Logarithmic spiral

Calculus Level 3

A particle moves along a spiral so that its path is given by r ⃗ ( t ) = e − α t ( cos ⁡ ( t ) sin ⁡ ( t ) ) , t ∈ [ 0 , ∞ ) \vec r(t) = e^{-\alpha t} \left( \begin{array}{c} \cos(t) \\ \sin(t) \end{array} \right), \quad t \in [0,\infty) with the time t t and a decay constant α = 0.1 \alpha = 0.1 . What is the total distance of the path from r ⃗ ( 0 ) = ( 1 , 0 ) \vec r(0) = (1,0) to r ⃗ ( ∞ ) = ( 0 , 0 ) \vec r(\infty) = (0,0) , that is covered by the particle for all time?


The answer is 10.05.

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

Markus Michelmann
Sep 24, 2017

The velocity of the particle is given by v ⃗ ( t ) = r ⃗ ˙ ( t ) = − α e − α t ( cos ⁡ ( t ) sin ⁡ ( t ) ) ⏟ e ⃗ r ( t ) + e − α t ( − sin ⁡ ( t ) cos ⁡ ( t ) ) ⏟ e ⃗ ϕ ( t ) = e − α t ( e ⃗ ϕ ( t ) − α e ⃗ r ( t ) ) \vec v(t) = \dot{\vec{r}} (t) = - \alpha e^{-\alpha t} \underbrace{\left( \begin{array}{c} \cos(t) \\ \sin(t) \end{array} \right)}_{\vec e_r(t)} + e^{-\alpha t} \underbrace{\left( \begin{array}{c} -\sin(t) \\ \cos(t) \end{array} \right) }_{\vec e_\phi(t)} = e^{-\alpha t} (\vec e_\phi(t) - \alpha \vec e_r(t)) with ∣ e ⃗ r ∣ = ∣ e ⃗ ϕ ∣ = 1 |\vec e_r| = |\vec e_\phi| = 1 and e ⃗ r ⋅ e ⃗ ϕ = 0 \vec e_r \cdot \vec e_\phi = 0 . Therefore, the absolute value of the velocity results to v ( t ) = ∣ v ⃗ ( t ) ∣ = v ⃗ ( t ) ⋅ v ⃗ ( t ) = 1 + α 2 ⋅ e − α t v(t) = |\vec v(t)| = \sqrt{\vec v(t) \cdot \vec v(t)} = \sqrt{1 + \alpha^2} \cdot e^{-\alpha t} By integrating the velocity over time, the path length is obtained: s = ∫ 0 ∞ v ( t ) d t = 1 + α 2 ∫ 0 ∞ e − α t d t = 1 + α 2 α ≈ 10.05 s = \int_0^\infty v(t) dt = \sqrt{1 + \alpha^2} \int_0^\infty e^{-\alpha t} dt = \frac{\sqrt{1 + \alpha^2}}{\alpha} \approx 10.05

4 Helpful 0 Interesting 0 Brilliant 0 Confused

We can actually do it directly without calculating the velocity , but that solution is more lengthy than yours . For a function y = f ( x ) y=f(x)

s = ∫ a b 1 + ( y ′ ) 2 d x s = \int_{a}^{b} \sqrt{1+\left(y^{'}\right)^2 } dx

Sabhrant Sachan - 3 years, 8 months ago

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In princple, you're right. I just thought it would be more obvious from a physical point of view to speak of a velocity. Furthermore, the spiral can not be uniquely represented by a function y = f ( x ) y = f(x) , since several (or even infinite) y y values ​​are associated with an x x -value.

Markus Michelmann - 3 years, 8 months ago

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you are absolutely correct , we cannot uniquely represent it as y = f ( x ) y=f(x) that is why we will integrate it in the parametric form .

d y = ( − α e − α t sin ⁡ t + e − α t cos ⁡ t ) d t d x = ( − α e − α t cos ⁡ t − e − α t sin ⁡ t ) d t dy = \left( -\alpha e^{-\alpha t} \sin{t} + e^{-\alpha t} \cos{t} \right) dt \\ dx = \left( -\alpha e^{-\alpha t} \cos{t} - e^{-\alpha t} \sin{t} \right) dt

The end result is the integral s = 1 + α 2 ∫ 0 ∞ e − α t d t s = \sqrt{1+\alpha^2} \int_{0}^{\infty} e^{-\alpha t} dt

Sabhrant Sachan - 3 years, 8 months ago

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