Question

If the square of $x$ varies as the cube of $y$, and $x=3$ when $y=4$, find the value of $y$ when $x=\frac{1}{\sqrt{3}}$.

Answer 1

As square of $x$ varies as cube of $y$ then let,

$x^2={ky}^3$

where $k$ is a constant, $x=3$ and $y=4$ satisfies the equation so,

$3^2=k{.4}^3$

$k=\frac{9}{64}$

$x^2=\frac{9}{64}{y}^3$

For $x=\frac{1}{\sqrt{3}}$

${\left(\frac{1}{\sqrt{3}}\right)}^2=\frac{9}{64}{y}^3$

$y^3=\frac{64}{9.3}=\frac{64}{27}$

$y=\frac{4}{3}=1\frac{1}{3}$