# Lesson Video: Vector-Valued Functions Mathematics • Higher Education

In this video, we will learn how to define, evaluate and graph vector-valued functions.

12:13

### Video Transcript

Until this point, weβve been most used to working with real-valued functions. These are functions whose range is a set of real numbers. Weβve worked with polynomials, exponentials, and trigonometric functions of this form. In this video, weβre 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 itβs usually used to represent time. And for the most part, we can evaluate these functions in the usual way.

Letβs 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. Itβs 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. Thatβs three tan of π by four. And of course, tan of π by four is simply one. So thatβs 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. Letβs 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.

Weβre 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. Letβs 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, weβre 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 itβs 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, weβll 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.

Letβs 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 weβre 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.

Letβs 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 weβre 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. Itβs 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. Itβs going to look a little something like this. Now, we arenβt 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 weβre going to take a couple of values of π‘. Letβs 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 weβre moving along this circle in a counterclockwise direction. And so, we add the arrows, as shown.

Letβs 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.

Letβs begin by recalling what this vector-valued function actually tells us. It takes a real number π‘, and it outputs a position vector. Itβs 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. Letβs 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 arenβt 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 letβs take a couple of values. Letβs 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 weβre moving on this curve from left to right. And now, weβve finished. Weβve sketched the graph of the vector-valued function π of π‘ equals π‘ cubed π plus π‘ π.

In our final example, weβre 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.

Letβs begin by setting up some parametric equations. This time weβre 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. Letβs see what that looks like.

Letβs 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‘. Itβs a little bit more difficult to spot whatβs 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 itβs moving around a cylinder. We can sketch the graph for π of π‘ then, as shown.

In this video, weβve 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.