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

Elaina Mullen

Elaina Mullen

Answered question

2023-02-09

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

Answer & Explanation

Matthias Jackson

Matthias Jackson

Beginner2023-02-10Added 5 answers

Step 1: Determine the general equation for the slope of the tangent
The derivative of the function yields the slope of a line.
Given f ( x ) = x 2 + 5
the slope is d f ( x ) d x = 2 x (using the exponent rule for exponents)
Step 2: Find the tangent's precise slope at the given position.
At ( 3 , 9 ) , x = 3
So the slope is d f ( 3 ) d x = 2 3 = 6
Step 3: Determine the unit vector with slope of the tangent
Consider a unit vector with a base at the origin and a slope of 6:
XXX y x = 6
XXX y = 6 x
and since it is a unit vector:
XXX x 2 + y 2 = 1
XXX x 2 + ( 6 x ) 2 = 1
XXX 37 x = 1
XXX x = 1 37
XXX y = 6 37
Unit vector parallel to the tangent: ( x , y ) = ( 1 37 , 6 37 )
Step 4: Find the unit vector that is parallel 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
XXX ( x , y ) = ( 1 37 , - 6 37 )

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