A curve y=f(x) passes through the point P(1,1). The normal to the curve at P is a(y-1)+(x-1)=0. If the slope of the tangent at any point on the curve is proportional to the ordinate of the point , determine the equation of the curve. Also obtain the area bounded by the y-axis , the curve and the normal to the curve at P.
Answers
Given : A curve y = f (x) passes through the point P (1, 1), the
normal to the curve at P is a (y - 1) + (x - 1) = 0.
the slope of the tangent at any point on the curve is proportional to the ordinate of that point,
To Find : Equation of the curve.
the area bounded by the y-axis, the curve and the normal to the curve at P
Solution:
slope of the tangent at any point on the curve is proportional to the ordinate of that point,
=> dy/dx ∝ y
=> dy/y = kdx
k is a constant
integrating both sides
=> ln y = kx + c
passes through (1 , 1)
=> 0 = k + c
=> c = - k
=> ln y = kx - k
=> ln y = k(x - 1)
or
normal to the curve at P is a (y - 1) + (x - 1) = 0
=> x + y = 2
=> slope = - 1
Slope of tangent = 1
dy/dx = ky = 1
y = 1
=> k(1) = 1
=> k = 1
Hence y = e⁽ˣ⁻¹⁾
equation of the curve is y = e⁽ˣ⁻¹⁾
y = e⁽ˣ⁻¹⁾
Area between curve and x axis
= ∫ydx = ∫e⁽ˣ⁻¹⁾dx x from 0 to 1
=> e⁽ˣ⁻¹⁾ x from 0 to 1
Area between curve and x axis = e⁽¹⁻¹⁾ - e⁽⁰⁻¹⁾ = 1 - 1/e
= 0.63212 sq units
Area by trapezium formed = (1/2) (2 + 1) * 1 = 3/2 ( ref figure)
area bounded by the y-axis, the curve and the normal to the curve at P
= 1.5 - 0.63212
= 0.86788 sq units
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