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Question
Find the equations of tangents and normals to the following curves at the indicated points on them : x sin 2y = y cos 2x at `(pi/4, pi/2)`
Solution
x sin 2y = y cos 2x
Differentiating both sides w.r.t. x, we get
`xd/dx (sin 2y) + sin 2y. d/dx(x) = y.d/dx(cos2x) + cos2x.dy/dx`
= `x.cos2y.d/dx(2y) + sin2y xx 1 = y.(-sin2x).d/dx(2x) + cos2x.dy/dx`
∴ `xcos2y xx 2 dy/dx + sin2y = -ysin2x xx 2 + cos2x.dy/dx`
∴ `(2x cos2y - cos 2x)dy/dx = -2y sin 2x - sin 2y`
∴ `dy/dx = (-2ysin2x - sin2y)/(2xcos2y - cos2x)`
∴ `(dy/dx)_("at" (pi/4, pi/2)) = (-2(pi/2)sin pi/(2) - sinpi)/(2(pi/4)cospi - cos pi/2)`
= `(-pi(1) - 0)/(pi/2(-1) - 0)`
= `(-pi)/(((-pi)/2)`
= 2
= slope of the tanget at `(pi/4, pi/2)`
∴ the equation of the tangent at `(pi/4, pi/2)` is
∴ `y - pi/(2) = 2(x - pi/4)`
∴ `y - pi/(2) = 2x - pi/2`
∴ 2x – y = 0
The slope of normal at `(pi/4, pi/2)`
= `(-1)/((dy/dx)_("at"(pi/4, pi/2)`
= `(-1)/(2)`
∴ The equation of the normal at `(pi/4, pi/2)` is
`y - pi/(2) = -(1)/(2)(x - pi/2)`
∴ `2y – pi = -x + pi/(4)`
∴ `8y - 4pi = -4x + pi`
∴ 4x + 8y – 5π = 0
Hence, the equation of the tangent and normal are
2x – y = 0 and 4x + 8y – 5π = 0 respectively.
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