Question 1 :
Differentiate the following
If y = sin-1 x/√(1-x2) show that (1-x2)y2 - 3xy1 - y = 0
Solution :
y = sin-1 x/√(1-x2)
√(1-x2) y = sin-1 x
√(1-x2) y' + y(1/2√(1-x2) )(-2x) = 1/√(1-x2)
√(1-x2) y' + y(-x/√(1-x2) ) = 1/√(1-x2)
((1-x2) y' - xy)/√(1-x2) = 1/√(1-x2)
(1-x2) y' - xy = 1
(1-x2) y'' + y'(-2x) - [xy' + y(1)] = 0
(1-x2) y'' - 2xy' - xy' - y = 0
(1-x2) y'' - 3xy' - y = 0
(1-x2) y2 - 3xy1 - y = 0
Hence proved.
Question 2 :
If x = a (θ + sin θ), y = a (1 - cos θ) then prove that at θ = π/2, y'' = 1/a
Solution :
x = a (θ + sin θ) dx/dθ = a (1 + cos θ)--(1) |
y = a (1 - cos θ) dy/dθ = a (0 + sin θ) = a sin θ---(2) |
dy/dx = (a sin θ) / a(1 + cos θ)
y' = dy/dx = sin θ / (1 + cos θ)
y' = 2sin (θ/2) cos (θ/2) / (2cos2 (θ/2))
y' = tan (θ/2)
d2y/dx2 = sec2(θ/2) (1/2) (dθ/dx)
By applying (1), we get
d2y/dx2 = sec2(θ/2) (1/2) (1/(a (1 + cos θ)))
= sec2(π/4) (1/2) /(a (1 + cos π/2)))
= ((√2)2/2) /(a (1 + 0))
= 1/a
Hence proved.
Question 3 :
If sin y = x sin(a + y), then prove that dy/dx = sin2 (a + y)/sin a , a ≠ nπ
Solution :
sin y = x sin(a + y)
x = sin y / sin(a + y)
Differentiate with respect to "y"
dx/dy = [sin (a + y) cos y - sin y cos (x + y)] / sin2(a + y)
dx/dy = sin (a + y - y)/ sin2(a + y)
dx/dy = sin a / sin2(a + y)
dy/dx = sin2(a + y) / sin a
Hence proved.
Question 4 :
If y = (cos-1x)2 prove that (1 -x2)(d2y/dx2) - x (dy/dx) - 2 = 0
Solution :
y = (cos-1x)2
dy/dx = 2cos-1x (-1/√(1-x2)
dy/dx = -2cos-1x /√(1-x2)
In order to remove square root, we may take squares on both sides.
(dy/dx)2 = (-2cos-1x /√(1-x2))2
(dy/dx)2 = 4(cos-1x)2 /(1-x2)
(dy/dx)2 = 4y2/(1-x2)
(1-x2)(dy/dx)2 = 4y
(1-x2) 2 (dy/dx) (d2y/dx2) + (dy/dx)2 (-2x) = 4(dy/dx)
Divide by 2 on both sides
(1-x2)(dy/dx) (d2y/dx2) - x(dy/dx)2 = 2(dy/dx)
Divide by dy/dx
(1-x2)(d2y/dx2) - x(dy/dx) = 2
(1-x2)(d2y/dx2) - x(dy/dx) - 2 = 0
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