What is the process for finding the unit normal at every point on a parabola?

[wahaj]

I am working on a fluid mechanics problem that has a parabolic gate with equation
[tex] y = x^2 [/tex]
To solve the problem I need two vectors namely [itex] \vec{r} \ and \ \hat{n} [/itex]. Assuming origin is at x = 0, the vector [itex] \vec{r} [/itex] is a vector corresponding to each point on the parabola. I calculated that to be [itex] \vec{r} = x \hat{i} + x^2 \hat{j} [/itex]. [itex] \hat{n} [/itex] is a unit vector normal to the parabola projecting into the fluid. I can find the unit normal at a point but how would I go about finding the unit normal at every point on the parabola. The parabolic gate lies in the first quadrant and the fluid lies to the left of the gate (it comes in from the second quadrant). The range for the gate is [0,2.5].

 

[SteamKing]

If you can find the tangent to the parabola at any point, you can also find the normal line at any point. Just like the position vector r can be described in terms of two parameters x and x^2, so can the tangent line and the normals to the curve. It takes a little calculus and a little algebra.

 

[berkeman]

I am working on a fluid mechanics problem that has a parabolic gate with equation
[tex] y = x^2 [/tex]
To solve the problem I need two vectors namely [itex] \vec{r} \ and \ \hat{n} [/itex]. Assuming origin is at x = 0, the vector [itex] \vec{r} [/itex] is a vector corresponding to each point on the parabola. I calculated that to be [itex] \vec{r} = x \hat{i} + x^2 \hat{j} [/itex]. [itex] \hat{n} [/itex] is a unit vector normal to the parabola projecting into the fluid. I can find the unit normal at a point but how would I go about finding the unit normal at every point on the parabola. The parabolic gate lies in the first quadrant and the fluid lies to the left of the gate (it comes in from the second quadrant). The range for the gate is [0,2.5].

Hi wahaj -- can you give us some context to this question? Is this for your work or for school?

 

[wahaj]

This is for school. I am practising for my midterm. As for context, I don't know what else to tell you unless you guys want me to start getting into fluid mechanics. I attached a crude picture of the gate if that helps. Basically I need to find the force that the water exert on the gate and for that I need the two vectors.
I haven't done this in a while so bear with me here. The tangent lines are linear so they have equation [itex] y = mx + b [/itex]. The slope [itex] m = y' = 2x [/itex] so the general equation for tangent lines becomes [itex] y = 2x^2 + b [/itex]. The negative reciprocal of the slope for tangent line is the slope of the normal lines so [itex] y = \frac {-1}{2x} * x + b = \frac{-1}{2} + b [/itex]. This does not look right. Not to mention I don't have a value for b.

 

[Chestermiller]

The unit tangent is:
[tex]\frac{\vec{i}+2x\vec{j}}{\sqrt{1+(2x)^2}}[/tex]
From that, you should be able to write down the unit normal vector upon inspection, since the dot product of the unit normal and unit tangent is zero.

Chet

 

[wahaj]

Great. I was able to solve this problem and hopefully all similar problems in the future. Thanks for the help