How do you find a unit vector (a) parallel to and (b) normal to the graph of f(x) at the indicated points given function: f(x)=25+x2 and point: (3,4)?

1 Answer
Jul 26, 2018

Unit vector parallel to the graph at (3,4) is is
n=45i+35j
Unit vector normal to the graph at (3,4) is is
n=35i45j

Explanation:

f(x)=25+x2
slope of the tangent indicates a vector parallel to the graph at a point
Differentiating
Let
y=f(x)
y=25+x2
Squaring both sides
y2=25+x2
y2x2=52
Now, Applyig chain rule and differentiating
2ydydx2x=0
dydx=xy
At
(x,y)(3,4);
dydx=34
Slope of a parallel line is
32+42=5
Unit vector parallel to the graph at (3,4) is is
n=45i+35j
Normal is perpendicular to the parallel.
Thus, slope of the normal is
m1m2=1
Slope of anormal line is
43
Unit vector normal to the graph at (3,4) is is
n=35i45j