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If x=f (t) and y=f (t) are differentiable functions of t, then prove that y is a differentiable function of x and $\dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{\dfrac{{{\text{dy}}}}{{{\text{dt}}}}}}{{\dfrac{{{\text{dx}}}}{{{\text{dt}}}}}},{\text{ where }}\dfrac{{{\text{dx}}}}{{{\text{dt}}}} \ne 0.$Hence find $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$if x = a${\text{co}}{{\text{s}}^2}{\text{t}}$ and y = a${\text{si}}{{\text{n}}^2}{\text{t}}$.

Last updated date: 25th Mar 2023
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Hint – Using the given data in the question, i.e. the values of x and y, we differentiate them. Then on the output, we apply a basic sine function formula to determine the answer.

x = a${\text{co}}{{\text{s}}^2}{\text{t}}$ and y = a${\text{si}}{{\text{n}}^2}{\text{t}}$.
$\dfrac{{{\text{dx}}}}{{{\text{dt}}}} = {\text{ aco}}{{\text{s}}^2}{\text{t}},{\text{ }}\dfrac{{{\text{dy}}}}{{{\text{dt}}}} = {\text{ asi}}{{\text{n}}^2}{\text{t}} \\ \Rightarrow \dfrac{{{\text{dx}}}}{{{\text{dt}}}} = {\text{ 2a cost }}\dfrac{{\text{d}}}{{{\text{dt}}}}{\text{cost}},{\text{ }}\dfrac{{{\text{dy}}}}{{{\text{dt}}}} = {\text{ 2a sint }}\dfrac{{\text{d}}}{{{\text{dt}}}}{\text{sint}} \\ \Rightarrow \dfrac{{{\text{dx}}}}{{{\text{dt}}}} = {\text{ 2a}}\left( {{\text{cost}})\times ({\text{ - sint}}} \right),{\text{ }}\dfrac{{{\text{dy}}}}{{{\text{dt}}}} = {\text{ a}}\left( {2{\text{sint}} \times {\text{cost}}} \right){\text{ }}\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{a}}{{\text{t}}^2}} \right) = {\text{2at}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{t}}} \right) \\ \Rightarrow \dfrac{{{\text{dx}}}}{{{\text{dt}}}} = {\text{ - 2a cost sint}},{\text{ }}\dfrac{{{\text{dy}}}}{{{\text{dt}}}} = {\text{ 2a sint cost }}\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\sin \theta } \right) = \cos \theta {\text{ and }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{cos}}\theta } \right){\text{ = - sin}}\theta {\text{ }}} \right) \\ \Rightarrow \dfrac{{{\text{dx}}}}{{{\text{dt}}}} = {\text{ - a sin2t}},{\text{ }}\dfrac{{{\text{dy}}}}{{{\text{dt}}}} = {\text{ a sin2t }}\left( {\sin 2\theta = 2{\text{sin}}\theta {\text{cos}}\theta } \right) \\$
$\dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{\dfrac{{{\text{dy}}}}{{{\text{dt}}}}}}{{\dfrac{{{\text{dx}}}}{{{\text{dt}}}}}},{\text{ where }}\dfrac{{{\text{dx}}}}{{{\text{dt}}}} \ne 0.$
$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$=$\dfrac{{{\text{asin2t}}}}{{{\text{ - asin2t}}}}$
$\Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}} = - 1$