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Find parametric equations to go around the unit circle with speed $e^t$ starting from $x=1$, $y=0$. When is the circle completed?

Why is the book leaving out the constant of integration when solving this problem, or what am I missing?

Here's my work:

parametric equations of the unit circle are:
$x = r\cos(t) = 1\cos(t)$ and
$y = r\sin(t) = 1\sin(t)$.

Because $\lvert v\rvert$ (speed) is $e^t$ I will have to adjust the parameter $t$ so I will set this new parameter as $w$ giving me my position function of: $$R(t) = \cos(w)i + \sin(w)j.$$ Hence, $$v = R'(t) = -(dw/dt)\sin(w)i + (dw/dt)\cos(w)j,$$ and so $$\lvert v\rvert = dw/dt,$$ which gives $dw/dt = e^t$

To find $w$ I integrate $dw/dt$ giving $w = e^t + c$. Plugging this into my position function gives: $$R(t) = \cos(e^t + c)i + \sin(e^t + c)j.$$ Since we know the starting position is $x=1$, $y=0$, we have $\cos(1+c)=1$ and $\sin(1+c)=0$. Hence, $1+c = 0$, so $c = -1$.

the final $R(t)$ should be: $$R(t) = \cos(e^t - 1)i + \sin(e^t - 1)j,$$ however, the book says the answer is $R(t) = \cos(e^t)i + \sin(e^t)j$.

When is the circle completed? Since we know $\lvert v\rvert = e^t$, we integrate it to find distance giving $$\int \lvert v\rvert = e^t + c$$ and $e^t + c = 0$. $c = -1$.

$e^t -1 = 2\pi$.

this should give: $$t = \ln(2\pi + 1).$$ however, the book says the answer is $t = \ln(2\pi)$.

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I'm finding hard to understand what you did: first, what do you need that weird $\,\omega\,$ for? You only need $\,e^t\,$ ! Second, beginning at $\,x=1,y=0\,$ just means beginning at $\,t=0\Longrightarrow x=\cos 0=1\,\,,\,y=\sin 0=0\,$ ...! –  DonAntonio Jan 8 '13 at 12:17
Your right you don't need the ω I just plugged it in there to make it easier for me, yep I got that part I should have been more clear that's how I got cos(1 + c) = 1 and sin(1+c) = 0 from cos(e^(1)+c) = 1 and sin(e^(1) + c) = 0 –  Chris Jan 8 '13 at 12:46

1 Answer 1

The book probably has a misprint. Unroll the circular path to the line segment $[0,2\pi]$ on the $x$-axis, we have $dx/dt=e^t$ and hence $x(t)=e^t+C$. As $x(0)=0$, we have $C=-1$ and hence $x(t)=e^t-1$. Therefore $x(t)=2\pi$ when $t=\ln(1+2\pi)$.

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