1

I need help with this problem. I'm calculating the length of the smooth simple arc of the portion of the parabola $y^2=16x$ which lies between the lines $x=0$ and $x=4$.

So I first parametrized the function like this: $t=x\Rightarrow y=\pm\sqrt{16t}$, thus $f(t)=(t, \pm\sqrt{16t})$. I first started with the positive square root. I calculated the derivative and the norm of the derivative: $f'(t)=(1,\frac{2}{\sqrt{t}})$ and $\Vert f'(t)\Vert=\sqrt{1^2+(\frac{2}{\sqrt{t}})^2}=\sqrt{\frac{t+4}{t}}$. After that, I used the formula of the length:$$l(t)=\int_{0}^{4} \sqrt{\frac{t+4}{t}}dt$$ the problem is that I don't know how to integrate that. Please help me.

davidllerenav
  • 833

5 Answers5

4

Try $t = u^2$. Following that, try $u=2\sinh x$.

More detailed solution:

Substitute $t = u^2$, note that $\mathrm{d}t = 2u\ \mathrm{d}u$:

$\displaystyle \int \sqrt{\frac{t+4}{t}}\ \mathrm{d}t = \int \sqrt{\frac{u^2+4}{u^2}}\cdot 2u\ \mathrm{d}u = 2\int\sqrt{u^2+4}\ \mathrm{d}u$

Second substitution, $u = 2\sinh x$, note that $\mathrm{d}u = 2\cosh x \ \mathrm{d}x$:

$\displaystyle 2\int\sqrt{u^2+4}\ \mathrm{d}u = 2\int \sqrt{4(\sinh^2 x + 1)}\cdot 2 \cosh x \ \mathrm{d}x = 4\int 2\cosh^2x \ \mathrm{d}x = 4\int (\cosh 2x + 1) \ \mathrm{d}x =4(\frac 12 \sinh 2x + x) + c = 2\sinh 2x + 4x + c$

and you can either work out the new bounds (easier) or try to express that in terms of the original variables (a little harder, but with the application of identities like $\displaystyle \sinh 2x = 2\sinh x \cosh x = 2\sinh x \sqrt{1+ \sinh^2 x}$, not that difficult.

Deepak
  • 27,479
  • For $t=u^2$ I ended up with $\int \frac{\sqrt{u^2+4}}{u}2du$. Then I did $u=2\sinh x$ as you said and I enden up with $\int \frac{\sqrt{4\sinh^2 x+4}}{2\sinh x}\cdot 2(2\cosh x dx)$, am I correct? What should I do next? – davidllerenav Mar 06 '19 at 23:08
  • @davidllerenav For $t = u^2$, don't forget $dt = 2udu$. Doesn't the denominator cancel out? – Deepak Mar 07 '19 at 01:03
  • I've edited my answer to be more detailed, I didn't have the time before. – Deepak Mar 07 '19 at 01:15
  • @davidllerenav I had made a numerical error in my sub earlier (wrote $4$ instead of $2$). All fixed now. I've verified this answer by working out the definite integral numerically and comparing the value I got. Spot on. – Deepak Mar 07 '19 at 07:01
  • Ok, thank you. Why the answer According to my book is $2\sqrt{2}+2\log(\sqrt{2}+1)$? – davidllerenav Mar 07 '19 at 07:22
  • The answer in your book is half the value of the definite integral; I can show that the integral reduces to $4 \sqrt 2 + 4 \log(\sqrt 2 + 1)$. So the integral has been evaluated correctly. Which means either that your book is wrong or you've forgotten a factor of half somewhere in your working for the arc length (remember that was the problem you were trying to solve, I only helped you work out the integral). – Deepak Mar 07 '19 at 08:07
  • @davidllerenav If you stated the question correctly, then your book's answer is wrong, no matter how you slice it. The parabola is mirror symmetric about the $x$-axis. It is uncertain if your question requires you to find the lengths both above and below the $x$-axis. But even for the portion above the axis, the arc length will be given by the integral. And the portion above and below will be double that. I don't see where the book answer is getting the factor of half from. I believe you're right and the book is wrong. – Deepak Mar 07 '19 at 09:49
  • I don't know. The question says: Sketch the following smooth simple arcs and calculate their length". I don't think I missed anything. Maybe the half below the x axis would do? – davidllerenav Mar 08 '19 at 05:02
2

For this kind of integral (rational function of $t$ and and the square root of a homographic function in $t$), the standard substitution is $$u=\sqrt{\frac{t+4}t}\iff u^2 =\frac{t+4}t,\quad u\ge 0,$$ to obtain the integral of a rational function in $u$; which one decomposes into partial fractions.

Bernard
  • 179,256
1

You can also set $u = 1+\frac4t$ to obtain $$\int_0^4 \sqrt{1+\frac4t}\,dt = \begin{bmatrix} u = 1+\frac4t \\ du = -4(u-1)^2\,dt \end{bmatrix} = \int_{2}^\infty \frac{\sqrt{u}}{4(u-1)^2}\,du$$ You can decompose the latter function as $$\frac{\sqrt{u}}{4(u-1)^2} = \frac1{16}\left(\frac1{(\sqrt{u}-1)^2}-\frac1{(\sqrt{u}+1)^2}\right)$$ Then notice that $$\int \frac{du}{(\sqrt{u}\pm1)^2} = \begin{bmatrix} z = \sqrt{u}\pm 1 \\ du = 2(z\mp 1)\,dz \end{bmatrix} = \int \frac{2(z\mp 1)}{z^2}\,dz$$ which should be solvable.

mechanodroid
  • 47,570
0

Hint

$$\dfrac4t=\cot^2u\iff t=?,0\le u\le\dfrac\pi4$$

Use Indefinite integral of secant cubed

lab bhattacharjee
  • 279,016
0

Letting $t=4\sinh^2 \theta$ yields $$ \begin{aligned} I&=\int_0^{\sinh (1)} \sqrt{\frac{4 \cosh ^2 \theta}{4 \sinh ^2 \theta}} \cdot 8 \sinh \theta \cosh \theta d \theta \\ & =8 \int_0^{\sinh ^{-1}(1)} \operatorname{cosh}^2 \theta d \theta \\ & =4 \int_0^{\sinh ^{-1}(1)}\left(1+\cosh (2 \theta)\right) d \theta \\ & =4\left[\theta+\frac{\sinh (2 \theta)}{2}\right]_0^{\sinh ^{-1}(1)} \\ & =4\left(\sinh ^{-1}(1)+\sqrt{2}\right) \\ OR&=4(\ln (1+\sqrt{2})+\sqrt{2}) \end{aligned}\\ $$

Lai
  • 31,615