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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives5 Marks
Question
If $\text{x}=\text{a}(\cos\text{t}+\text{t}\sin\text{t})\ \text{and}\ \text{y}=\text{a}(\sin\text{t}-\text{t}\cos\text{t}),$ find the value of $\frac{\text{d}^2\text{y}}{\text{dx}^2}\ \text{at}\ \text{t}=\frac{\pi}{4}.$
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Answer

We have,
$\text{x}=\text{a}(\cos\text{t}+\text{t}\sin\text{t})\ \text{and}\ \text{y}=\text{a}(\sin\text{t}-\text{t}\cos\text{t}),$
On differentiating with respect to t, we get
$\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[\text{a}(\cos\text{t}+\text{t}\sin\text{t})]=-\text{a}\sin\text{t}+\text{a}\sin\text{t}+\text{at}\cos\text{t}=\text{at}\cos\text{at}$
and
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[\text{a}(\sin\text{t}-\text{t}\cos\text{t})]=\text{a}\cos\text{t}-\text{a}\cos\text{t}+\text{at}\sin\text{t}=\text{at}\sin\text{t}$
Now, $\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{at}\sin\text{t}}{\text{at}\cos\text{t}}=\tan\text{t}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}(\tan\text{t})$
$=\frac{\text{d}}{\text{dt}}(\tan\text{t})\times\frac{\text{dt}}{\text{dx}}=\sec^2\text{t}\times\frac{1}{\text{at}\cos\text{t}}$
$=\frac{1}{\text{at}\cos^3\text{t}}$
$\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)_{\text{t}=\frac{\pi}{4}}=\frac{1}{\text{a}\Big(\frac{\pi}{4}\Big)\cos^3\Big(\frac{\pi}{4}\Big)}=\frac{8\sqrt{2}}{\text{a}\pi}$
Hence, $\text{at}\ \text{t}=\frac{\pi}{4},\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{8\sqrt{2}}{\text{a}\pi}$

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