Velocity of a particle on a Parabola

Question

A particle moves along the top of the parabola $y^2 = 2x$ from left to right at a constant speed of 5 units per second. Find the velocity of the particle as it moves through the point $(2, 2)$.

So I isolate $y$, giving me $y=\sqrt{2x}$. I then find the derivative of $y$, which is $1/\sqrt{2x}$. And $\sqrt{2x}=y$ so the derivative of also equal to $1/y$. At $(2,2)$ the derivative is 0.5. Not sure where to go from there though. Any help is appreciated.

Answer 1

In that case is useful to use parametric equation that is

$$y^2=2x \implies \vec s(t)=(2c^2t^2,2ct) \implies \vec v(t)=s'(t)=(4c^2t,2c)$$

and since the point $(2,2)$ is reached at $t=1/c$, that is $\vec s(1/c)=(2,2)$, the velocity at that point is $\vec v(1/c)=(4c,2c)$ and since the speed is $5$ we have

$$\|v(1/c)\|=\sqrt{16c^2+4c^2}=5 \implies c=\frac{\sqrt 5}2$$

therefore the velocity at $(2,2)$ is

$$v(t)=\left(2\sqrt 5,\sqrt 5\right)$$

As an alternative

$$y^2=2x \implies 2ydy=2dx \implies \frac{dy}{dx}=\frac1y$$

therefore at the point $(2,2)$ the tangent vector is $(4,2)$ and then the velocity vector is $k(4,2)$ and form the condition on the speed we find

$$5=k\sqrt{16+4}\implies k=\frac{\sqrt 5}2$$

Answer 2

$(2y) dy/dx=2$; $dy/dx =1/y$.

Slope at $(2,2)$: $dy/dx =1/2= \tan \alpha$.

$\cos \alpha =2/√5$, $\sin \alpha =1/√5.$ (Pythagoras).

$v_x= v \cos \alpha$; $v_y = v \sin \alpha$, where $v = |\vec v|$ .

$\vec v = (v_x,v_y) $.

Answer 3

in cartesian co-ordinate sysytem :

$(speed)^2=\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2=25$ also,

$\left(\dfrac{dy}{dt}\right)=\dfrac{1}{y}\left(\dfrac{dx}{dt}\right) $ put it in above equation

to get ,

$\left(\dfrac{dx}{dt}\right)=\dfrac{5y}{\sqrt{1+y^2}}$ and $\left(\dfrac{dy}{dt}\right)=\dfrac{5}{\sqrt{1+y^2}}$

thus, at point (2,2) speed will be

$\sqrt{ (2\sqrt5)^2+(\sqrt 5)^2}=5 $ unit per second

in above calculations gravity is not taken into account

Answer 4

Consider the curve in a parametric form $\vec{z}=(x(t),y(t))$, in that case we know that

$$ \vec{v}=\frac{\mathrm{d}\vec{z}}{\mathrm{d}t}=(x'(t),y'(t)) $$

Additionally, we know that the speed is $\|\vec{v}\|=5$. Therefore, we need to find the direction of $\vec{v}$. Given the equation of the curve

$$ y^2=2x $$ we can write the parametric from in terms of $y$ alone: $$ \vec{z}=\left(\frac{y(t)^2}{2},y(t) \right) $$ Moreover, we can pick the parameter $t$ such that $y(t)=ct$, where $c$ is a constant. Then: $$ \vec{z}=\left(\frac{c^2t^2}{2},ct \right) $$ Then the velocity is $$ \vec{v}=(c^2t,c) $$ At the point $\vec{z}=(2,2)$, $t=2/c$. Therefore,

$$ \vec{v}=(2c,c) \Rightarrow \|\vec{v}\|=\sqrt{4c^2+c^2}=5 $$ Then, $c=\sqrt{5}$. Lets check our solution, $$ \begin{align} \vec{z}(t)=&\left(\frac{5t^2}{2},\sqrt{5}t\right)\\ \vec{v}(t)=&\left(5t,\sqrt{5}\right) \end{align} $$ at $t=\frac{2}{\sqrt{5}}$: $$ \begin{align} \vec{z}\left(\frac{2}{\sqrt{5}}\right)=&\left(\frac{5}{2}\cdot\frac{4}{5},\sqrt{5}\frac{2}{\sqrt{5}}\right)=(2,2)\\ \vec{v}\left(\frac{2}{\sqrt{5}}\right)=&\left(5\frac{2}{\sqrt{5}},\sqrt{5}\right)=(2\sqrt{5},\sqrt{5}) \end{align} $$