Question

The equation of the normal to the curve y=sinX at the point (pie,0)​

Answer 1

Answer :

y = x - π

Solution :

Here ,

The given curve is y = sinx ----(1)

Now ,

Differentiating eq-(1) both sides with respect to x , we get ;

=> dy/dx = d(sinx)/dx

=> dy/dx = cosx

=> Slope of tangent = cosx

At (π , 0)

=> Slope of tangent = cosπ

=> Slope of tangent = -1

Also ,

We know that , the product of mutually perpendicular lines = -1

Thus ,

=> Slope of tangent • Slope of normal = -1

=> -1 • Slope of normal = -1

=> Slope of normal = -1/-1

=> Slope of normal = 1

Also ,

We know that , the equation of a straight line passing through point (x1 , y1) with slope m is given by ;

(y - y1) = m(x - x1)

Thus ,

The equation of the normal line at point (π , 0) and having the slope = 1 will be given as ;

=> (y - 0) = 1•(x - π)

=> y = x - π

Hence ,

Required equation of normal is y = x - π