Find a unit vector (a) parallel to and (b) normal to the graph....

In summary, the conversation discusses finding a unit vector parallel and normal to the graph of a function at a given point. The process involves finding the slope of the tangent using differentiation, then making it a unit vector and rotating it 90 degrees for the normal vector. The conversation also touches on the concept of perpendicular vectors in 3 dimensions.
  • #1
NBAJam100
146
0

Homework Statement


Find a unit vector (a) parallel to and (b) normal to the graph of f(x) at the given point.

F(x)= -x^2 + 5
Point= (1,4)

Homework Equations



no equations really.

The Attempt at a Solution



Im not exactly sure what find a unit vector to the parallel and to the normal even means. I sketched the graph and plotted the points but I am not sure where to go from there. I know a unit vector is of length one. If i draw a unit vector off of a parabola, in the normal (assuming the normal is straight out perpendicular of the parabola). I would get a triangle (45,45,90) and sqrt(2)/2 sides and a 1 hypot. I am not really sure where to go from there...
 
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  • #2
You know how to find the slope of a tangent using differentiation, right? Then if i and j are the x and y unit vectors, a tangent vector is a vector a*i+b*j that has the same slope. So find one. Now make it unit. Then rotate it 90 degrees to find the normal.
 
  • #3
Dick said:
You know how to find the slope of a tangent using differentiation, right? Then if i and j are the x and y unit vectors, a tangent vector is a vector a*i+b*j that has the same slope. So find one. Now make it unit. Then rotate it 90 degrees to find the normal.


To find the slope of the tangent i would just take the derivative and plug in the value of x where the point is correct? I would then just use that slope to find the x and y components of the tangent? I could be totally wrong here. sorry, i am just having trouble visualizing this problem
 
  • #4
NBAJam100 said:
To find the slope of the tangent i would just take the derivative and plug in the value of x where the point is correct? I would then just use that slope to find the x and y components of the tangent? I could be totally wrong here. sorry, i am just having trouble visualizing this problem

Yes. Try it out. Don't just worry about it.
 
  • #5
Dick said:
Yes. Try it out. Don't just worry about it.


Ok, so the first step is to take the deriv. of -x^2+5, which is -2x. At the point (1,4) the slope is -2. So that means the norm is sqrt(5). So plotting the point from that point would be... P= (1,4) and Q= (2,2). When i put it into vector component form i get PQ= <1,-2>. To turn it into a unit vector i divide the vector by the norm, which is sqrt(5). So <1,-2>/sqrt(5). Which gives <1/sqrt(5), -2/sqrt(5)> for the final vector at the origin.
 
  • #6
NBAJam100 said:
Ok, so the first step is to take the deriv. of -x^2+5, which is -2x. At the point (1,4) the slope is -2. So that means the norm is sqrt(5). So plotting the point from that point would be... P= (1,4) and Q= (2,2). When i put it into vector component form i get PQ= <1,-2>. To turn it into a unit vector i divide the vector by the norm, which is sqrt(5). So <1,-2>/sqrt(5). Which gives <1/sqrt(5), -2/sqrt(5)> for the final vector at the origin.

Perfect. What about the normal?
 
  • #7
Dick said:
Perfect. What about the normal?


It would be the same except the V2 would turn from negative to positive because it is rotated above the x axis... So the normal would = <1/sqrt(5), 2/sqrt(5)> correct?
 
  • #8
NBAJam100 said:
It would be the same except the V2 would turn from negative to positive because it is rotated above the x axis... So the normal would = <1/sqrt(5), 2/sqrt(5)> correct?

Nope. To rotate 90 degrees you interchange x and y and then make one of them negative. Draw a picture.
 
  • #9
Dick said:
Nope. To rotate 90 degrees you interchange x and y and then make one of them negative. Draw a picture.


Alright, i went back and found where i went wrong. So it would just switch the two values and change the sign of one... So <2/sqrt(5), 1/sqrt(5). I am not 100% sure where you got the negative from because both the x and y components of the vector are in quadrant one.
 
  • #10
The point of "switch the two values and change the sign of one" is because two vectors are perpendicular if and only if their dot product is 0. If we start with <a, b>, "switch the two values and change the sign of one" we get either <-b, a> or <b, -a>. It doesn't matter which because <a, b>.<-b, a>= -ab+ ab= 0 and <a, b>.<b, -a>= ab- ab= 0.
 
  • #11
HallsofIvy said:
The point of "switch the two values and change the sign of one" is because two vectors are perpendicular if and only if their dot product is 0. If we start with <a, b>, "switch the two values and change the sign of one" we get either <-b, a> or <b, -a>. It doesn't matter which because <a, b>.<-b, a>= -ab+ ab= 0 and <a, b>.<b, -a>= ab- ab= 0.


OHHH! Wow, i totally forgot about that, thanks a bunch for all of the help ivy and dick!
 
  • #12
Finding perpendiculars in 3 dimensions is a lot more complicated! At a point on a line there is a whole plane of perpendicular vectors.
 
  • #13
HallsofIvy said:
Finding perpendiculars in 3 dimensions is a lot more complicated! At a point on a line there is a whole plane of perpendicular vectors.

Ahh, sounds fun! Maybe one day you'll be guiding me through a problem involving 3-d perpendiculars! haha!
 

FAQ: Find a unit vector (a) parallel to and (b) normal to the graph....

What is a unit vector?

A unit vector is a vector with a magnitude of 1. This means that it has a length of 1 unit and is often used to describe the direction of a vector.

How do you find a unit vector parallel to a graph?

To find a unit vector parallel to a graph, you first need to determine the slope of the graph at a given point. Then, you can use the formula (1, m) or (m, 1) to create a vector that is parallel to the graph. Finally, you can normalize the vector by dividing each component by the vector's magnitude to get a unit vector.

How do you find a unit vector normal to a graph?

To find a unit vector normal to a graph, you need to determine the slope of the graph at a given point. Then, you can use the formula (1, -1/m) or (-1/m, 1) to create a vector that is perpendicular to the graph. As with finding a unit vector parallel to a graph, you can normalize the vector to get a unit vector.

Why is finding a unit vector parallel and normal to a graph important?

Finding a unit vector parallel and normal to a graph is important because it allows us to describe the direction of a vector in a mathematical way. This is useful in many scientific fields, such as physics and engineering, where understanding the direction of a vector is crucial in solving problems and making predictions.

Can a unit vector be negative?

No, a unit vector cannot be negative. As mentioned earlier, a unit vector has a magnitude of 1 and represents a direction rather than a magnitude. Therefore, all components of a unit vector must be positive in order to maintain a magnitude of 1.

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