Compute spiral length from parametric curve.

Question

Problem

Compute spiral length when $0\le t \le 2\pi$. Spiral is given in parametric form as:

$$ r(t)=\begin{bmatrix} e^{-t}\cos(t) \\ e^{-t}\sin(t) \end{bmatrix} $$

Attempt to solve

One way to do this would be to compute spiral length in segments and to get accurate result the segment size will $\rightarrow 0$ (i call segment size as $dt$). Since we want to compute length of the curve summing segments of $||r(t)||$ wouldn't be sufficient. Instead we sum segments of tangent length $||r'(t)||$. Now i compute derivative of each component with chain rule.

$$ r'(t)=\begin{bmatrix} -e^{-t}\cos(t)-e^{-t}\sin(t) \\ -e^{-t} \sin(t)+e^{-t}\cos(t)\end{bmatrix} $$

$$||r'(t)||=\sqrt{(-e^{-t}\cos(t)-e^{-t}\sin(t))^2+(-e^{-t} \sin(t)+e^{-t}\cos(t))^2}$$

The spiral length in $ [0,2\pi]$ can be computed with riemann integral.

$$ \int_{0}^{2\pi}||r'(t)||dt $$

Only problem is that the current expression for $r'(t)$ doesn't look nice. Maybe there is possibility that the form could be reduced to much nicer one ?

One thing that we can noted right away is that both sides of the $+$ sign inside square root are the same except other one has negative $-e^{-t}\cos(t)$ other side $+e^{-t}\cos(t)$. Since we are squaring both sides the $-$ sign doesn't change anything. we could as well use expression:

$$ ||r'(t)|| = \sqrt{2(e^{-t}\cos(t)+e^{-t}\sin(t))^2} $$ $$ ||r'(t)|| = \sqrt{(e^{-t}(2\cos(t)+2\sin(t)))^2} $$

Now if we assume $(r \in \mathbb{R}| r> 0)$. This should be fine since we want to define length in $[0,2\pi]$ which meets the condition.

$$ ||r'(t)|| = e^{-t}(2\cos(t)+2\sin(t))$$

I don't know if it is possible to get more compact from here. (probably not)

$$ \int_{0}^{2\pi} e^{-t}(2\cos(t)+2\sin(t)) dt= (2(\int e^{-t}\cos(t)+\int e^{-t}\sin(t) ))\Big|_{0}^{2\pi}$$

If someone knows better notation for this what i am trying to do please tell.

Now we need to solve two integrals by parts. $\int fdg = fg - \int gdf$ $$ 2\int e^{-t}\cos(t)=2(\frac{1}{2}e^{-t}(\sin(t)-\cos(t))) $$ $$ =e^{-t}(\sin(t)-\cos(t)) $$ And the other one.

$$ 2\int e^{-t}\sin(t)=2(-\frac{1}{2}e^{-t}(\sin(t)+\cos(t))) $$ $$ =-e^{-t}(\sin(t)+\cos(t)) $$

Now combining these

$$= ( e^{-t}(\sin(t)-\cos(t))-e^{-t}(\sin(t)+\cos(t))) \Big|_{0}^{2\pi}$$ $$=-(e^{-2\pi}+1)e^{-2\pi}\approx -1.87\cdot 10^{-3} $$

I get a small negative value for length now this cannot possibly be right answer but i can't find the error.

Answer 1

Hint:

Review $\left\|r'(t)\right\|$, \begin{align*} \left\|r'(t)\right\|&=\sqrt{\left(-e^{-t}\cos t-e^{-t}\sin t\right)^2+\left(-e^{-t}\sin t+e^{-t}\cos t\right)^2}\\ &=\sqrt{2e^{-2t}\cos^2t+2e^{-2t}\sin^2 t}\\ &=\sqrt{2e^{-2t}\left(\cos^2 t+\sin^2 t\right)}\\ &=\sqrt 2 e^{-t} \end{align*}

Answer 2

There is Euler's Formula $e^{ix}=\cos x+i \sin x$ that simplifies the calculations by exchanging $\cos$ and $\sin$ with the real and imaginary part of $e^{it}$.

$$r(t)= \begin{bmatrix} e^{-t}\cos t \\ e^{-t}\sin t \end{bmatrix} = \frac{1}{2} \begin{bmatrix} e^{-t(1+i)} + e^{-t(1-i)} \\ i(e^{-t(1+i)} - e^{-t(1-i)}) \end{bmatrix} = \begin{bmatrix} e^{-zt} + e^{- \overline{z}t} \\ i(e^{-zt} - e^{- \overline{z}t}) \end{bmatrix} $$

$$r'(t)=-\frac{1}{2} \begin{bmatrix} ze^{-zt} + \overline{z}e^{-\overline{z}t} \\ i(ze^{-zt} - \overline{z}e^{-\overline{z}t}) \end{bmatrix} $$

$$||r'(t)||^2 = \frac{1}{4} [(ze^{-zt} + \overline{z}e^{-\overline{z}t})^2 - (ze^{-zt} - \overline{z}e^{-\overline{z}t})^2]=2 e^{-2t}$$

The anti derivative of $$||r'(t)||= \sqrt{2}e^{-t}$$ is $$F(t)=-||r'(t)||$$

$$L(t)=\int_0^{t}-F(\tau)d\tau=F(t)-F(0)=\sqrt{2}(1-e^{-t})$$

And $L(2 \pi)=\sqrt{2}(1-e^{-2 \pi})$