Sketching a vector valued function

In summary: So if we take any point on that line of intersection, we can find the value of t that corresponds to that point, and we can use that t value to determine the equation of the line.
  • #1
270
13

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 separate lines, where obviously there is just a single line
 
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  • #2
You have to make a picture that shows the curve in three dimensions.

Do you know what kind of curve this is?
 
  • #3
icesalmon said:
I end up getting two separate 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
it's a line
 
  • #5
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
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
icesalmon said:

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 separate 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
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
icesalmon said:
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.
 

What is a vector valued function?

A vector valued function is a mathematical function that maps a set of input values to a set of output values, where each output value is a vector. In other words, it is a function that takes in multiple variables and produces a vector as its output.

Why is it important to sketch a vector valued function?

Sketching a vector valued function helps visualize the behavior of the function and understand its properties. It also allows for easier interpretation and analysis of the function's behavior.

What are the steps for sketching a vector valued function?

The steps for sketching a vector valued function include identifying the domain and range, plotting points to determine the shape of the curve, and considering any transformations or restrictions on the function.

How do you interpret the graph of a vector valued function?

The graph of a vector valued function shows the path of the vector as it changes with respect to the input variables. The direction and magnitude of the vector at each point on the graph can provide information about the behavior of the function.

Can a vector valued function have a three-dimensional graph?

Yes, a vector valued function can have a three-dimensional graph since it maps multiple input variables to a vector output, which can have three components. In fact, many vector valued functions are used in three-dimensional applications such as physics and engineering.

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