Courses
Courses for Kids
Free study material
Free LIVE classes
More
LIVE
Join Vedantu’s FREE Mastercalss

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
VerifiedVerified
362.4k+ views
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.
Last updated date: 01st Oct 2023
Total views: 362.4k
Views today: 3.62k
Trending doubts