Question

The curvature of a circle depends inversely upon its radius r.

(a) True

(b) False

I had been asked this question in exam.

The doubt is from Curvature topic in portion Differential Calculus of Engineering Mathematics

Answer 1

Right answer is (a) True

For explanation I would say: Using parametric form of circle  x = r.cos(t) : y = r.sin(t)

Using curvature in parametric form k=\(\left |\frac{y”x’-y’x”} {((x’)^2+(y’)^2)^{\frac{3}{2}}}\right |\) we have

k=\(\left |\frac{(-rsin(t))(-rsin(t))-(-rcos(t))(rcos(t))} {((-rsin(t))^2+(rcos(t))^2)^{\frac{3}{2}}}\right |\)

k=\(\left |\frac{r^2(sin^2(t)+cos^2(t))} {r^3(sin^2(t)+cos^2(t))^{\frac{3}{2}}}\right |\)

k=\(\left |\frac{1}{r}\right |\)