KCET · Maths · Application of Derivatives
The length of the subtangent at \(t\) on the curve \(\mathrm{x}=\mathrm{a}(\mathrm{t}+\sin \mathrm{t}), \mathrm{y}=\mathrm{a}(1-\cos \mathrm{t})\) is
- A \(a \sin \mathrm{t}\)
- B \(2 \mathrm{a} \sin \left(\frac{\mathrm{t}}{2}\right) \tan \left(\frac{\mathrm{t}}{2}\right)\)
- C \(2 \mathrm{a} \sin \frac{\mathrm{t}}{2}\)
- D \(2 a \sin ^{3}\left(\frac{t}{2}\right) \sec \left(\frac{t}{2}\right)\)
Answer & Solution
Correct Answer
(A) \(a \sin \mathrm{t}\)
Step-by-step Solution
Detailed explanation
Given, \(x=a(t+\sin t), y=a(1-\cos t)\)
\[
\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a}(1+\cos \mathrm{t}), \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a}(\sin \mathrm{t})
\]
\[
\therefore \quad \frac{d y}{d x}=\frac{a \sin t}{a(1+\cos t)}=\tan \frac{t}{2}
\]
\[
\begin{aligned}
\therefore \text { Length of subtangent } &=\frac{y}{d y / d x} \\
&=\frac{a(1-\cos t)}{\tan \frac{t}{2}} \\
&=2 a \sin \frac{t}{2} \cos \frac{t}{2} \\
&=a \sin t
\end{aligned}
\]
\[
\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a}(1+\cos \mathrm{t}), \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a}(\sin \mathrm{t})
\]
\[
\therefore \quad \frac{d y}{d x}=\frac{a \sin t}{a(1+\cos t)}=\tan \frac{t}{2}
\]
\[
\begin{aligned}
\therefore \text { Length of subtangent } &=\frac{y}{d y / d x} \\
&=\frac{a(1-\cos t)}{\tan \frac{t}{2}} \\
&=2 a \sin \frac{t}{2} \cos \frac{t}{2} \\
&=a \sin t
\end{aligned}
\]
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