If x= 3sint-sin3t, y= 3cost-cos3t, find d^2y/dx^2 at t=pi/3 - Maths - Continuity and Differentiability

Question

If x= 3sint-sin3t, y= 3cost-cos3t, find d^2y/dx^2 at t=pi/3

Vimala · Accepted Answer

We have

$x = 3 \sin t - \sin 3 t \Rightarrow \frac{ⅆ x}{ⅆ t} = 3 \cos t - 3 \cos 3 t ...... (1) y = 3 \cos t - \cos 3 t \Rightarrow \frac{ⅆ y}{ⅆ t} = - 3 \sin t + 3 \sin 3 t ...... (2)$

Therefore,

$\frac{ⅆ y}{ⅆ x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{- 3 \sin t + 3 \sin 3 t}{3 \cos t - 3 \cos 3 t} = \frac{\sin 3 t - \sin t}{\cos t - \cos 3 t} = \frac{2 \cos 2 t \sin t}{2 \sin 2 t \sin t} = \cot 2 t$

Therefore,

$\frac{ⅆ^{2} y}{ⅆ x^{2}} = - 2 \cos e c^{2} 2 t \cdot \frac{ⅆ t}{ⅆ x} = - \frac{2 \cos e c^{2} 2 t}{\frac{d x}{d t}} = - \frac{2 \cos e c^{2} 2 t}{3 (\cos t - \cos 3 t)}$

Thus,

${(\frac{ⅆ^{2} y}{ⅆ x^{2}})}_{(t = \frac{π}{3})} = - \frac{2 \cos e c^{2} 2 (\frac{π}{3})}{3 (\cos (\frac{π}{3}) - \cos 3 (\frac{π}{3}))} = - \frac{2 \cos e c^{2} 2 (\frac{π}{3})}{3 (\cos (\frac{π}{3}) - \cos 3 (\frac{π}{3}))} = - 2 \times {(\frac{2}{\sqrt{3}})}^{2} \cdot \frac{1}{3 (\frac{1}{2} + 1)} = (- 2 \times \frac{4}{3} \times \frac{2}{9}) = \frac{- 16}{27}$

Thus, ${(\frac{ⅆ^{2} y}{ⅆ x^{2}})}_{(t = \frac{π}{3})} = \frac{- 16}{27}$