### Video Transcript

Until this point, weβve been most
used to working with real-valued functions. These are functions whose range is
a set of real numbers. Weβve worked with polynomials,
exponentials, and trigonometric functions of this form. In this video, weβre going to
evaluate and graph vector-valued functions, functions whose range is a vector or a
set of vectors and whose domain is a subset of the real numbers. These are hugely important, as they
can be used to measure many things such as velocity, acceleration, and force.

We are most interested in vector
functions π whose values are three-dimensional vectors such that π of π‘ equals π
of π‘, π of π‘, β of π‘, or π of π‘ π plus π of π‘ π plus β of π‘ π. And we use the letter π‘ to
represent our independent variable since itβs usually used to represent time. And for the most part, we can
evaluate these functions in the usual way.

Letβs see what this might look
like.

For the given function π of π‘
equals two csc two π‘ π plus three tan π‘ π, evaluate π of π by four.

Here, π is a vector-valued
function. Itβs a function whose range is a
set of vectors. And we can evaluate π of π by
four in much the usual way by substituting π‘ equals π by four into each component
function. For the horizontal component, the
component of π, we get two csc two times π by four. Of course, we know that csc π₯ is
equal to one over sin π₯ and two times π by four is equal to π over two. So this becomes two over sin of π
by two. And since sin π by two is simply
one, we end up with two divided by one, which is two. We repeat this for the vertical
component for π. Thatβs three tan of π by four. And of course, tan of π by four is
simply one. So thatβs three times one, which is
three. That means that π of π by four is
two π plus three π. Contextually, this gives us the
position of a point in the plane when π‘ is equal to π by four.

Another important skill is to be
able to identify the domain of vector-valued functions. Letβs see how that might work.

Find the domain of the
vector-valued function π of π‘ equals two π‘ squared π plus root π‘ minus one π
plus five over two π‘ plus four π.

Weβre looking to find the domain of
our vector-valued function. Now, each component function in our
vector-valued function will have its own domain, the set of values it can take. The domain of our vector-valued
function π will be the intersection of these three domains. So our job is to begin by
identifying the domain of each component function. Letβs begin by finding the domain
of the horizontal component two π‘ squared.

This is a polynomial. Now, we know that the domain of a
polynomial is simply the set of all real numbers. So the domain of this component
function is indeed the set of real numbers. And what about the domain of our π
component, the domain of the square root of π‘ minus one? Well, weβre interested in the
values the square root of π‘ minus one can take. And we know that the square root of
a negative number is not a real number. This means we need the expression
inside our square root π‘ minus one to be either equal to zero or greater than
zero. Solving for π‘ by adding one to
both sides and we obtain that π‘ must be greater than or equal to one. And we found the domain of this
component function.

Our final component function is
five over two π‘ plus four. We recall that the domain of the
quotient of two functions is equal to the intersection of the domain of each, but
where the denominator is not equal to zero. Our numerator and denominator are
polynomials. So the domain is all real
numbers. But we must ensure that two π‘ plus
four cannot be equal to zero. By subtracting four from both
sides, we see that two π‘ cannot be equal to negative four. And then dividing by two, we find
that π‘ cannot be equal to negative two.

So now, we have the domain of each
of our component functions. Remember, the domain of our
vector-valued function is the intersection of these. So itβs real numbers greater than
or equal to one, but not including negative two. Of course, negative two is less
than one. So our domain is simply real
numbers greater than or equal to one. And we can write this using
interval notation, as shown.

In our next example, weβll look how
we can sketch the graphs of vector-valued functions.

Sketch the graph of the
vector-valued function π of π‘ equals five cos π‘ π plus five sin π‘ π.

Letβs begin by thinking about what
this vector-valued function actually tells us. It takes a real number π‘ and the
output is a position vector. The horizontal component is five
cos π‘ and its vertical component is five sin π‘. We could say that in the
π₯π¦-plane, the π₯-value of any coordinate on our graph would be given by five cos
π‘ and the π¦-value would be given by five sin π‘. And then, once weβre armed with
this information, we could do one of two things. We could construct a table and try
inputting various values of π‘ and plotting the π₯- and π¦-coordinates. Alternatively, we can try
manipulating our equations to eliminate π‘ and see if we get something we
recognise.

Letβs look at that latter
method. We begin by spotting that we have a
function in cos of π‘ and one in sin of π‘. Now, we know that cos squared π‘
plus sin squared π‘ equals one. So weβre going to square each
expression for π₯ and π¦ and find their sum. We obtain π₯ squared to be equal to
five squared cos squared π‘ or 25 cos squared π‘. And similarly, π¦ squared is 25 sin
squared π‘. Then, we see that π₯ squared plus
π¦ squared is 25 cos squared π‘ plus 25 sin squared π‘. We factor by 25. And on the right-hand side, our
expression becomes 25 times cos squared π‘ plus sin squared π‘. Of course, cos squared π‘ plus sin
squared π‘ is equal to one. So we find that π₯ squared plus π¦
squared equals 25.

And at this stage, you might
recognise this equation. Itβs the equation for a circle
whose centre is at the origin and whose radius is the square root of 25 or five
units. And now, we have enough information
to be able to sketch our graph. Itβs going to look a little
something like this. Now, we arenβt actually quite
finished. Notice how we created an equation
in π₯ and π¦. These are parametric equations. And when we plot a parametric
graph, we must consider the direction in which the curve is sketched. So weβre going to take a couple of
values of π‘. Letβs take π‘ equals zero and π‘
equals one.

