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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If $\text{x}=\text{a}\sin\text{t}-\text{b}\cos\text{t},\text{y}=\text{a}\cos\text{t}+\text{b}\sin\text{t},$ Prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{x}^2+\text{y}^2}{\text{y}^2}$ 
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Answer

We have

$\text{x}=\text{a}\sin\text{t}-\text{b}\cos\text{t},\text{y}=\text{a}\cos\text{t}+\text{b}\sin\text{t},$

On differentiating with respect to t, we get

$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\sin\text{t}-\text{b}\cos\text{t})=\text{a}\cos\text{t}+\text{b}\sin\text{t}$

and

$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\cos\text{t}+\text{b}\sin\text{t})=-\text{a}\sin\text{t}+\text{b}\cos\text{t}$

$\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-\text{a}\sin\text{t}+\text{b}\cos\text{t}}{\text{a}\cos+\text{b}\sin\text{t}}$

Therefore

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}\Big(\frac{-\text{a}\sin\text{t}+\text{b}\cos\text{t}}{\text{a}\cos\text{t}+\text{b}\sin\text{t}}\Big)$

$=\frac{\text{d}}{\text{dt}}\Big(\frac{-\text{a}\sin\text{t}+\text{b}\cos\text{t}}{\text{a}\sin\text{t}+\text{b}\sin\text{t}}\Big)\times\frac{\text{dt}}{\text{dx}}$

$=\frac{(\text{a}\cos\text{t}+\text{b}\sin\text{t})\frac{\text{d}}{\text{dt}}(-\text{a}\sin\text{t}+\text{b}\cos\text{t})-(-\text{a}\sin\text{t}+\text{b}\cos\text{t})\frac{\text{d}}{\text{dt}}(\text{a}\cos\text{t}+\text{b}\sin\text{t})}{(\text{a}\cos\text{t}+\text{b}\sin\text{t})^2}$

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

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

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

$=\frac{-\text{y}^2-\text{x}^2}{\text{y}^3}$

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