NCERT Class 12 Maths · Chapter 5 Continuity And Differentiability · Exercise 5.6 Question 6

Question

Without eliminating parameter t (θ in the trigonometric source forms), find dy/dx for x=$a \left(t - \sin{\left(t \right)}\right)$, y=$a \left(\cos{\left(t \right)} + 1\right)$.

Accepted Answer

Final Answer: $\frac{\sin{\left(t \right)}}{\cos{\left(t \right)} - 1}$ where the real parametrisation is defined and dx/dt≠0. Step 1: Differentiate both coordinates: dx/dt=$a \left(1 - \cos{\left(t \right)}\right)$, dy/dt=$- a \sin{\left(t \right)}$. Step 2: Divide the two rates: dy/dx=(dy/dt)/(dx/dt)=$- \frac{\sin{\left(t \right)}}{1 - \cos{\left(t \right)}}$. Step 3: Simplify to $\frac{\sin{\left(t \right)}}{\cos{\left(t \right)} - 1}$. 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 6

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