Answer to Question #135760 in Calculus for Promise Omiponle

Question

1. For the function f(x,y) = (25 − x^2 − y^2)^(−1/2), please sketch the level curves of the function, and then list the domain and the range of the function, write down a curve that represents the boundary, and state whether the domain is an open or closed region.

Answer 1

$f(x,y) = \dfrac{1}{\sqrt{25-x^2-y^2}} = \dfrac{1}{\sqrt{5^2-(x^2+y^2)}} .$

Let us determine the level curve $f(x,y) = \dfrac15,$ $\dfrac15 = \dfrac{1}{\sqrt{5^2-(x^2+y^2)}} \Rightarrow x^2+y^2 = 0, \; x=y=0.$

Let us determine the level curve $f(x,y) = \dfrac13,\;\; \dfrac19 = \dfrac{1}{{5^2-(x^2+y^2)}} \Rightarrow x^2+y^2 = 4^2.$

The square root may be calculated only for non-negative numbers, so $5^2-(x^2+y^2) \ge 0.$ Moreover, the square root is placed in the denominator, so it should not be equal to 0. Therefore, $x^2+y^2 < 5^2,$ so all (x,y) are situated in the circle with boundary $x^2+y^2=5^2$ and the domain is open.

Next, the largest value of f(x,y) can be obtained if the denominator is smallest, namely if $x^2+y^2 = 0,$ so $f(x,y) = f(0,0) = \dfrac{1}{\sqrt{5^2-0}}= \dfrac{1}{5}.$

But if $x^2+y^2$ is very close to $5^2$ , the denominator is close to 0, so f(x,y) tends to infinity. Therefore, the range of function is $\left[ \dfrac15, +\infty \right).$