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