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 Post subject: Graph of a curve and line integralsPosted: Sat Jan 16, 2016 1:31 am

Joined: Sat Nov 07, 2015 6:12 pm
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Location: Larisa
Let $\gamma$ be a curve defined as $\gamma(t) = e^{-t} (\cos t, \sin t ), \;\; t \geq 0$.
1. Sketch the graph of the curve $\gamma( [0, +\infty))$.
2. Evaluate the line integrals:
$$\begin{matrix} & \displaystyle ({\rm i})\; \oint_{\gamma}\left ( x^2 +y^2 \right )\, {\rm d}s & & ({\rm ii}) \displaystyle \oint_{\gamma} (-y, x)\cdot {\rm d}(x, y) \end{matrix}$$

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 Post subject: Re: Graph of a curve and line integralsPosted: Sat Jan 16, 2016 1:38 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Hello Tolaso

The graph of the curve is presented at the following image:
Attachment:

The curve $\displaystyle{\gamma}$ is an infinitely many times differentiable map of $\displaystyle{\left[0,+\infty\right)}$

to $\displaystyle{\mathbb{R}^2}$ and we have that :

$$\gamma^\prime(t)=-e^{-t}\,\left(\cos\,t+\sin\,t,\,\sin\,t-\cos\,t\right)\,,t\geq 0$$

and
$$||\gamma^\prime(t)||=e^{-t}\,\sqrt{\left(\cos\,t+\sin\,t\right)^2+\left(\sin\,t-\cos\,t\right)^2}=\sqrt{2}\,e^{-t}\,,t\geq 0$$

If $\displaystyle{x(t)=e^{-t}\,\cos\,t\,,y(t)=e^{-t}\,\sin\,t\,,t\geq 0}$, then

$$x^2(t)+y^2(t)=e^{-2\,t}\,\cos^2\,t+e^{-2\,t}\,\sin^2\,t=e^{-2\,t}\,,t\geq 0$$

Now,

\begin{aligned} \int_{\gamma} \left(x^2+y^2\right)\,\mathrm{d}s&=\int_{0}^{\infty}\left(x^2(t)+y^2(t)\right)\,\||\gamma^\prime(t)||\,\mathrm{d}t\\&=\int_{0}^{\infty}e^{-2\,t}\,\sqrt{2}\,e^{-t}\,\mathrm{d}t\\&=\sqrt{2}\,\int_{0}^{\infty}e^{-3\,t}\,\mathrm{d}t\\&=\left[-\dfrac{\sqrt{2}}{3}\,e^{-3\,t}\right]_{0}^{\infty}\\&=\dfrac{\sqrt{2}}{3}\end{aligned}

and

\begin{aligned} \int_{\gamma} \left(-y,x\right)\cdot d(x,y)&=\int_{0}^{\infty}\left(-y(t),x(t)\right)\cdot \gamma^\prime(t)\,\mathrm{d}t\\&=\int_{0}^{\infty}-e^{-t}\,\left(-e^{-t}\,\sin\,t,e^{-t}\,\cos\,t\right)\cdot \left(\cos\,t+\sin\,t,\sin\,t-\cos\,t\right)\,\mathrm{d}t\\&=\int_{0}^{\infty}-e^{-t}\,\left[-e^{-t}\,\sin\,t\,\cos\,t-e^{-t}\,\sin^2\,t+e^{-t}\,\cos\,t\,\sin\,t-e^{-t}\,\cos^2\,t\right]\,\mathrm{d}t\\&=\int_{0}^{\infty}-e^{-t}\cdot (-e^{-t})\,\mathrm{d}t\\&=\int_{0}^{\infty}e^{-2\,t}\,\mathrm{d}t\\&=\left[-\dfrac{e^{-2\,t}}{2}\right]_{0}^{\infty}\\&=\dfrac{1}{2}\end{aligned}

P.S: The graph is part of the logarithmic spiral.

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