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In section 13.4 problem 33 of Calculus third edition early transcedentals by Jon Rogawski and Colin Adams, it states

Let $$s(t)=\int_{-\infty}^t\|r'(u)\| \ du$$ for Bernoulli spiral $$r(t)= \langle e^t\cos(4t),e^t\sin(4t) \rangle$$ Show that the radius of curvature is porportional to $s(t)$

Though my textbook states the curvature $k(t)$ is equal to

$$\frac{\|T'(t)\|}{\|r'(t)\|}$$

which is also (for arbitrary regular parametrizations)

$$\frac{\|r'(t) \times r''(t)\|}{\|r'(t)\|^3}$$

I get different answers applying both methods.

Method one

$$T(t)=\frac{r'(t)}{\|r'(t)\|}$$

$$r'(t)=\langle e^t\cos{\left(4t\right)}-4e^t\sin{\left(4t\right)}, e^t\sin(4t)+4e^t \cos(4t) \rangle$$

$$\|r'(t)\|=\sqrt{17}e^t$$

$$T(t)=\frac{1}{\sqrt{17}e^t}{\langle e^t\cos{\left(4t\right)}-4e^t\sin{\left(4t\right)}, e^t\sin(4t)+4e^t \cos(4t) \rangle}=\frac{1}{\sqrt{17}}{\langle \cos{\left(4t\right)}-4\sin{\left(4t\right)}, \sin(4t)+4 \cos(4t) \rangle}$$

$$T'(t)=\frac{1}{\sqrt{17}}{\langle -4\sin{\left(4t\right)}-16\cos{\left(4t\right)}, 4\cos(4t)-16 \sin(4t) \rangle}$$

$$\|T'(t)\|=\frac{1}{\sqrt{17}}\sqrt{16+256}=\frac{\sqrt{272}}{\sqrt{17}}=4$$

$$k(t)=\frac{\|T'(t)\|}{\|r'(t)\|}=\frac{4}{\sqrt{17}e^{2t}}$$

The eqution $s(t)$ is

$$\int_{-\infty}^{t}\|r'(t)\| \ du=\int_{-\infty}^{t}\sqrt{17}e^t=\sqrt{17}e^t$$

Hence the ratio of $s(t)$ to the radius of curvature $k(t)$ is

$$\frac{\sqrt{17}e^t}{4\sqrt{17}e^t}=1/4$$

Method Two

In the second method we calculate

$$\frac{\|r'(t) \times r''(t)\|}{\|r'(t)\|^3}$$

We already know

$$\|r'(t)\|=\sqrt{17}e^t$$

$$\|r'(t)\|^3=17\sqrt{17}e^{3t}$$

Calcuating $$r''(t)$$, we get

$$r''(t)=\langle -15e^t\cos(4t)-8e^t\sin(4t), -15e^t\sin(4t)+8e^t\cos(4t) \rangle $$

Hence $\|r'(t)\times r''(t)\|$ is equals to

$$\sqrt{68}e^t=2\sqrt{17}e^{t}$$

Hence $$\frac{\|r'(t) \times r''(t)\|}{\|r'(t)\|^3}=\frac{2}{17 e^{2t}}$$

With this we know the ratio of $s(t)$ to radius of curvature is

$$\frac{\sqrt{17}e^t}{\frac{2}{17} e^{2t}}=\frac{17\sqrt{17}}{2}e^t$$

Reasons for different answers

One possibility is in the textbook. They state,

"In practice we compute the curvature using the following formula, which is valid for abitrary regular paramterizations

$$\frac{\|r'(t) \times r''(t)\|}{\|r'(t)\|^3}$$

However I'm not sure what "arbitrary regular parmaterization" means. How does this give us two different answers?

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  • 2
    $\begingroup$ Method one, shouldn't it be $T(t)=\frac{r'(t)}{||r'(t)||}$? $\endgroup$ – Jakobian Aug 1 '18 at 23:05
  • $\begingroup$ Method two works only in $R^3$, as you can guess from the fact that you have a cross product in the numerator. The correct formula for planar curves (i.e. curves in R^2) is $\kappa = \frac{\vert \det(r', r'') \vert}{\Vert r' \Vert^3}$. $\endgroup$ – MSobak Aug 1 '18 at 23:35
  • $\begingroup$ I found that $||r'(t)\times r''(t)|| = 68e^{2t}$, and we get the same result. $\endgroup$ – Jakobian Aug 1 '18 at 23:41
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    $\begingroup$ @Sobi while I agree, we can say that whenever a 2d vector $(x, y)$ is involved in our vector product, we replace it with $(x, y, 0)$. That way we don't have to memorize 2 formulas, but just 1 $\endgroup$ – Jakobian Aug 2 '18 at 0:20
  • $\begingroup$ @Sobi I used your formula and still get the same answer. $\endgroup$ – Arbuja Aug 2 '18 at 3:47
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The problem reduces to a simple miscalculation in end of the second section. In the calculation of the cross product’s norm, I assume there was a calculation error, as it should be $\sqrt{(68e^{2t})^2}$, which becomes $68e^{2t}$. Therefore our second curvature becomes $\frac{4}{\sqrt{17}}e^-t$, which is the same as your curvature by computing it the first way. (In one line in the first part there is a type where you write $e^{-2t}$ however your calculations and your work leading up to it doesn’t use that so I will assume it is a mistake).