When π‘ is equal to zero, we know
that π₯ is equal to five times cos of zero, which is just five. Similarly, π¦ is equal to five sin
of zero, which is zero. And so, we begin by plotting the
coordinate of five, zero. Similarly, when π‘ is equal to one,
π₯ is equal to five times cos of one, which is zero, and π¦ is equal to five sin of
one, which is five. So we move from five, zero to zero,
five. This tells us weβre moving along
this circle in a counterclockwise direction. And so, we add the arrows, as
shown.

Letβs have a look at another
example, where forming a pair of parametric equations and eliminating the parameter
can help us sketch the graph.

Sketch the graph of the
vector-valued function π of π‘ equals π‘ cubed π plus π‘ π.

Letβs begin by recalling what this
vector-valued function actually tells us. It takes a real number π‘, and it
outputs a position vector. Itβs horizontal component is π‘
cubed and its vertical component is π‘. So we could say that in the
π₯π¦-plane, the π₯-value of any coordinate on our graph would be given by π‘ cubed
and the π¦-value would be given by π‘. And there are two things we could
do next. We could try forming a table and
inputting values of π‘ and plotting the π₯- and π¦-coordinates. Alternatively, we can manipulate
our equations for π₯ and π¦ to see if we can eliminate our parameter and get
something we recognise. Letβs look at that latter
method.

We are told that π¦ is equal to
π‘. We can, therefore, replace π‘ with
π¦ in our equation for π₯. And we find that π₯ is equal to π¦
cubed. We could alternatively say that π¦
is equal to the cube root of π₯. So how do we sketch this graph? Well, we know how to sketch the
graph of π¦ equals π₯ cubed. But we also know that π¦ is equal
to the cube root of π₯ is the inverse function of π¦ equals π₯ cubed. We then recall that to sketch the
graph of an inverse function, we reflect the graph of the original function in the
line π¦ equals π₯. And we obtain the graph of π¦ is
equal to the cube root of π₯ or π₯ is equal to π¦ cubed, as shown.

Now, of course, we arenβt actually
quite finished. Remember, we created a pair of
parametric equations. And we know that when we plot a
parametric graph, we must consider the direction in which the curve is sketched. So letβs take a couple of
values. Letβs consider π‘ equals zero and
π‘ equals one. When π‘ is equal to zero, π₯ is
equal to zero cubed or zero and π¦ is equal to zero. Similarly, when π‘ is equal to one,
π₯ is equal to one cubed, which is, of course, one and π¦ is also equal to one. So we start at the point zero, zero
and we move up to the point one, one. This means weβre moving on this
curve from left to right. And now, weβve finished. Weβve sketched the graph of the
vector-valued function π of π‘ equals π‘ cubed π plus π‘ π.

In our final example, weβre going
to look at how we might go about sketching a curve defined by a three-dimensional
vector-valued function.

Sketch the curve whose
vector-valued function is given by π of π‘ equals sin π‘ π plus cos π‘ π plus π‘
π.

Letβs begin by setting up some
parametric equations. This time weβre working in three
dimensions. So we set these as π₯, π¦, and π§,
where π₯ is sin π‘, π¦ is cos π‘, and π§ is π‘. We could go about just substituting
some values of π‘ in. Letβs see what that looks like.

Letβs try values of π‘ at zero, π
by four, π by two, three π by four, and π. When π‘ is zero, π₯ is sin of zero,
which is zero, π¦ is cos of zero, which is one, and π§ is zero. Then, when π‘ is equal to π by
four, sin of π‘ and cos of π‘ and therefore π₯ and π¦ are equal to root two over
two. Correct to three significant
figures at 0.707, π§ is equal to π by four, which is 0.785. By substituting π by two, π‘
equals three π by four, and π‘ equals π into each of our equations, we get the
values shown. So we can see that the values of π§
increase at the same rate as the values of π‘. Itβs a little bit more difficult to
spot whatβs happening in the π₯π¦-plane though. We do, however, know that cos
squared π‘ plus sin squared π‘ is equal to one. So we can say that π₯ squared plus
π¦ squared must be equal to one. This means in this plane we have a
circle centred at the origin with a radius of one unit.

We can also look at the values in
our table. And we see that this starts at the
point with coordinates zero, one and then moves to the point with coordinates one,
zero. In this plane, it must be moving in
a clockwise direction. As it does, according to our
π§-coordinates, it spirals upwards, almost as if itβs moving around a cylinder. We can sketch the graph for π of
π‘ then, as shown.

In this video, weβve learned how to
work with vector-valued functions. These are of the form π of π‘
equals π of π‘ π plus π of π‘ π plus β of π‘ π. We saw that we can find the domain
of a vector-valued function by looking for the intersection of the domains of each
of the component functions. We also saw that specifically when
working in two dimensions, if we form parametric equations of the form π₯ equals π
of π‘ and π¦ equals π of π‘ and then eliminate the parameter π‘, that can help us
to sketch the curves of these functions, but that we must also represent the
direction in which the curve is being sketched by using a set of arrows.