How do you find a unit vector a) parallel to and b) normal to the graph of f(x)=-(x^2)+5 at given point (3,9)?

1 Answer
Jan 6, 2017

Parallel unit vector: #(vec(1/sqrt(37),6/sqrt(37)))#

Perpendicular unit vector: #(vec(6/sqrt(37),-1/sqrt(37)))#

Explanation:

Step 1: Determine the general equation for the slope of the tangent
The slope of a line is given by the derivative of the function.
Given #f(x)=x^2+5#
the slope is #(df(x))/(dx)=2x# (using the exponent rule for exponents)

Step 2: Determine the specific slope of the tangent at the given point
At #(3,9)#, #x=3#
So the slope is #(df(3))/(dx)=2 * 3=6#

Step 3: Determine the unit vector with slope of the tangent
Consider a unit vector with slope of #6# and a base at the origin:
#color(white)("XXX")y/x=6#

#color(white)("XXX")rarr y = 6x#

and since it is a unit vector:
#color(white)("XXX")sqrt(x^2+y^2)=1#

#color(white)("XXX")rarrsqrt(x^2+(6x)^2)=1#

#color(white)("XXX")rarr sqrt(37)x=1#

#color(white)("XXX")rarr x=1/sqrt(37)#

#color(white)("XXX")rarr y=6/sqrt(37)#

Unit vector parallel to the tangent: #(vec(x,y)) = (vec(1/37,6/37))#

Step 4: Determine the unit vector perpendicular to the tangent
Remember that if a line has a slope of #m# then all lines perpendicular to it will have a slope of #-1/m#.
So we are looking for a unit vector with slope #(-1/6)#
Using the same logic as in Step 3, we have
#color(white)("XXX")(vec(x,y))=(vec(1/sqrt(37),-6/sqrt(37)))#