Welcome to a video that will show you how to determine a vector-valued function from a rectangular equation.
So for example if we have the rectangular equation y = 0.5 x squared minus 3.
As we see graphed here we should be able to express this using a vector-valued function.
The way we do this is express this as a
set of parametric equations and then set it up as a vector valued function .
So for example, on this equation:
if we let X = t, we can perform substitution here for X
and replace it with t, so we can say that y
is equal to 0.5 t squared.
Minus 3. So now that we have the parametric equations
for this curve we can now express this as a vectored valued function. Where
Vector t of t is going to be = to
t times the unit vector I plus,
0.5 T squared minus 3.
Times unit vector J or in component form.
We would have T 0.52 to the second -3.
Now remember when we express a curve as vectored valued function it gives the curve orientation.
Meaning as t increases this curve a sketched in a specific direction. So notice that
as T increases X also increases.
Which means that this curve would be sketched moving from left to right.
Again as T increases,
the curve is moving toward the right. Let's take a look at one that's a little bit more involved.
Here we want to write, here want to write the
equation X squared divided by 25 + y squared over 4 = 1.
As a vector-valued function , first we should recognize this as an ellipse as we see sketched here:
The next thing we have to remember is the trig identity that sine squared theta,
plus cosine squared theta is equal to 1.
So using this identity, if we can let
x equal some function in terms of t.
Or 4x squared over 25 equals either sine Squared t
or cosine squared t and then that y equals some function in terms of t.
So that y squared over 4 is either sine squared T or cosine squared T.
It would then equal 1. So a way we can do that is we can let
X equal 5 cosine t.
Notice if we square 5 cosine t we would have 25 cosine squared.
The 25's would simplify out, and then if we let y
equal since this is 4 let y = 2 sine t.
Well if we square y were going to have 4 sine squared t.
This would result in sine squared t and I'll go ahead and show how this works out.
Again if we have X squared over 25 we square this were going to have 25
cosine squared T over 25,
plus Y squared over 4 that would be 4 sine Squared t.
Equals 1 and again the 25 over 25 the 4 over 4.
Simplifies nicely and this is now a true equation in terms of t
So these parametric equations are correct.
Which make the transition to a vectored valued function pretty easy,
Where we're going to have 5 cosine t
Times a unit vector I
plus 2 sine t
times a unit vector J or in component form.
However now that it's a vector-valued function it does have an orientation.
So if we let t equal 0 the x component would be 5
And the y component would be zero. So we would be at this point here:
and then at pi over 2.
The x component would be zero and the y component would be at 2.
So we would be at here which means we are graphing this counter clockwise.
Based upon how we define the vector-valued function.
Okay that's gonna do it for this video. Thank you!