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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – davidllerenav Mar 6 at 23:08
  • $\begingroup$ @davidllerenav For $t = u^2$, don't forget $dt = 2udu$. Doesn't the denominator cancel out? $\endgroup$ – Deepak Mar 7 at 1:03
  • $\begingroup$ I've edited my answer to be more detailed, I didn't have the time before. $\endgroup$ – Deepak Mar 7 at 1:15
  • $\begingroup$ @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. $\endgroup$ – Deepak Mar 7 at 7:01
  • $\begingroup$ Ok, thank you. Why the answer According to my book is $2\sqrt{2}+2\log(\sqrt{2}+1)$? $\endgroup$ – davidllerenav Mar 7 at 7:22
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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.

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  • $\begingroup$ Then du=$-\frac{2}{t^2\sqr{\frac{t+4}{t}}}dt$? $\endgroup$ – davidllerenav Mar 7 at 0:42
  • $\begingroup$ It's simpler than that: express $t$ in function of $u$, and calculate $\mathrm dt$ in function of $\mathrm d u$ (and $u$, of course). $\endgroup$ – Bernard Mar 7 at 0:56
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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.

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Hint

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

Use Indefinite integral of secant cubed

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