If x = 2 cos t − cos 2t, y = 2 sin t − sin 2t, find \[\frac{d^2 y}{d x^2}\text{ at } t = \frac{\pi}{2}\] ?

Here, \[x = 2 \cos t - \cos2t \text { and }y = 2 \sin t - \sin2t\] \[\text{ Differentiating w . r . t . t, we get} \] \[\frac{d x}{d t} = - 2 \sin t + 2 \sin2t \text { and }\frac{d y}{d t} = 2 \cos t - 2 \cos2t\] \[ \therefore \frac{d y}{d x} = \frac{2 \cos t - 2 \cos2t}{- 2 \sin t + 2 \sin2t} = \frac{\cos t - \cos2t}{- \sin t + \sin2t}\] \[\text { Differentiating w . r . t . x, we get}\] \[\frac{d^2 y}{d x^2} = \frac{\left( - \sin t + 2 \sin2t \right)\left( - \sin t + \sin2t \right) - \left( \cos t - \cos2t \right)\left( - \cos t + 2 \cos2t \right)}{\left( - \sin t + \sin2t \right)^2}\frac{dt}{dx}\] \[ = \frac{\left( - \sin t + 2 \sin2t \right)\left( - \sin t + \sin2t \right) - \left( \cos t - \cos2t \right)\left( - \cos t + 2 \cos2t \right)}{\left( - \sin t + \sin2t \right)^2 \left( - 2 \sin t + 2 \sin2t \right)}\] \[\text { At } t = \frac{\pi}{2}: \] \[\frac{d^2 y}{d x^2} = \frac{\left( - 1 + 0 \right)\left( - 1 + 0 \right) - \left( 0 + 1 \right)\left( - 0 - 2 \right)}{\left( - 1 + 0 \right)^2 \left( - 2 + 0 \right)} = \frac{1 + 2}{- 2} = \frac{- 3}{2}\]

If X = 2 Cos T − Cos 2t, Y = 2 Sin T − Sin 2t, Find D 2 Y D X 2 at T = π 2 ? - Mathematics

प्रश्न

If x = 2 cos t − cos 2t, y = 2 sin t − sin 2t, find $\frac{d^{2} y}{d x^{2}} at t = \frac{π}{2}$ ?

उत्तर

Here,

$x = 2 \cos t - \cos 2 t and y = 2 \sin t - \sin 2 t$

$Differentiating w . r . t . t, we get$

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

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

$Differentiating w . r . t . x, we get$

$\frac{d^{2} y}{d x^{2}} = \frac{(- \sin t + 2 \sin 2 t) (- \sin t + \sin 2 t) - (\cos t - \cos 2 t) (- \cos t + 2 \cos 2 t)}{{(- \sin t + \sin 2 t)}^{2}} \frac{d t}{d x}$

$= \frac{(- \sin t + 2 \sin 2 t) (- \sin t + \sin 2 t) - (\cos t - \cos 2 t) (- \cos t + 2 \cos 2 t)}{{(- \sin t + \sin 2 t)}^{2} (- 2 \sin t + 2 \sin 2 t)}$

$At t = \frac{π}{2} :$

$\frac{d^{2} y}{d x^{2}} = \frac{(- 1 + 0) (- 1 + 0) - (0 + 1) (- 0 - 2)}{{(- 1 + 0)}^{2} (- 2 + 0)} = \frac{1 + 2}{- 2} = \frac{- 3}{2}$

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Simple Problems on Applications of Derivatives

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Q 36 Q 35 Q 37

पाठ 12: Higher Order Derivatives - Exercise 12.1 [पृष्ठ १७]

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आरडी शर्मा Mathematics [English] Class 12

पाठ 12 Higher Order Derivatives
Exercise 12.1 | Q 36 | पृष्ठ १७

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Simple Problems on Applications of Derivatives

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If X = 2 Cos T − Cos 2t, Y = 2 Sin T − Sin 2t, Find D 2 Y D X 2 at T = π 2 ? - Mathematics

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