Question

Find the radius of curvature of
the curve r=a(1+cosΘ) at the pointΘ=π/2
​

Answer 1

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Answer 2

Given:

r = a(1 + cosθ)

θ = π/2

To find:

Radius of curvature of  the curve.

Solution:

$K = \frac{|r^{2} +2(r')^{2}-rr'' |}{[r^{2}+(r')^{2} ]^{3/2} }$

r = a(1 + cosθ)

r' = -asinθ

r'' = -acosθ

Substituting r, r', and r''

$K = \frac{|(a(1+cos\theta))^{2} +2(-asin\theta)^{2}-a(1+cos\theta)(-acos\theta) |}{[(a(1+cos\theta))^{2}+(-asin\theta)^{2} ]^{3/2} }$

$K= \frac{a^{2} +a^{2} cos^{2}\theta + 2a^{2}cos\theta + 2a^{2} sin^{2}\theta + a^{2} cos\theta + a^{2} cos^{2}\theta}{[a^{2} + a^{2}cos^{2}\theta + 2a^{2}\theta + a^{2}sin^{2}\theta ]^{3/2} }$

Since, we know that sin²θ + cos²θ = 1

Using this identity in above equation, we get

$K = \frac{3[a^{2} + a^{2}cos\theta ]}{[2(a^{2} + a^{2}cos\theta)]^{3/2} }$

$K = \frac{3[a^{2} + a^{2}cos\theta ]}{2^{3/2} (a^{2} + a^{2}cos\theta)^{3/2} }$

$K = \frac{3}{2^{3/2} a(1 + cos\theta)^{1/2} }$

Now, substituting the θ = π/2

$K = \frac{3}{2^{3/2} a(1 + cos(\pi /2))^{1/2} }$

We know that cos(π/2) = 0

$K = \frac{3}{2^{3/2} a(1 +0)^{1/2} }$

$K = \frac{3}{2^{3/2} a }$

$K = \frac{3}{2\sqrt{2} a }$

Therefore, radius of curvature of curve at the point θ = π/2 is 3/2√2a