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Ratling [72]
3 years ago
13

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)

Mathematics
1 answer:
cupoosta [38]3 years ago
4 0
<span>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)

</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>
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