Question

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)?

Answer 1

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`