Problem about curves. A particle is running along circumference $x^2+y^2=25$

Question

I'm considering a problem about curves. A particle is running along circumference $$x^2+y^2=25$$ with a costant modulus speed compliting a turn in 2 second. I need to determinate the acceleration in the point (3,4) starting from the equality $$\gamma(t)=5(cos\theta(t),sin\theta(t))$$.

I know the equation of the accelaration should be: $$\gamma(t)''=5(-\theta^2 cos\theta(t),-\theta^2 sin\theta(t))$$ I thought about determinate the angle from polar coordinates, however I'don't know whether it is a good idea:

\begin{cases} p=\sqrt {x^2+y^2} &&y=psin\theta &&x=pcos\theta \end{cases}

\begin{cases} p=\sqrt {3^2+4^2} &&4=5sin\theta &&3=5cos\theta \end{cases} In this way $$ \theta=53^° $$. But now i tried a lot of ways but I'm stucked. someone can help me. Thanks

Answer 1

You have $\gamma(t)=5(\cos at,\sin at)$, where $t$ is the time. Since $\gamma(2)=\gamma(0)=(5,0)$, then $a=\pi$. Thus $\gamma(t)=5(\cos \pi t,\sin \pi t)$. Thus $\gamma^{\prime\prime}(t)=-5\pi^2(\cos\pi t,\sin\pi t)$.

Now, for point $(3,4)$, we have $\cos\pi t_1=3/5$ and $\sin\pi t_1=4/5$ for some time $t_1\in[0,2]$. Then $\gamma^{\prime\prime}(t_1)=-5\pi^2(\cos\pi t_1,\sin\pi t_1)=-5\pi^2=-\pi^2(3,4)$.

Answer 2

The parametrization is just

\begin{align} \gamma(t)&=5(\cos (at), \sin (at)) \end{align}

But what is $a$?

Well, we know that $\gamma(0)=(5,0)=\gamma(2)$, so

$$ \gamma(2)=5(\cos 2a, \sin 2a)=(5,0) $$

Thus, $$ \cos 2a= 1\\ \sin 2a= 0 $$

And if we look a nice $a$: one between $0$ and $2\pi$, we see that $a=\pi$ works.

The acceleration is then

$$ \frac {d^2}{dt^2}\gamma(t)=5\pi^2(-\cos \pi t,-\sin \pi t)\tag 1 $$

Now, we want to figure out to which $t$ it corresponds the point $(3,4)$, but

$$ x=5\cos \pi t\\ y=5 \sin \pi t $$

Setting $(x,y)=(3,4)$, we get

$$ \frac 1 \pi \cos ^{-1} \frac 3 5=\frac 1 \pi \sin^{-1}\frac 4 5=t_0 $$

Plug that in $(1)$ to get your result.