Suppose x varies inversely as y, y varies directly as the square of t and x = 1, when t = 3. Find x when t = \(\frac{1}{3}\).
\(x \propto \frac{1}{y}\)
\(x = \frac{k}{y}\)
\(y \propto t^{2}\)
\(y = ct^{2}\)
k and c are constants.
\(x = \frac{k}{ct^{2}}\)
Let \(\frac{k}{c} = d\) (a constant)
\(x = \frac{d}{t^{2}}\)
\(1 = \frac{d}{3^{2}} \implies d = 9\)
\(\therefore x = \frac{9}{t^{2}}\)
\(x = 9 \div (\frac{1}{3})^{2} \)
= \( 9 \div \frac{1}{9} = 9 \times 9 = 81\)
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