T^2}= 0+ \frac{1}{4T^2}$
so $x^2+ (y+ \frac{1}{2T})^2= \frac{1}{4T^2}$ .

That is a circle with center at $(0, -\frac{1}{2T})$ and radius $1/2T$ so it is tangent to the x-axis at (0,0).

$T= x^2- y^2+ 8y$ is clearly a hyperbola. Complete the square in y again: $y^2- 8y= y^2- 8y+ 16- 16= (y- 4)^2- 16$ so adding 16 to both sides give $x^2- (y- 4)^2= T+ 16$ .