## Thomas' Calculus 13th Edition

$2 \sqrt 5 i+\sqrt 5 j$
Consider the statement that the velocity vector is tangent to the graph of $y^2=2x$ at the point$(2,2)$ and having length $5$ The direction of the velocity vector is given by $y^2=2x$ and $2y\dfrac{dy}{dx}=2$ So, we have the slope at point$(2,2)$ will be $\dfrac{dy}{dx}=\dfrac{1}{2}$ Here, we have the tangent vector lies in the direction of the vector , that is, $i+\dfrac{1}{2} y$ Now, velocity $v=5(\dfrac{i+\dfrac{1}{2} y}{\sqrt{1+1/4}})$ Thus $v=2 \sqrt 5 i+\sqrt 5 j$