Class 12 Continuity And Differentiability · Long Answer3 Marksparametric_derivatives

Class 12 Maths · Continuity And Differentiability · Long Answer #74

For x=cos t(3−2cos²t), y=sin t(3−2sin²t), find dy/dx at t=π/4, interpreting the derivative as the limiting slope of the traced curve.

✅ Step-by-Step Solution

Final Answer: 1

Differentiate: dx/dt=3sin t cos2t and dy/dt=3cos t cos2t.
For nearby t≠π/4, their ratio is cot t. At π/4 both parameter derivatives vanish, so direct cancellation at that point is not itself a proof.
The difference quotient [y(t)−y(t₀)]/[x(t)−x(t₀)] has a 0/0 form. Twice applying l’Hôpital, or using the second-order Taylor terms, gives y″(t₀)/x″(t₀)=(−3√2)/(−3√2)=1.
Therefore the limiting curve slope at t=π/4 is 1.

[0.5 marks] Differentiate: dx/dt=3sin t cos2t and dy/dt=3cos t cos2t.
[1 mark] For nearby t≠π/4, their ratio is cot t. At π/4 both parameter derivatives vanish, so direct cancellation at that point is not itself a proof.
[1 mark] The difference quotient [y(t)−y(t₀)]/[x(t)−x(t₀)] has a 0/0 form. Twice applying l’Hôpital, or using the second-order Taylor terms, gives y″(t₀)/x″(t₀)=(−3√2)/(−3√2)=1.
[0.5 marks] Therefore the limiting curve slope at t=π/4 is 1.

Practise interactive questions, flashcards, visual mind maps, and timed board tests for Continuity And Differentiability.