We want to find the domain
of the given vector valued function.
The domain of a vector valued function is the intersection
of the domains of X of T, Y of T, and Z of T,
were X of T, Y of T, and Z of T are the
X Y and Z components of the vector valued function.
Going back to our example,
notice how X of T is equal to T cubed,
Y of T is equal to the square root
of the quantity T plus four,
and the Z of T is equal to the square root
of the quantity five minus T.
Now we'll find the domain of X of T, Y of T, and Z of T,
and then we will find the intersection of all three domains,
which will give us the domain of the vector valued function.
To help us do this, we will look at the graph
of X of T, Y of T, and Z of T.
Here's the graph of X of T equals T cubed,
where the domain is set of all possible values of T.
Notice how T can be any real number, and therefore,
the domain of X of T would be all reals.
If we wanted to use interval notation,
we would have the open interval from
negative infinity to positive infinity.
Looking at the graph of Y of T,
notice how the possible values of T would be
all of the values of T that are
greater than or equal to negative four.
The domain of Y of T, again, is
T greater than or equal to negative four.
If we wanted to solve this algebraically, without a graph,
we would have to recognize that the quantity T plus four
can't be negative, which means it would
have to be greater than or equal to zero.
If we solve this for T, notice how, again,
we would have T is greater than or equal to negative four.
Using interval notation, we would have the interval
from negative four to infinity,
where the interval would be closed on negative four,
and open on infinity.
Then, finally, for Z of T, again, analyzing the graph,
notice how the possible values of T
would have to be less than or equal to positive five.
So, the domain of Z of T,
again, would be T less than or equal to positive five.
Algebraically, we would have to recognize
that the quantity five minus T must be
greater than or equal to zero,
so if we add T to both sides, notice how we
would have five is greater than or equal to T,
which is equivalent to T is less than or equal to five.
Again, using interval notation,
we would have an interval from negative infinity
to positive five, where the interval
is closed on positive five.
Now, let's go ahead and graph all three
intervals on the same number line,
and determine the intersection of these three intervals,
which will give us the domain of the vector valued function.
Let's let five be here, and negative four be here.
Well, first, the domain of X of T was all real numbers,
or all reals, which will be the entire number line.
The domain of Y of T, T is greater than or equal to
negative four, so we would have a closed point
on negative four, arrow to the right.
The domain of Z of T is T is
less than or equal to positive five,
so we would have a closed point on
positive five, and an arrow to the left.
Again, we are looking for the intersection of
these three intervals, or the values
that are in all three intervals.
Notice how this would be the
closed interval from negative four to positive five.
All of these values are in all three domains, and therefore,
this interval is the domain of the vector valued function.
So, using set builder notation, the domain would be T,
such that T is greater than or equal to negative four,
and less than or equal to positive five.
If we wanted to use interval notation,
we would have the close interval
from negative four to positive five.
Because the interval is closed, we use square brackets.
Before we go, let's take a look at the graph of R of T.
With the given vector valued function,
we would trace out the space curve.
I hope you found this helpful.