Question

If x= 2cost + cos2t, y= 2sint + sin2t then \[\dfrac{{dy}}{{dx}}{\text{ at t = }}\dfrac{\pi }{4}\] is:
A. 1-$\sqrt 2 $
B. – (1+$\sqrt 2 $)
C. $\sqrt 2 $
D. $\dfrac{1}{{\sqrt 2 }}$

Answer 1

Hint: In this type of question where x and y are given in terms of other variables and we have to find the derivative of y with respect to x. The method is to first find the derivative of ‘x’ with respect to ‘t’ and similarly calculate the derivative of ‘y’ with respect to ‘t’ then divide them to get the derivative of ‘y’ with respect ‘x’.

Complete step-by-step answer:
In the question, we have to calculate $\dfrac{{dy}}{{dx}}$ but ‘y’ is not given in terms of ‘x’.
‘x’ and ‘y’ are given in terms of another variable ‘t’.
x = 2cost + cos2t
y = 2sint + sin2t
But, we know that:
\[\dfrac{{dy}}{{dx}} = \dfrac{{\left( {\dfrac{{dy}}{{dt}}} \right)}}{{\left( {\dfrac{{dx}}{{dt}}} \right)}}.\] -----(1)
So, we will first calculate the derivative of ‘y’ and ‘x’ with respect to ‘t’.
y = 2sint + sin2t
Using chain rule of differentiation, we get:
\[\dfrac{{dy}}{{dt}}{\text{ = }}\dfrac{{d\left( {2\sin {\text{t - sin2t}}} \right)}}{{dt}}\] = 2cost + 2cos2t.

And

x = 2cost + cos2t
Using chain rule of differentiation, we get:
\[\dfrac{{dx}}{{dt}}{\text{ = }}\dfrac{{d\left( {2\cos {\text{t + cos2t}}} \right)}}{{dt}}\] = -2sint – 2sin2t

Putting the values of $\dfrac{{dy}}{{dt}}{\text{ and }}\dfrac{{dx}}{{dt}}$ in equation 1, we get:
$ \dfrac{{dy}}{{dx}} = \dfrac{{\left( {\dfrac{{dy}}{{dt}}} \right)}}{{\left( {\dfrac{{dx}}{{dt}}} \right)}} = \dfrac{{2\cos {\text{t - 2cos2t}}}}{{ - 2\sin {\text{t - 2sin2t}}}} = \dfrac{{\cos {\text{t - cos2t}}}}{{ - \sin {\text{t - sin2t}}}}. \\
  {\text{Now, we can calculate }}\dfrac{{dy}}{{dx}}{\text{ at t = }}\dfrac{\pi }{4}.{\text{ It is given as:}} \\
  {\left( {\dfrac{{dy}}{{dx}}} \right)_{{\text{t = }}\dfrac{\pi }{4}}}
= \dfrac{{\cos \dfrac{\pi }{4}{\text{ - cos2}} \times \dfrac{\pi }{4}}}{{ - \sin \dfrac{\pi }{4}{\text{ - sin2}} \times \dfrac{\pi }{4}}}
 = \dfrac{{\cos \dfrac{\pi }{4}{\text{ - cos}}\dfrac{\pi }{2}}}{{ - \sin \dfrac{\pi }{4}{\text{ - sin}}\dfrac{\pi }{2}}}
= \dfrac{{\dfrac{1}{{\sqrt 2 }} - 0}}{{ - \dfrac{1}{{\sqrt 2 }} - 1}}
= \dfrac{1}{{ - \left( {\sqrt 2 + 1} \right)}}
= - \left( {\sqrt 2 - 1} \right)
= 1 - \sqrt 2 . \\ $
So, option A is correct.

Note: In this question, the dependent variable y is not given terms of independent variable x but both of them are given terms of another variable ‘t’. So, here you should use the formula given by equation 1. For calculating the differentiation, you must remember the chain rule of differentiation.