I need help with this problem:
Sketch the following smooth simple arcs in $\mathbb{R}^2$ and calculate their length.
- The curve parametrized by $f(t)=(e^t\cos t, e^t\sin t), t\in [1, 2]$.
- The graph (caternary) $y=\cosh x = \frac{1}{2}(e^x+e^{-x})$ between $x=-1$ and $x=1$.
- 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?
(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:
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.