How do I get two different answers? "A particle moves along the top of a parabola $y^2=2x$ from left to right at a constant speed of 5 units per seconds. What is the velocity of the particle at the point (2,2)?"
Method one: Let $t=y$. Then, $x=t^2/2$. So, $\vec{r}(t) = (t^2/2,t)$. Taking the derivative will give us the velocity, and since The point (2,2) corresponds to $t=2$, the velocity vector will be (2,1).
Method two: If the particle is moving at a constant speed in the $x$ direction, $dx/dt=5$. Then, $dy/dt = 5/y$ and at the point (2,2), $dy/dt = 5/2$. So, the velocity vector is (5,5/2).
What am I doing wrong in the first method? 
 A: The problem with your method one is that "taking the derivative" is not taking each coordinate separately but is taking the derivative on $s = \sqrt{x^2 + y^2} = {\sqrt {{t^4\over 4} + t^2}}$. Have you noticed that you did not even use the fact $v=5$ in this method?
EDIT: the real problem with this first method is that you cannot let $t=y$ by assuming the displacement in $y$ direction changes at constant speed.
Your second method is wrong too. $v = {ds\over dt} = 5$ but not ${dx\over dt} = 5$. In fact since $s = \sqrt{x^2 + y^2}$, this means 
$$v = {ds\over dt} = {x\over{\sqrt{x^2+y^2}}}\cdot {dx\over dt} + {y\over{\sqrt{x^2+y^2}}}\cdot {dy\over dt} = {\sqrt 2 \over 2}\cdot ({dx\over dt} + {dy\over dt})$$ at $(2,2)$
And ${dx\over dy} = y = 2$ means $dx = 2dy$ at the point so
$$5 =v =  {\sqrt 2 \over 2}\cdot ({dx\over dt} + {1\over 2}\cdot{dx\over dt})$$
which implies ${dx\over dt} = {10\sqrt2 \over 3}$ and therefore $dy\over dt$ = $5\sqrt 2\over 3$ at the point.
A: In your "method one" the parameter $t$ is not the time! it's just a renaming of $y$ and in the "method two" you cannot use $\frac{\mathrm dx}{\mathrm dt}=5\,\text{ms}^{-1}$ because we kwnow that the speed is constant not the speed in the x direction! So both methods give wrong answers.
The trajectory is $\pmb r(t)=\left(x(t), y(t)\right)$ and the velocity is $\dot{\pmb r}(t)=\pmb v(t)=\left(\dot x(t), \dot y(t)\right)$ with speed $$v(t)=\sqrt{\dot x^2(t)+ \dot y^2(t)}$$
Recall that if $y=f(x)$ the curve can be parametrized by $x=\lambda t$ and $y=f(\lambda t)$. So for $y^2=2x$ we have 
$$
\pmb r(t)=\left(\lambda t,\sqrt{2\lambda t}\right)\quad\text{and}\quad\pmb v(t)=\left(\lambda,\frac{\lambda}{\sqrt{2\lambda t}}\right)
$$
The point $\pmb r(t^*)=(2,2)$ is on the curve, so we have $\lambda t^*=2 $ and from
$$
v(t)=\sqrt{\lambda^2+ \frac{\lambda^2}{{2\lambda t}}}=5
$$
for all $t$ and in particular for $t^*$ we find 
$$
v(t^*)=\sqrt{\lambda^2+ \frac{\lambda^2}{{2\lambda t^*}}}=\lambda\sqrt{1+ \frac{1}{{4}}}=\lambda \frac{\sqrt 5}{2}=5\quad\Longrightarrow\quad\lambda=2 \frac{5}{\sqrt 5}=2\sqrt{5}
$$
Thus $t^*=\frac{2}{2\sqrt 5}=\frac{\sqrt 5}{5}$ and

$$
\pmb v\left(\frac{\sqrt 5}{5}\right)=\left(\lambda,\frac{\lambda}{\sqrt{2\lambda t^*}}\right)=\left(2\sqrt{5},\sqrt{5}\right)
$$

Another method is the following.
The velocity of the particle along parabola is in the direction of its tangent and can be represented as
$$
\pmb v(\theta)=\left(v\cos\theta,\,v\sin\theta\right)
$$
where $v(\theta)=||\pmb v(\theta)||=5\,\text{ms}^{-1}$ and $\theta=\theta(t)$ is the angle between $\pmb v$ and the $x$ axis varying with the time $t$.
From 
$$
\tan\theta=\frac{\mathrm d y}{\mathrm d x}=\frac{1}{y}
$$
At the angle $\theta^*$ corresponnding to the point $(2,2)$ we have
$$
\tan\theta^*=\frac{1}{2}\quad\Longrightarrow \sec\theta^*=\sqrt{1+\tan^2\theta^*}=\sqrt{1+\frac{1}{4}}=\frac{\sqrt 5}{2}
$$
so we have
$$
\cos\theta^*=\frac{1}{\sec\theta^*}=\frac{2}{\sqrt 5}\quad\text{and}\quad \sin\theta^*=\sqrt{1-\cos^2\theta^*}=\sqrt{1-\frac{4}{5}}=\frac{1}{\sqrt 5}
$$
Finally the velocity at $\theta^*$ corresponding to the point $(2,2)$ is
$$
\pmb v(\theta^*)=\left(v\cos\theta^*,\,v\sin\theta^*\right)=\left(5\frac{2}{\sqrt 5},\,5\frac{1}{\sqrt 5}\right)=\left(2\sqrt{5},\sqrt{5}\right)
$$
A third way may be the following.
If a planar curve in $\Bbb R^2$ is defined by the equation $y=f(x)$, where $f$  is continuously differentiable, then the arc length is given by:
$$
 s=\int _{a}^{b}{\sqrt {1+\left({\frac {\mathrm dy}{\mathrm dx}}\right)^{2}}}\mathrm dx
$$
So for $y^2=2x$, $y=\sqrt{2x}$, for $x,\,y\ge 0$ and the arc length is given by
$$
 s(t)=\int _{0}^{x(t)}{\sqrt {1+\left({\frac{\mathrm dy}{\mathrm dx}}\right)^{2}}}\mathrm dx=\int _{0}^{x(t)}\sqrt {1+\frac{1}{2x}}\;\mathrm dx
$$
and
$$
v(t)=\dot s(t)=\sqrt {1+\frac{1}{2x}}\dot x(t)
$$
This can be also derived immediately from
$$v(t)=\sqrt {\dot x^2(t)+\dot y^2(t)}=\sqrt{\dot x^2(t)+\frac{\dot x^2(t)}{2x(t)}}=\sqrt {1+\frac{1}{2x(t)}}\dot x(t)$$
We have $v(t)=5\,\text{ms}^{-1}$ and in particular for $t^*$ such that $(x(t^*),y(t^*))=(2,2)$ we have
$$\sqrt {1+\frac{1}{2x(t^*)}}\dot x(t^*)=\frac{\sqrt 5}{2}\dot x(t^*)=5$$
that is $$\dot x(t^*)=5\frac{2}{\sqrt 5}=2\sqrt{5}$$
From $v(t)=\sqrt {\dot x^2(t)+\dot y^2(t)}$ we find
$$
\dot y(t^*)=\sqrt { v^2(t^*)-\dot x^2(t^*)}=\sqrt { 5^2-(2\sqrt{5})^2}=\sqrt 5
$$
so we have
$$
\pmb v(t^*)=\left(\dot x(t^*),\dot y(t^*)\right)=\left(2\sqrt{5},\sqrt 5\right)
$$
