Kinematics and the parametrization of a curve

An object moves along a path given by the equation $$y(x)=2x^{2}-3x-11$$ with a constant speed of 5m/s. Find the velocity at x=2.

My approach: We know that the speed of the object is given by $$25=v^{2} = v_{x}^2+v_y^2$$ Therefore $$25 =v_x^2(1+(\frac{v_x}{v_y})^2)=(v_x^2(1+(\frac{dy/dt}{dx/dt})^2)=(v_x^2(1+(\frac{dy}{dx})^2)$$ now, we know that $$\frac{dy}{dx}=4x-3$$ so $$v_x^{2}=\frac{25}{(1+(4x-3)^2)}$$ and $$v_x=\sqrt\frac{25}{26}$$ now $$v_y=\sqrt{25-\frac{25}{26}}$$ and finally $$\vec v (2)=\frac{5}{\sqrt{26}}(\vec i + 5\vec j)$$ is my approach correct? Also, let's take $$x=f(t)$$ is there any way to parametrize the curve with f so that we can take $$\vec v = df/ dt$$ as the velocity?

I think it is much simpler than what you are doing. As you mention, $$\frac{dy}{dx} = 4x-3$$. So we know that the tangent vector to the curve (no matter how it is parameterized) points in the direction of the vector $$(1,4x-3)$$, since this will be tangent to the curve. When $$x=2$$, this is $$(1,5)$$. So the velocity vector points in this direction. We just have to re-scale it so the length is $$5$$. The length of $$(1,5)$$ is $$\sqrt{26}$$, so the velocity should be $$\frac{5}{\sqrt{26}}(1,5)$$.
• When you solved for $v_x^2$, you have $(1+(4x-3))^2$ in the denominator, instead of $1+(4x-3)^2$. – Nick May 20 at 0:44