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I have this question:

Set up an integral that represents the length of the curve.Use a calculator to find the length correct to 4 places.

$$y^2 = lnx$$ and $-1 \leq y \leq 1$

so implicitly differentiating:

$$2y \frac{dy}{dx} = \frac{1}{x}$$

$$\frac{dy}{dx} = \frac{1}{x2y}$$

and $y = \sqrt{lnx}$

So the curve is

$$\int_{-1}^1 \sqrt{1+(\frac{1}{2x\ln{x}})^2}$$

$$\int_{-1}^1 \sqrt{1+(\frac{1}{4x^2\ln{x}^2})}$$

Is that right? I can just plug that into a CAD right?

Wolfram exceeded the time allotted... did I set this up incorrectly?

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    $\begingroup$ I'm not sure, but shouldn't that be a $ln(x)$ instead of a $ln^2(x)$ in the last line, because you are squaring the square root of the log? $\endgroup$ – Seth Mar 19 '19 at 22:00
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The limits of integration in the OP are not correct. In fact, inasmuch as $\log(x)$ is not defined for $x\le 0$, the integral in the OP is also not defined.

To proceed correctly, we note that $y=\sqrt{\log(x)}$ for $y\in[0,1]$ ($x\in [1,e]$) and $y=-\sqrt{\log(x)}$ for $y\in [-1,0]$ ($x\in [1,e]$). Then, we see that

$$\begin{align} \text{Length of Curve}&=2\int_{1}^e \sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\\\\ &=2\int_{1}^e \sqrt{1+\frac{1}{4x^2\log(x)}}\,dx\\\\ &\approx. 4.25523282937328 \end{align}$$

Alternatively, we have $x=e^{y^2}$ so the

$$\begin{align} \text{Length of Curve}&=\int_{-1}^1 \sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy\\\\ &=2\int_0^1 \sqrt{1+4y^2e^{2y^2}}\,dy\\\\ & \approx 4.25523282937328 \end{align}$$

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  • $\begingroup$ Did you change the limit of integration and forget to multiply by 2 at the end? $\endgroup$ – Jwan622 Mar 19 '19 at 22:15
  • $\begingroup$ I think you took the integral but forgot to multiply it by 2 right? $\endgroup$ – Jwan622 Mar 20 '19 at 15:23
  • $\begingroup$ @Jwan622 Ah, now I understand. Yes, I did forget the factor of $2$. I've edited accordingly. $\endgroup$ – Mark Viola Mar 20 '19 at 16:07
  • $\begingroup$ I think I misunderstood something about what limits of integration to use... if the arc length formula that I use uses dy, then the limits of integration have to be for y. Conversely, if the arc length formula I use uses dx, then the limits have to be for x right? $\endgroup$ – Jwan622 Mar 22 '19 at 19:05
  • $\begingroup$ @Jwan622 Yes, that is correct. And that is, in fact, what you will see in the two approaches that I presented herein. $\endgroup$ – Mark Viola Mar 22 '19 at 19:15
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What would make this easier is to integrate with respect to $y$, and not $x$. You can re-write this as $x=e^{y^2}$. Now $\frac{dx}{dy}=e^{y^2}.2y$. Hence, the length of the curve can be written as $$\int_{-1}^1 \sqrt{4y^2e^{2y^2}+1}dy$$ You can now plug this into the calculator.

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  • $\begingroup$ @MarkViola- Ah if you meant the $4y^2$ factor, I have now added it $\endgroup$ – Anju George Mar 19 '19 at 21:59
  • $\begingroup$ How do you get to $x = e^y^2$ again? What's the rule? YOu can just take everything as an exponent to e? $\endgroup$ – Jwan622 Mar 19 '19 at 22:11

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