In general, both formulas for curvature will hold, and be equal. The only time I can think of would be when it is not a curve embedded in $R^3$ in order to use the cross product, however if you use the formula provided by Sobi you can avoid this problem. The derivation of the cross product definition of curvature begins with the classic definition, so one will never work without the other.

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There seems to be only a small mistake in OP's calculation.

We can check the validity of the formula
\begin{align*} \frac{\|T^\prime(t)\|}{\|r^\prime(t)\|}=\frac{\|r^\prime(t)\times r^{\prime\prime}(t)\|}{\|r^\prime(t)\|^3}\tag{1} \end{align*} by taking real-valued functions $x=x(t), y=y(t)$ and setting \begin{align*} r^{\prime}(t)=\langle x(t),y(t)\rangle \end{align*}

We obtain \begin{align*} \|r^\prime(t)\|&=\sqrt{x^2(t)+y^2(t)}\\ T(t)&=\frac{r^\prime(t)}{\|r^\prime(t)\|}\\ &=\left\langle \frac{x(t)}{\sqrt{x^2(t)+y^2(t)}},\frac{y(t)}{\sqrt{x^2(t)+y^2(t)}}\right\rangle\\ T^\prime(t)&=\left\langle\frac{y(t)\left(y(t)x^\prime(t)-x(t)y^\prime(t)\right)}{\left(x^2(t)+y^2(t)\right)^{3/2}},\right.\\ &\left.\qquad\qquad-\frac{x(t)\left(y(t)x^\prime(t)-x(t)y^\prime(t)\right)}{\left(x^2(t)+y^2(t)\right)^{3/2}}\right \rangle\\ &=\frac{y(t)x^\prime(t)-x(t)y^\prime(t)}{\|r^\prime(t)\|^3}\Big\langle y(t),-x(t)\Big\rangle\\ \|T^\prime(t)\|&=\frac{|y(t)x^\prime(t)-x(t)y^\prime(t)|}{\|r^\prime(t)\|^3}\sqrt{x^2(t)+y^2(t)}\\ &=\frac{|y(t)x^\prime(t)-x(t)y^\prime(t)|}{\|r^\prime(t)\|^2}\tag{2}\\ r^\prime(t)\times r^{\prime\prime}(t)&=\langle x(t),y(t),0 \rangle\times\langle x^\prime(t),y^\prime(t),0\rangle\\ &=\langle 0,0,y(t)x^\prime(t)-x(t)y^\prime(t)\rangle\\ \|r^\prime(t)\times r^{\prime\prime}(t)\|&=|y(t)x^\prime(t)-x(t)y^\prime(t)|\tag{3} \end{align*}

From (2) and (3) we obtain \begin{align*} \color{blue}{\frac{\|T^\prime(t)\|}{\|r^\prime(t)\|}}&=\frac{|y(t)x^\prime(t)-x(t)y^\prime(t)|}{\|r^\prime(t)\|^3} \color{blue}{=\frac{\|r^\prime(t)\times r^{\prime\prime}(t)\|}{\|r^\prime(t)\|^3}} \end{align*} and the claim (1) follows.

Now we consider the special case \begin{align*} r^{\prime}(t)&=\langle x(t),y(t)\rangle\\ &=e^t\langle \cos(4t)-4\sin(4t),\sin (4t)+4\cos (4t)\rangle \end{align*}

We calculate \begin{align*} \|r^\prime(t)\|&=\sqrt{x^2(t)+y^2(t)}\\ &=e^t\sqrt{17}\\ r^{\prime\prime}(t)&=e^t\langle -8\sin(4t)-15\cos(4t),8\cos(4t)-15\sin(4t)\rangle\\ \|r^\prime(t)\times r^{\prime\prime}(t)\|&=|y(t)x^\prime(t)-x(t)y^\prime(t)|\\ &=e^{2t} |(\sin(4t)+4\cos(4t))(-8\sin(4t)-15\cos(4t))\\ &\qquad\qquad-(\cos(4t)-4\sin(4t))(8\cos(4t)-15\sin(4t))|\\ &=68e^{2t}\\ \|T^\prime(t)\|&=\frac{68e^{2t}}{17e^{2t}}\\ &=4 \end{align*} We finally obtain \begin{align*} \frac{\|T^\prime(t)\|}{\|r^\prime(t)\|}&\color{blue}{=\frac{4}{e^t\sqrt{17}}}\\ \frac{\|r^\prime(t)\times r^{\prime\prime}(t)\|}{\|r^\prime(t)\|^3}&=\frac{68e^{2t}}{\left(e^t\sqrt{17}\right)^3}\color{blue}{=\frac{4}{e^t\sqrt{17}}} \end{align*}

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