The derivative gives us the slope at 3 to be: -2x ------------ at x=3: -3/4
2sqrt(25-x^2)
</span><span>so we have a vector that is parallel to the slope of the tangent line is: <4,-3> </span> <span>the mag = 5 so; unit tangent = <4/5 , -3/5> </span> <span>since perpendicular lines have a -1 product between slopes we get the normal to be...
<3/5,4/5> </span> <span>It is <4,-3> because it is rise over run. Rise is y component of vector and run is x component of vector.</span>