Find a unit vector parallel to and normal to the graph of f(x) at the indicated point. f(x) = sqrt(25-x^2) point (3,4)

1 answer:

4 0

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.

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