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) = sqrt(25+x^2) and point: (3,4)?

1 Answer
Jul 26, 2018

Unit vector parallel to the graph at (3,4) is is
n^->=4/5i +3/5j^
Unit vector normal to the graph at (3,4) is is
n^->=3/5i-4/5j

Explanation:

f(x)=sqrt(25+x^2)
slope of the tangent indicates a vector parallel to the graph at a point
Differentiating
Let
y=f(x)
y=sqrt(25+x^2)
Squaring both sides
y^2=25+x^2
y^2-x^2=5^2
Now, Applyig chain rule and differentiating
2ydy/dx-2x=0
dy/dx=x/y
At
(x,y)-=(3,4);
dy/dx=3/4
Slope of a parallel line is
sqrt(3^2+4^2)=5
Unit vector parallel to the graph at (3,4) is is
n^->=4/5i +3/5j^
Normal is perpendicular to the parallel.
Thus, slope of the normal is
m_1m_2=-1
Slope of anormal line is
-4/3
Unit vector normal to the graph at (3,4) is is
n^->=3/5i-4/5j