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)

Question

amistre64 · 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)

amistre64 · Answer

so we have a vector that is parallel to the slope of the tangent line is:  <4,-3> right?

amistre64 · Answer

the mag = 5 so; unit tangent = <4/5 , -3/5>

amistre64 · Answer

since perp lines have a -1 product between slopes we get the normal to be...

<3/5,4/5>

anonymous · Answer

Correct! It is <4,-3> because it is rise over run. Rise is y component of vector and run is x component of vector. Thank you.

amistre64 · Answer

youre welcome :)