Question

A thin metal plate, located in the xy-plane, has temperature T(x, y) at the point (x, y). Sketch some level curves (isothermals) if the temperature function is given by
T(x, y)= 100/1+x^2+2y^2

Answer 1

Answer:

Step-by-step explanation:

Given that:

$T(x,y) = \dfrac{100}{1+x^2+y^2}$

This implies that the level curves of a function(f) of two variables relates with the curves with equation f(x,y) = c

here c is the constant.

$c = \dfrac{100}{1+x^2+2y^2} \ \ \--- (1)$

By cross multiply

$c({1+x^2+2y^2}) = 100$

$1+x^2+2y^2 = \dfrac{100}{c}$

$x^2+2y^2 = \dfrac{100}{c} - 1 \ \ -- (2)$

From (2); let assume that the values of c > 0 likewise c < 100, then the interval can be expressed as 0 < c <100.

Now,

$\dfrac{(x)^2}{\dfrac{100}{c}-1 } + \dfrac{(y)^2}{\dfrac{50}{c}-\dfrac{1}{2} }=1$

This is the equation for the  family of the eclipses centred at (0,0) is :

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

$a^2 = \dfrac{100}{c} -1 \ \ and \ \ b^2 = \dfrac{50}{c}- \dfrac{1}{2}$

Therefore; the level of the curves are all the eclipses with the major axis:

$a = \sqrt{\dfrac{100 }{c}-1}$  and a minor axis $b = \sqrt{\dfrac{50 }{c}-\dfrac{1}{2}}$  which satisfies the values for which 0< c < 100.

The sketch of the level curves can be see in the attached image below.