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I need help with this problem:

Sketch the following smooth simple arcs in $\mathbb{R}^2$ and calculate their length.

  1. The curve parametrized by $f(t)=(e^t\cos t, e^t\sin t), t\in [1, 2]$.
  2. The graph (caternary) $y=\cosh x = \frac{1}{2}(e^x+e^{-x})$ between $x=-1$ and $x=1$.
  3. The portion of the parabola $y^2=16x$ which lies between the lines $x=0$ and $x=4$.

Ok, I did the first one like this:

$$f'(t)=(-e^t\sin t+e^t \cos t, e^t \cos t+e^t \sin t)$$

$$\Vert f'(t)\Vert = \sqrt{(-e^t\sin t+e^t \cos t)^2+(e^t \cos t+e^t \sin t)^2}=\sqrt{e^{2t}-2e^{2t}\sin t\cos t+e^{2t}+2e^{2t}\cos t\sin t}=e^t\sqrt{2}$$

$$l(t)=\int_{1}^{2} e^t\sqrt{2}dt=\sqrt{2}\int_{1}^{2}e^tdt=\sqrt{2}(e^2-e)$$Am I correct?

For the second one, I did the same process: $f'(t)=\frac{1}{2}(e^x-e^{-x})=\Vert f'(t)\Vert$, then I ended up with the lenght being $l(t)=\int_{-1}^{1}\frac{1}{2}(e^x-e^{-x})dt=0$, what am I doing wrong?

For the third one, I parametrized the function with$x=t\Rightarrow y=\sqrt{16t}\Rightarrow f(t)=(t, 4\sqrt{t})$. Thus $$f'(t)=\left(1,\frac{2}{t^{1/2}}\right)$$ $$\Vert f'(t)\Vert = \sqrt{(1)^2+\left(\frac{2}{\sqrt{t}}\right)^2}=\sqrt{1+\frac{4}{t}}=\sqrt{\frac{t+4}{t}}$$ Am I correct?

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(1) Yes, that's a correct calculation of arclength.

(2) The points are $(x,\cosh x)$. It looks like you forgot to add the term from differentiating the first coordinate.
Also, its an arclength; we should always take the positive square root. If you get something like $\sqrt{\sinh^2 x}$, that should be $|\sinh x|$.

(3) Be careful - that's only half the curve. You've parametrized the half with positive $y$, but not the half with negative $y$. Also, you still have to do the integration.

Also, this question asked you to sketch the curves. Here are all three graphs drawn on the same axes:

Figure 1

The curve in part (1) is in blue. The curve in part (2) is in red. The curve in part (3) is in green and purple; the part you found is in purple, and the other half is in green. The axes have ticks at each integer, with a large tick every 4.

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  • $\begingroup$ I don't understand why the points of (2) are $(x, \cosh x). Is that the parametrization of (2)? For (3) it would be (t, \pm\sqrt{16t}), right? How do I integrate that? I don't know how. $\endgroup$ – davidllerenav Mar 6 at 20:03
  • $\begingroup$ Also, how do you manage to sketch the graphs? I tried to but the first and second where pretty hard, I had to look the graph on desmos. $\endgroup$ – davidllerenav Mar 6 at 20:06
  • $\begingroup$ Yes, that's a parametrization of (2); the points $(x,g(x))$ are the standard parametrization of the graph of $g$. For (3), if you insist on using $x=t$ as the parameter, we'll need two functions - and the arclength of the whole curve is the sum of the arclengths of the two parts. Writing the two functions as if they were one with a $\pm$ sign doesn't help; split them apart and use both possibilities $(t,\sqrt{16t})$ and $(t,-\sqrt{16t})$. $\endgroup$ – jmerry Mar 6 at 20:14
  • $\begingroup$ I drew the graphs here with the programming language Asymptote; it's a 14-line program, and two of those lines are blank. Two preamble lines to import the right package and set the size, one line for each function, one line for each axis and one each for each colored graph. To do it by hand? Find a few key points, use the derivative to get tangent directions, and interpolate something smoothish. The more details you calculate, the more accurate it'll be. $\endgroup$ – jmerry Mar 6 at 20:18
  • $\begingroup$ So the parametrization would be $t=x$ and $y=\cosh t$. The I just need to do the same process I did for (1), right? For (3) then I integrate that and then I integrate for $negative$? How do I integrate that square root? $\endgroup$ – davidllerenav Mar 6 at 20:25

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