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Provide Solution for RD Sharma Class 12 Chapter 11 Higher Order Derivatives Exercise Fill in the blanks Question 2

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Answer:  \frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2}} \sec ^{3} \theta

Hint: Differentiate x & y w.r.t  \theta. Use   \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}.   Again, differentiate w.r.t x.

Given: y=b \cos \theta \& \quad x=a \sin \theta


It is given that: x=a \sin \theta \& y=b \cos \theta

Differentiating y w.r.t t,

\frac{d y}{d \theta}=-b \sin \theta                                                                               …(i)                                         

Differentiating y w.r.t t,

\frac{d x}{d \theta}=a \cos \theta                                                                   …(ii)

So, \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}

\frac{d y}{d x}=\frac{-b \sin \theta}{a \cos \theta}                        …(iii)

Again, differentiate (iii) w.r.t x:

\begin{aligned} &\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(\frac{-b \sin \theta}{a \cos \theta}\right) \\ & \end{aligned}

\frac{d^{2} y}{d x^{2}}=\frac{d}{d \theta}\left(\frac{-b \sin \theta}{a \cos \theta}\right) \cdot \frac{d \theta}{d x}

\begin{aligned} &=\frac{d}{d \theta}\left(\frac{-b}{a} \tan \theta\right) \cdot \frac{1}{a \cos \theta} \\ & \end{aligned}                                      [ using (ii) ]               

=\frac{-b}{a} \sec ^{2} \theta \cdot\left(\frac{1}{a \cos \theta}\right) \\ 

=\frac{-b}{a^{2}} \sec ^{3} \theta                

Thus,  \frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2}} \sec ^{3} \theta.


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