# What is this length of this curve using a calculator?

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?

• 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? – Seth Mar 19 '19 at 22:00

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}

• Did you change the limit of integration and forget to multiply by 2 at the end? – Jwan622 Mar 19 '19 at 22:15
• I think you took the integral but forgot to multiply it by 2 right? – Jwan622 Mar 20 '19 at 15:23
• @Jwan622 Ah, now I understand. Yes, I did forget the factor of $2$. I've edited accordingly. – Mark Viola Mar 20 '19 at 16:07
• 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? – Jwan622 Mar 22 '19 at 19:05
• @Jwan622 Yes, that is correct. And that is, in fact, what you will see in the two approaches that I presented herein. – Mark Viola Mar 22 '19 at 19:15

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.

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