If x varies inversely as the cube root of y and x = 1 when y = 8, find y when x = 3
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Correct Answer: Option C
Explanation:
\(x \propto \frac{1}{\sqrt[3]{y}} \implies x = \frac{k}{\sqrt[3]{y}}\)
When y = 8, x = 1
\(1 = \frac{k}{\sqrt[3]{8}} \implies 1 = \frac{k}{2}\)
\(k = 2\)
\(\therefore x = \frac{2}{\sqrt[3]{y}}\)
When x = 3,
\(3 = \frac{2}{\sqrt[3]{y}} \implies \sqrt[3]{y} = \frac{2}{3}\)
\(y = (\frac{2}{3})^{3} = \frac{8}{27}\)
\(x \propto \frac{1}{\sqrt[3]{y}} \implies x = \frac{k}{\sqrt[3]{y}}\)
When y = 8, x = 1
\(1 = \frac{k}{\sqrt[3]{8}} \implies 1 = \frac{k}{2}\)
\(k = 2\)
\(\therefore x = \frac{2}{\sqrt[3]{y}}\)
When x = 3,
\(3 = \frac{2}{\sqrt[3]{y}} \implies \sqrt[3]{y} = \frac{2}{3}\)
\(y = (\frac{2}{3})^{3} = \frac{8}{27}\)