# Show that the length of an arc of the exponential spiral `r=ae^(btheta)` is proportional to the difference of the radial coordinates of its endpoints. Here `a > 0` and` b in RR` , `b ne0` .

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### 1 Answer

Suppose the endpoints are `theta_1` and `theta_2` .

Then the radial coordinates of the endpoints are `r_1 = ae^(btheta_1)` and `r_2 = ae^(btheta_2)` .

The length of arc connecting these points is

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

Calculate and simplify the square root in this expression first:

`(dr)/(d theta) = (ae^(btheta))' = abe^(btheta)`

`r^2 + ((dr)/(d theta))^2 = (ae^(btheta))^2 + (abe^(btheta))^2 = a^2(1 + b^2)e^(2btheta)`

The square root of this expression is `asqrt(1+b^2)e^(btheta)` .

The length of the arc then will be

`L=asqrt(1 + b^2) int_(theta_1) ^ (theta_2) e^(btheta) d theta = `

`=asqrt(1 + b^2) 1/b e^(btheta) |_(theta_1) ^ (theta_2) = `

`a/b sqrt(1 + b^2) (e^(btheta_2) - e^(btheta_1))` .

Substituting the expressions for `r_1 ` and `r_2 ` (see above) into this expression for L, we can see that the length of arc is in fact proportional to the difference of the radial coordinates of the endpoints:

`L = (sqrt(1+b^2))/b (r_2 - r_1)` .