The values of and are always positive, and and vary inversely. If is 10 when is 2, then find when is 4000.
The values of $x$ and $y$ are always positive, and $x^2$ and $y$ vary inversely. If $y$ is 10 when $x$ is 2, then find $x$ when $y$ is 4000.
$y = \frac{k}{x^2}, k \in \mathbb{R}$
$k = x^2 \cdot y = 2^2 \cdot 10 = 40$
$y = \frac{40}{x^2}$
$4000 = \frac{40}{x^2}$
$100 = \frac{1}{x^2}$
$100x^2 = 1$
$x^2 = \frac{1}{100}$
$x = \frac{1}{10}$
My solution was this
Since x^2 and y are inversely proportional, their product is constant. So,
2^2 times 10=x^2 times 4000,
x=1/10