Answer:
Step-by-step explanation:
Given that:
[tex]T(x,y) = \dfrac{100}{1+x^2+y^2}[/tex]
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.
[tex]c = \dfrac{100}{1+x^2+2y^2} \ \ \--- (1)[/tex]
By cross multiply
[tex]c({1+x^2+2y^2}) = 100[/tex]
[tex]1+x^2+2y^2 = \dfrac{100}{c}[/tex]
[tex]x^2+2y^2 = \dfrac{100}{c} - 1 \ \ -- (2)[/tex]
From (2); let assume that the values of c > 0 likewise c < 100, then the interval can be expressed as 0 < c <100.
Now,
[tex]\dfrac{(x)^2}{\dfrac{100}{c}-1 } + \dfrac{(y)^2}{\dfrac{50}{c}-\dfrac{1}{2} }=1[/tex]
This is the equation for the family of the eclipses centred at (0,0) is :
[tex]\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1[/tex]
[tex]a^2 = \dfrac{100}{c} -1 \ \ and \ \ b^2 = \dfrac{50}{c}- \dfrac{1}{2}[/tex]
Therefore; the level of the curves are all the eclipses with the major axis:
[tex]a = \sqrt{\dfrac{100 }{c}-1}[/tex] and a minor axis [tex]b = \sqrt{\dfrac{50 }{c}-\dfrac{1}{2}}[/tex] which satisfies the values for which 0< c < 100.
The sketch of the level curves can be see in the attached image below.
