Question

Let r=r(theta), a<=theta<=b, be a curve C given in polar coordinates. Let f(x,y) be a function. Can you derive the formula for the ∫fds over C in terms of theta then use your formula to find the arc-length of the cardiod r = 1+costheta?

Answer 1

`s = int_(theta_1)^(theta_2) sqrt(r^2 + ((dr)/(d theta))^2)d theta`

For `r = 1 + costheta`,

`s = 8`.

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Like how we have `y = f(x)`, we now have `r = f(theta)`. We know that:

`x = rcostheta` `" "" "" "` `y = rsintheta`

DERIVING ARC LENGTH IN TERMS OF THETA

Define distance `D = sqrt((Deltax)^2 + (Deltay)^2)`. Since a straight arbitrary line length is preserved in all coordinate systems, we suppose `D = D(theta)`, and sum over all intervals `Delta theta`.

Then, if we multiply by each `(Delta theta)^2/(Delta theta)^2`, define `s` as the arc length due to summing each distance `D_i`:

`s -= sum_i D_i ~~ sum_i sqrt((Deltax)^2/(Delta theta)^2 (Delta theta)^2 + (Deltay)^2/(Delta theta)^2 (Delta theta)^2)`

`~~ sum_i sqrt(((Deltax)/(Delta theta))^2 + ((Deltay)/(Delta theta))^2)Delta theta`

where all `Deltax_i`, `Deltay_i`, and `Deltatheta_i` are the same for each `i`.

For infinitesimally short distances `d theta`, we thus take the limit as `Deltatheta -> 0` to get:

`s = int_(theta_1)^(theta_2) sqrt(((del x)/(del theta))^2 + ((dely)/(del theta))^2)d theta`

We thus will need to `(delx)/(del theta)` and `(dely)/(del theta)`, square each of them, and add them together. For convenience, let us find `s` in terms of `(dr)/(d theta)` and `r`.

Remember that `r = f(theta)` as well, so use the product rule and chain rule!

`(delx)/(del theta) = del/(del theta)[rcos theta] = -rsintheta + costheta(dr)/(d theta)`

`(dely)/(del theta) = del/(del theta)[rsin theta] = rcostheta + sintheta(dr)/(d theta)`

Square each of these to get:

`((delx)/(del theta))^2 = r^2sin^2theta - 2rsinthetacostheta(dr)/(d theta) + ((dr)/(d theta))^2cos^2theta`

`((dely)/(del theta))^2 = r^2cos^2theta + 2rsinthetacostheta(dr)/(d theta) + ((dr)/(d theta))^2sin^2theta`

Adding these, we obtain:

`((delx)/(del theta))^2 + ((dely)/(del theta))^2`

`= r^2sin^2theta - cancel(2rsinthetacostheta(dr)/(d theta)) + ((dr)/(d theta))^2cos^2theta + r^2cos^2theta + cancel(2rsinthetacostheta(dr)/(d theta)) + ((dr)/(d theta))^2sin^2theta`

`= r^2(sin^2theta + cos^2theta) + ((dr)/(d theta))^2(sin^2theta + cos^2theta)`

`= r^2 + ((dr)/(d theta))^2`

Therefore,

`color(blue)(s = int_(theta_1)^(theta_2) sqrt(r^2 + ((dr)/(d theta))^2)d theta)`

Now we have a formula in terms of only `r` and `theta`!

ARC LENGTH EXAMPLE

We now use this formula to find the arc length of `r = 1 + costheta`.

`(dr)/(d theta) = -sintheta => ((dr)/(d theta))^2 = sin^2theta`

Note that one cycle is `[0,2pi]`. The function will be symmetric since `cos` is even, so twice the arc length from `0` to `pi` would be the arc length from `0` to `2pi` (if you do the integral from `0` to `2pi`, you may think the arc length is zero, which it is not!).

Therefore:

`s = 2int_(0)^(pi) sqrt((1 + costheta)^2 + sin^2theta)d theta`

`= 2int_(0)^(pi) sqrt(1 + 2costheta + cos^2theta + sin^2theta)d theta`

`= 2int_(0)^(pi) sqrt(2 + 2costheta)d theta`

`= 2sqrt2int_(0)^(pi) sqrt(1 + costheta)d theta`

To help us do this integral, recall that:

`cos^2theta = (1 + cos2theta)/2`

Therefore, `2cos^2(theta/2) = 1 + costheta`. This will simplify to give:

`color(blue)(s) = 2sqrt2 int_(0)^(pi) sqrt(2cos^2(theta/2))d theta`

`= 4 int_(0)^(pi) cos(theta/2)d theta`

`= 4|[2sin(theta/2)]|_(0)^(pi)`

`= 8[sin(pi/2) - sin(0/2)]`

`= color(blue)(8)`