1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.