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 $t = \ln(2\pi)$.