Asked by Anamika
if tangent to the curve y=2sinx +sin2x are drawn at p(x=60degree) then find the area of the quadrilateral formed by the tangent the normal at P and the coordinate axis
Answers
Answered by
Steve
y'(x) = 2cosx+2cos2x
y'(pi/3) = 1-1 = 0
y(pi/3) = √3+√3/2 = 3√3/2
so, we have the axes, and the lines y=3√3/2 and x=pi/3
the rectangle has area pi√3/2
y'(pi/3) = 1-1 = 0
y(pi/3) = √3+√3/2 = 3√3/2
so, we have the axes, and the lines y=3√3/2 and x=pi/3
the rectangle has area pi√3/2
Answered by
Anamika
plse read the question carefully...its find the area of quadrilateral not the rectangle.
Answered by
Steve
it happens that the quadrilateral formed by those lines is a rectangle.
the tangent is horizontal and the normal is vertical.
the tangent is horizontal and the normal is vertical.
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