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#