(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the equation of normal to curve x=cos b , y=sin b at the point where b=pi/4 . Find the coordinates of the point on the curve where the normal meets the curve again .

2. Relevant equations

3. The attempt at a solution

Gradient of normal :

dx/db=-sin b

dy/db=cos b

dy/dx=-cot b

gradient of normal = tan b , when b=pi/4 , gradient of normal =1

The normal cuts the curve at point (1/sqrt(2) , 1/sqrt(2))

equation of normal is y-1/sqrt(2)=1(x-1/sqrt(2))

y=x --- 1

The cartesian equation of the curve is cos^(-1) x=sin^(-1) y

sin y=cos x --- 2

combine 1 and 2 , sin x=cos x

tan x=0

x=0 , pi , 2pi

Since the domain is not mentioned , i take it as 0<=x<=2pi

Therefore , the normal would cut the curve at points (0,pi/2) , (pi,3pi/2), (2pi, 5pi/2)

Am i correct ?

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# Homework Help: Parametric differentiation

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