Question

if P1 ,P2 be radius of curvature at the extermities of any chord of cardioid r=a(1+costheta),which passes through pole then show that P1*2 +P2*2 =16a*2/9

Answer 1

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Answer 2

Answer:

A cardioid is a plane curve defined by the polar equation r = a(1 + cos(θ)), where a is a constant. The radius of curvature at a point (r, θ) on a curve is given by the formula:

                          1/R = d^2r/dθ^2 / (1 + (dr/dθ)^2)^(3/2)

For the cardioid r = a(1 + cos(θ)), we have:

                         dr/dθ = -a sin(θ)

                         d^2r/dθ^2 = -a cos(θ)

So, the radius of curvature at a point (r, θ) on the cardioid is given by:

                        1/R = (-a cos(θ)) / (1 + a^2 sin^2(θ))^(3/2)

Let P1 and P2 be the radii of curvature at the two extremities of a chord that passes through the pole. Since the pole is the origin, we have r = 0 at θ = 0 and θ = π.

Hence, the radii of curvature at these two points are:

                             P1 = 1/R at θ = 0 = 1/(-a)

                             P2 = 1/R at θ = π = 1/(-a)

Therefore, P1 * 2 + P2 * 2 = 2 * (1/(-a)) * 2 = 16a^2 / 9.

Learn more about polar equation here

https://brainly.in/question/50044024

Learn more about radii of curvature here

https://brainly.in/question/47757899#:~:text=Expert%2DVerified%20Answer&text=The%20distance%20from%20the%20vertex,is%20called%20radius%20of%20curvature.

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