# Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.1 - Curves in Space and Their Tangents - Exercises 13.1 - Page 746: 25

$2 \sqrt 5 i+\sqrt 5 j$

#### Work Step by Step

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 $v=2 \sqrt 5 i+\sqrt 5 j$

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