# Simple Unit Vector Propblem

1. Jul 7, 2014

### hagobarcos

1. The problem statement, all variables and given/known data

Find the unit vector normal to the function f(x) = -3x^2 + 5 and the point (1,2)
and find a unit vector parallel, using f(x) and the point (1,2).

2. Relevant equations

How do I find the slope here? I attempted to just use the point (0,5), and then the given point (1,2),
resulting in a slope of -3/1 or -3.

And since a vector is sometimes a line between two points, used that to develop a form.
So vector parallel equals <1i, -3j>

The perpendicular would simply be pointing 90 degrees out.
So, after setting the dot product equal to zero, vector perp. equals <3,1>

Is this the right track?

3. The attempt at a solution

Then to find the unit vectors for the parallel bit and the perpendicular bit I divided by their magnitudes, sqrt(10).
Attached is a photo of what I did.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Jul 7, 2014

### Fredrik

Staff Emeritus
(1,2) is a point on the graph. The slope at that point is the derivative of the function at 1. Your way of calculating the slope only yields an approximation of the true slope (i.e. the derivative of f at 1). The approximation is better if you choose the second point closer to (1,2). But you don't want an approximation, you want the exact value. You could choose an arbitrary point (x,f(x)) compute the approximate slope as (f(x)-2)/(x-1) and then let x go to 1, but instead of actually doing that, you should recognize that the result of this limit operation is by definition f'(1). You should calculate f'(1) the way you usually find derivatives.

I think it's easier to find a vector parallel to the graph at (1,2) than to find one that's perpendicular to it. So I would find the parallel one first, and then use it to find the perpendicular one.

Last edited: Jul 7, 2014
3. Jul 7, 2014

### hagobarcos

Ohhhhhh! Right. *face palm* :D