Class 12 Continuity And Differentiability · Long Answer3 Marksparametric_derivatives

Class 12 Maths · Continuity And Differentiability · Long Answer #152

For x=\(a \left(t - \sin{\left(t \right)}\right)\), y=\(a \left(\cos{\left(t \right)} + 1\right)\), find d²y/dx². Assume a≠0 and 1−cos t≠0.

✅ Step-by-Step Solution

Final Answer: \(\frac{1}{a \left(\cos{\left(t \right)} - 1\right)^{2}}\)

Differentiate in t: dx/dt=\(a \left(1 - \cos{\left(t \right)}\right)\) and dy/dt=\(- a \sin{\left(t \right)}\).
Thus dy/dx=(dy/dt)/(dx/dt)=\(- \frac{\sin{\left(t \right)}}{1 - \cos{\left(t \right)}}\).
Use d²y/dx²=[d(dy/dx)/dt]/(dx/dt)=\(\frac{1}{a \left(\cos{\left(t \right)} - 1\right)^{2}}\).
Therefore d²y/dx²=\(\frac{1}{a \left(\cos{\left(t \right)} - 1\right)^{2}}\).

[0.5 marks] Differentiate in t: dx/dt=\(a \left(1 - \cos{\left(t \right)}\right)\) and dy/dt=\(- a \sin{\left(t \right)}\).
[1 mark] Thus dy/dx=(dy/dt)/(dx/dt)=\(- \frac{\sin{\left(t \right)}}{1 - \cos{\left(t \right)}}\).
[1 mark] Use d²y/dx²=[d(dy/dx)/dt]/(dx/dt)=\(\frac{1}{a \left(\cos{\left(t \right)} - 1\right)^{2}}\).
[0.5 marks] Therefore d²y/dx²=\(\frac{1}{a \left(\cos{\left(t \right)} - 1\right)^{2}}\).

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