NCERT Class 12 Maths · Chapter 5 Continuity And Differentiability · Exercise 5.6 Question 5

Question

Without eliminating parameter t (θ in the trigonometric source forms), find dy/dx for x=$\cos{\left(t \right)} - \cos{\left(2 t \right)}$, y=$\sin{\left(t \right)} - \sin{\left(2 t \right)}$.

Accepted Answer

Final Answer: $- \frac{- \cos{\left(t \right)} + 2 \cos{\left(2 t \right)}}{\left(4 \cos{\left(t \right)} - 1\right) \sin{\left(t \right)}}$ where the real parametrisation is defined and dx/dt≠0. Step 1: Differentiate both coordinates: dx/dt=$- \sin{\left(t \right)} + 2 \sin{\left(2 t \right)}$, dy/dt=$\cos{\left(t \right)} - 2 \cos{\left(2 t \right)}$. Step 2: Divide the two rates: dy/dx=(dy/dt)/(dx/dt)=$\frac{\cos{\left(t \right)} - 2 \cos{\left(2 t \right)}}{- \sin{\left(t \right)} + 2 \sin{\left(2 t \right)}}$. Step 3: Simplify to $- \frac{- \cos{\left(t \right)} + 2 \cos{\left(2 t \right)}}{\left(4 \cos{\left(t \right)} - 1\right) \sin{\left(t \right)}}$. This quotient requires dx/dt≠0; any cancelled zero requires a separate limiting or curve check.

NCERT Class 12 Maths · Chapter 5 Continuity And Differentiability · Exercise 5.6 Question 5

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