The length of the normal at point - t - of the curve x-at-sin t- y-a1-cos t is -RPET 2001-A- a sin tB- 2 a sin 3 t - 2 t - 2C- 2 a sin t - 2 tan t - 2D- 2 a sin t - 2

The correct option is B 2a sin^3 (t/2) sec(t/2)y = ax^n+1 + bx^ n⇒dy/dx = (n+1)ax^n nbx^ n 1 ⇒d^2y/dx^2 = n(n+1)ax^n 1 n(n+1)bx^ n 2 ⇒ x^2 d^2y/dx^2 = n(n ...

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Standard XII

Mathematics

Length of Normal

The length of...

Question

The length of the normal at point ‘t’ of the curve x = a(t + sin t), y = a (1 - cos t) is
[RPET 2001]

a sin t

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2 a s i n^{3} (t / 2) s e c (t / 2)

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2a sin(t / 2) tan(t/2)

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2a sin (t / 2)

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Solution

The correct option is B $2 a s i n^{3} (t / 2) s e c (t / 2)$
$y = a x^{n + 1} + b x^{- n} \Rightarrow \frac{d y}{d x} = (n + 1) a x^{n} - n b x^{- n - 1}$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = n (n + 1) a x^{n - 1} - n (n + 1) b x^{- n - 2}$
$\Rightarrow x^{2} \frac{d^{2} y}{d x^{2}} = n (n + 1) y .$

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The length of the normal at point ‘t’ of the curve x = a(t + sin t), y = a (1 - cos t) is [RPET 2001]

The correct option is B 2a sin3(t/2)sec(t/2)y=axn+1+bx−n⇒dydx=(n+1)axn−nbx−n−1 ⇒d2ydx2=n(n+1)axn−1−n(n+1)bx−n−2 ⇒x2 d2ydx2=n(n+1)y.

The length of the normal at point ‘t’ of the curve x = a(t + sin t), y = a (1 - cos t) is
[RPET 2001]