The length of the normal at point -t- of the curve x - a-t - sin t- y - a -1 - cos t-is -RPET 2001-2a -sin-3 -t-2- sec-t-2-2a sin-t - 2- tan-t-2-2a sin -t - 2-a sin t

The correct option is A 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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Similar questions

Q. Find the equations of the tangent and the normal to the following curves at the indicated points.

(i) x = θ + sinθ, y = 1 + cosθ at θ =

\frac{π}{2}

(ii)

x = \frac{2 a t^{2}}{1 + t^{2}}, y = \frac{2 a t^{3}}{1 + t^{2}} at t = \frac{1}{2}

(iii) x = at², y = 2at at t = 1
(iv) x = asect, y = btant at t
(v) x = a(θ + sinθ), y = a(1 − cosθ) at θ
(vi) x = 3cosθ − cos³θ, y = 3sinθ − sin³θ [NCERT EXEMPLAR]

Q. Find the equations of the tangent and the normal to the following curves at the indicated points.

(i) x = θ + sin θ, y = 1 + cos θ at θ = π/2
(ii)

x = \frac{2 a t^{2}}{1 + t^{2}}, y = \frac{2 a t^{3}}{1 + t^{2}} at t = 1 / 2

(iii) x = at², y = 2at at t = 1
(iv) x = a sec t, y = b tan t at t
(v) x = a (θ + sin θ), y = a(1 − cos θ) at θ

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

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

Q. Show that the equation of normal at any point

t

on the curve

x = 3 cos t - {cos}^{3} t

and

y = 3 sin t - {sin}^{3} t

4 (y {cos}^{3} t - x {sin}^{3} t) = 3 sin 4 t

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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]