Question

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${\text{si}}{{\text{n}}^2}{\text{t}}$.

Answer 1

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.

Complete step-by-step answer:
Given data,
x = a${\text{co}}{{\text{s}}^2}{\text{t}}$ and y = a${\text{si}}{{\text{n}}^2}{\text{t}}$.
Differentiating x and y with respect to t, we get
$
  \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) \\
$
Therefore,
$\dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{\dfrac{{{\text{dy}}}}{{{\text{dt}}}}}}{{\dfrac{{{\text{dx}}}}{{{\text{dt}}}}}},{\text{ where }}\dfrac{{{\text{dx}}}}{{{\text{dt}}}} \ne 0.$
$\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$=$\dfrac{{{\text{asin2t}}}}{{{\text{ - asin2t}}}}$
$ \Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}} = - 1$
Hence, the answer.

Note – In order to solve questions of this type the key is to differentiate the given terms precisely. General knowledge of differentials of basic trigonometric functions is required. Then the value obtained is converted into the desired form using formulae of trigonometric functions.