Question

find the radius of curvature of the cardiod r=a(1-cos(theta))​

Answer 1

Answer:

4/3.a.sin(theta/2)

Step-by-step explanation:

1.find dr/d@

2.find d^2r/d@^2

3.Then apply the polar form of radius of curvature

Answer 2

Given:

r= a(1-cos∅)

To Find:

the radius of curvature

Solution:

    r= a(1-cos∅)

   dr/d∅= a(sin∅) = a sin∅

   d²r/d∅²= a cos ∅

⇒  ∫= [r² + (dr/d∅)]$^{3/2}$/r² + 2(dr/d∅)²- r× d²r/d∅

 ⇒  [r² + (dr/d∅)]$^{3/2}$/ r²+ 2(dr/d∅)²- r(d²r/d∅²)

  ⇒ [a² (1-cos∅)² + a²sin²∅]$^{3/2}$/a²(1-cos∅)²+ 2a²sin²∅-a(1-cos∅)a cos∅

  ⇒[a²- (1+ cos²∅- 2cos∅)+ a²sin²∅]$^{3/2}$/a²(1-cos∅)²+ 2a²sin²∅-a²(1-cos∅)cos∅

  ⇒ [a²+ a²cos²∅- 2a²cos∅+a²sin²∅]$^{3/2}$/a²(1-cos∅)²+ 2sin²∅- (1-cos∅)cos∅

  ⇒ [a²+a²-2a²cos∅]$^{3/2}$/a²(1+cos²∅-2cos∅+2sin²∅-(cos∅-cos²∅)

 ⇒  [2a²(1-cos∅]$^{3/2}$]/a²(1+ cos²∅-2cos∅+2sin²∅-cos∅+cos∅)

  ⇒ a³[2(1-cos∅)$^{3/2}$/a²(3-3cos∅)

   ⇒a[2(1-cos∅)]$^{3/2}$/(1-cos∅)

  ⇒ 2√2a/3  (1-cos∅)$^{1/2}$

   ⇒2√2a.(2 sin²∅/2)$^{1/2}$/3

  4/3a sin∅/2