CURVATURE AND RADIUS OF CURVATURE RADIUS OF CURVATURE

CURVATURE AND RADIUS OF CURVATURE

RADIUS OF CURVATURE Let P be any point on the curve C. Draw the tangent at P to the circle. The circle having the same curvature as the curve at P touching the curve at P, is called the circle of curvature. It is also called the osculating circle. The centre of the circle of the curvature is called the centre of curvature. The radius of the circle of curvature is called the radius of curvature and is denoted by ‘ρ’. Note : 1. If k (> 0) is the curvature of a curve at P, then the radius of curvature of the curve of ρ is 1/k. This follows from the definition of radius of curvature and the result that the curvature of a circle is the reciprocal of its radius Note : 2. If for an arc of a curve, ψ decreases as s increases, then dψ/ds is negative, i. e. , k is negative. But the radius of a circle is nonnegative. So to take ρ = 1/k(in magnitude)=ds/dψ some authors regard k also as non-negative i. e. , k= dψ/ds

The sign of dψ/ds indicates the convexity and concavity of the curve in the neighbourhood of the point. Many authors take ρ =ds/dψ and discard negative sign if computed value is negative. ∴ Radius of curvature ρ =1/ |k| ·

Radius of Curvature in Cartesian Form Suppose the Cartesian equation of the curve C is given by y = f (x) and A be a fixed point on it. Let P(x, y) be a given point on C such that arc AP = s. Then we know that dy /dx = tan ψ. . . (1) where ψ is the angle made by the tangent to the curve C at P with the x-axis and ds/dx ={ 1+(dy/dx)^2}^1/2……(2) Differentiating (1) w. r. t x, we get d^2 y/dx^2 = sec^2ψ ⋅ dψ /dx = (1 + tan^2 ψ )dψ /ds. ds/dx =[1+(dy/dx)^2] 1/ ρ [1+(dy/dx)^2] ^1/2

[By using the (1) and (2)] Radius of Curvature in Parametric Form Let x = f (t) and y = g (t) be the Parametric equations of a curve C and P (x, y) be a given point on it Then And dy/dx= dy/dt / dx/dt d^2 y/dx^2 = d/dt{dy/dt / dx/dt}. dt/dx

=dx/dt. d^2 y/dt^2 -dy/dt. d^2 x/dt^2 ______________. 1 / dx/dt [dx/dt]^2 dx/dt. d^2 y/dt^2 -dy/dt. d^2 x/dt^2 = _____________ [dx/dt]^3 Substituting the values of dy/dx and d^2 y/dx^2 in the Cartesian form of the radius of curvature of the curve y = f (x) . Therefore. =[1+(dy/dt /dx/dt)^2] ^3/2 ______________ dx/dt. d^2 y/dt^2 -dy/dt. d^2 x/dt^2/dx/dt^3

{(dx/dt)^2+(dy/dt)^2}^3/2 ρ= _______________ dx/dt. d^2 y/dt^2 -dy/dt. d^2 x/dt^2 This is the cartesian form of the radius of curvature in parametric form. Examples 1. Find the radius of curvature at any point on the curve y = a log sec x/a 2. For the curve y = c cos h (x /c), show that ρ =y^2/c 3. Find the radius of curvature at (1, – 1) on the curve y = x 2 – 3 x + 1. 4. Find the radius of curvature at (a, 0) on y = x^3 (x – a). 5. Find the radius of curvature at x =πa/4 on y = a sec (x/a). 6. Find ρ at any point on x = a (θ + sinθ) and y = a (1 – cosθ).

Radius of Curvature in Pedal Form Let polar form of the equation of a curve be r = f (θ) and P(r, θ) be a given point on it. Let the tangent to the curve at P subtend an angle ψ with the initial side. If the angle between the radius vector OP and the tangent at P is φ then we have ψ = θ + φ Let p denote the length of the perpendicular from the pole O to the tangent at P. Then from the figure, sin φ =OM/OP =p/r Hence, p= r sin φ ………………(1) Therefore This is the Pedal form of the radius of curvature

Radius of Curvature in Polar Form Let r = f (θ) be the equation of a curve in the polar form and p(r, θ) be a point on it. Then This is the formula for the radius of curvature in the polar form.

Examples for pedal and polar forms 1. Find the radius of the curvature of each of the following curves: (i) r^3 = 2 ap^2 (Cardiod) (ii) p^2 = ar 2. Find the radius of curvature of the cardiod r = a (1 + cos θ) at any point (r, θ) on it. Also prove that ρ^2/r is a constant 3. Show that for the curve rn = an cos nθ the radius of curvature is A^n/ (n+1) r^n-1 4. Find the radii of curvature of the following curves (i) r = a e^θ cot α (ii) r (1 + cos θ) = a

CENTER OF CURVATURE Let P(x, y) be any point on the curve and let PT be the tangent at P making an angle ψ with the positive direction of x-axis. Let c(α, β) be the center of curvature corresponding to P(x, y). Then eqation of circle of curvature is (X-α)^2 +(y- β)^2 = p^2

Length of chords of curvature parallel to x-axis and y- axis Parallel to x –axis = 2 dy/dx[1+(dy/dx)2] _________ d^2 y/dx^2 Parellel to y-axis 2[1+(dy/dx)2] ________ = d^2 y/dx^2 Example- If y =a log sec x/a , then prove that the chord of curvature parallel tp y – axis is of constant length. -

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