Sketching a vector valued function

  • Thread starter icesalmon
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  • #1
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Homework Statement


sketch r(t) = (-T+1)i + (4T+2)j + (2T+3)k


The Attempt at a Solution


I have a vector <-1,4,2> and a point (-1,-2,-3)
x= -1(t)+(-1)
y = 4(t)+(-2)
z = 2(t) +(-3)
and solve for t in terms of x
x = -t -1
t = -1-x
and rewrite the y and z expressions in terms of x
y = 4(-1-x) -2
z = 2(-1-x) -3
y= -4x-6
z = -2x -5

I graph y = -4x-6
and z = -2x -5 (i have no idea how to get a single expression out of these or how to address the problem of graphing r(t) without going through this process (should I just pick values for T (e.g t = 0, t = -1, t =1 etc... this is a rather tedious way to do this)
I end up getting two seperate lines, where obviously there is just a single line
 

Answers and Replies

  • #2
vela
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You have to make a picture that shows the curve in three dimensions.

Do you know what kind of curve this is?
 
  • #3
Office_Shredder
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I end up getting two seperate lines, where obviously there is just a single line
The graph of z = -2x-5 is actually a plane in three dimensions. What you got were two equations that have to hold simultaneously which is different from just plotting each of them on their own.


I agree with vela, the key to this problem is to identify from r(t) what kind of shape the graph will be, and then calculate the key properties that you need to draw the graph
 
  • #4
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it's a line
 
  • #5
Office_Shredder
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So to graph it what do you need to know about the line? Then think about how you can calculate those things.

For example if you wanted to graph a line in two dimensions, you would probably want to know its slope and its y-intercept, or something along those lines.
 
  • #6
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a vector parallel to that line (determined by the two points) and one of the points.
p(x,y,z) q(x1,y1,z1)
pq = <x-x1,y-y1,z-z1)
v = pq = <a,b,c>
x = x0 + at
y = y0 + at
z = z0 + at
this seems like it would be a good way to go, except I already have all the work done for me.
 
  • #7
HallsofIvy
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Homework Statement


sketch r(t) = (-T+1)i + (4T+2)j + (2T+3)k


The Attempt at a Solution


I have a vector <-1,4,2> and a point (-1,-2,-3)
x= -1(t)+(-1)
y = 4(t)+(-2)
z = 2(t) +(-3)
and solve for t in terms of x
x = -t -1
t = -1-x
and rewrite the y and z expressions in terms of x
y = 4(-1-x) -2
z = 2(-1-x) -3
y= -4x-6
z = -2x -5

I graph y = -4x-6
and z = -2x -5
(i have no idea how to get a single expression out of these or how to address the problem of graphing r(t) without going through this process (should I just pick values for T (e.g t = 0, t = -1, t =1 etc... this is a rather tedious way to do this)
I end up getting two seperate lines, where obviously there is just a single line
These are the equations of planes. They are both satisified on the line of intersection of the two planes.
This is a line and a line is determined by two points. When x= 0, we have y= -6, z= -5, the point (0, -6, -5). When x= -4, we have y= 16- 6= 10 and z= 8- 5= 3, the point (-4, 10, 3).

Draw the line through those two points.
 
  • #8
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8
are you saying y = 4(-1-x) - 2 and z = 2(-1-x) -3 are plane equations? If so, how can that be possible if r(t) is a line? or maybe it's referring to the intersection of the two parameter equations which, if they are planes, would be a line. You're being clear, i'm sure I don't understand the idea enough though.
 
  • #9
vela
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are you saying y = 4(-1-x) - 2 and z = 2(-1-x) -3 are plane equations? If so, how can that be possible if r(t) is a line?
Yes, they're plane equations because you're working in three dimensions. When you said you graphed y=4(-1-x)-2, I assume you meant you graphed it in the xy-plane, but by doing so, you're assuming z=0. You're adding that equation to make it into a line. Without the z=0 constraint, you have a plane.

Similarly, when you graphed z=2(-1-x)-3, you assumed y=0. Without that additional constraint, you again get a plane.

or maybe it's referring to the intersection of the two parameter equations which, if they are planes, would be a line. You're being clear, i'm sure I don't understand the idea enough though.
Right. At any point in the intersection of those two planes, the values of x, y, and z satisfy the parameterized equation for some value of t.
 

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