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If x = a (θ + sinθ), y = a(1 - cosθ), then at θ = π/2, the length of the normal is
- a)2a
- b)a/2
- c)a√2
- d)a/√2
Correct answer is option 'C'. Can you explain this answer?
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If x = a (θ + sinθ), y = a(1 - cosθ), then at &theta...
Given:
x = a(θ + sinθ)
y = a(1 - cosθ)
At θ = π/2:
x = a(π/2 + sin(π/2)) = a(π/2 + 1) = a(π/2 + 1)
y = a(1 - cos(π/2)) = a(1 - 0) = a
Equation of the normal:
The equation of the normal at any point (x,y) on a curve given by y = f(x) is given by:
y - f(x) = -1/f'(x)(x - x)
Calculating the slope of the tangent:
Given y = a(1 - cosθ) = a(1 - cosθ), differentiate y with respect to θ to get the slope of the tangent:
dy/dθ = a*sinθ
At θ = π/2, dy/dθ = a
Therefore, the slope of the tangent is a.
Calculating the slope of the normal:
The slope of the normal will be the negative reciprocal of the slope of the tangent at the given point.
Therefore, the slope of the normal = -1/a
Calculating the length of the normal:
The length of the normal is given by the formula:
Length = √(x2 + y2)
Substitute x = a(π/2 + 1) and y = a into the formula:
Length = √((a(π/2 + 1))^2 + a^2)
= √(a^2(π^2/4 + π + 1) + a^2)
= √(a^2(π^2/4 + π + 1 + 1))
= √(a^2(π^2/4 + π + 2))
= a√(π^2/4 + π + 2)
= a√(π^2 + 4π + 8)/4
= a√(π^2 + 4π + 4 + 4)/4
= a√(π + 2)^2/4
= a(π + 2)/2
= aπ/2 + a
Therefore, the length of the normal at θ = π/2 is a(π/2 + 1), which simplifies to a√2. Hence, the correct answer is option 'C'.
x = a(θ + sinθ)
y = a(1 - cosθ)
At θ = π/2:
x = a(π/2 + sin(π/2)) = a(π/2 + 1) = a(π/2 + 1)
y = a(1 - cos(π/2)) = a(1 - 0) = a
Equation of the normal:
The equation of the normal at any point (x,y) on a curve given by y = f(x) is given by:
y - f(x) = -1/f'(x)(x - x)
Calculating the slope of the tangent:
Given y = a(1 - cosθ) = a(1 - cosθ), differentiate y with respect to θ to get the slope of the tangent:
dy/dθ = a*sinθ
At θ = π/2, dy/dθ = a
Therefore, the slope of the tangent is a.
Calculating the slope of the normal:
The slope of the normal will be the negative reciprocal of the slope of the tangent at the given point.
Therefore, the slope of the normal = -1/a
Calculating the length of the normal:
The length of the normal is given by the formula:
Length = √(x2 + y2)
Substitute x = a(π/2 + 1) and y = a into the formula:
Length = √((a(π/2 + 1))^2 + a^2)
= √(a^2(π^2/4 + π + 1) + a^2)
= √(a^2(π^2/4 + π + 1 + 1))
= √(a^2(π^2/4 + π + 2))
= a√(π^2/4 + π + 2)
= a√(π^2 + 4π + 8)/4
= a√(π^2 + 4π + 4 + 4)/4
= a√(π + 2)^2/4
= a(π + 2)/2
= aπ/2 + a
Therefore, the length of the normal at θ = π/2 is a(π/2 + 1), which simplifies to a√2. Hence, the correct answer is option 'C'.
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If x = a (θ + sinθ), y = a(1 - cosθ), then at θ = π/2, the length of the normal isa)2ab)a/2c)a√2d)a/√2Correct answer is option 'C'. Can you explain this answer?
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