Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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

seo-qna
Last updated date: 17th Jul 2024
Total views: 448.8k
Views today: 9.48k
Answer
VerifiedVerified
448.8k+ 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.