Question

Derivation for:-
ds=r.d.theta

(Simple and clear please)

Answer 1

Answer:

I'm writing theta as A to reduce error.

There are several ways of doing it. I'll explain one method.

Consider a rt angled triangle having sides as x and y inclined at an angle A and the hypotenuse be s. Say, there is a small displacement on sides dx and dy causing a small change in angle dA.

Hence, by Pythagoras theorem, small change in hypotenuse, ds is given by:

ds^2 = dx^2 + dy^2.......................................(i)

In polar coordinates, ds becomes the arc length of rotation, and x and y are expressed as rcosA and rsinA respectively, where r is the radius swept. Considering, small change in dA also causes small change in dr.

Also, dx = -rsindA + drcosA and dy = rcosdA + drsinA

If dA is very small, and dA--->0, then sindA is approx. dA, sinA = 1 and hence, dx = rdA as dr is very small as compared to r. Also, cosdA = 0, cosA = 1 and hence, dy = dr.

Substituting in eq (I) we get,

ds^2 = (rdA)^2 + (dr)^2

Or, ds^2 = r^2dA^2 + dr^2

Again as dr is very small than r, so, ds^2 = r^2dA^2 i.e ds = rdA.