WEBVTT
00:00:01.140 --> 00:00:05.340
Until this point, weβve been most used to working with real-valued functions.
00:00:05.640 --> 00:00:08.430
These are functions whose range is a set of real numbers.
00:00:08.690 --> 00:00:13.360
Weβve worked with polynomials, exponentials, and trigonometric functions of this form.
00:00:13.880 --> 00:00:24.560
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.
00:00:24.950 --> 00:00:31.620
These are hugely important, as they can be used to measure many things such as velocity, acceleration, and force.
00:00:32.160 --> 00:00:45.910
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.
00:00:46.320 --> 00:00:52.520
And we use the letter π‘ to represent our independent variable since itβs usually used to represent time.
00:00:53.010 --> 00:00:56.520
And for the most part, we can evaluate these functions in the usual way.
00:00:56.790 --> 00:00:58.200
Letβs see what this might look like.
00:00:59.080 --> 00:01:07.590
For the given function π of π‘ equals two csc two π‘ π plus three tan π‘ π, evaluate π of π by four.
00:01:08.170 --> 00:01:11.010
Here, π is a vector-valued function.
00:01:11.180 --> 00:01:13.660
Itβs a function whose range is a set of vectors.
00:01:13.960 --> 00:01:23.940
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.
00:01:24.390 --> 00:01:30.600
For the horizontal component, the component of π, we get two csc two times π by four.
00:01:30.940 --> 00:01:37.470
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.
00:01:37.730 --> 00:01:40.610
So this becomes two over sin of π by two.
00:01:40.980 --> 00:01:45.550
And since sin π by two is simply one, we end up with two divided by one, which is two.
00:01:46.070 --> 00:01:48.840
We repeat this for the vertical component for π.
00:01:49.090 --> 00:01:51.070
Thatβs three tan of π by four.
00:01:51.510 --> 00:01:54.030
And of course, tan of π by four is simply one.
00:01:54.190 --> 00:01:56.120
So thatβs three times one, which is three.
00:01:56.660 --> 00:02:01.680
That means that π of π by four is two π plus three π.
00:02:02.650 --> 00:02:08.020
Contextually, this gives us the position of a point in the plane when π‘ is equal to π by four.
00:02:08.780 --> 00:02:13.480
Another important skill is to be able to identify the domain of vector-valued functions.
00:02:13.720 --> 00:02:15.150
Letβs see how that might work.
00:02:15.710 --> 00:02:25.540
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.
00:02:26.170 --> 00:02:29.280
Weβre looking to find the domain of our vector-valued function.
00:02:29.510 --> 00:02:36.530
Now, each component function in our vector-valued function will have its own domain, the set of values it can take.
00:02:36.670 --> 00:02:42.640
The domain of our vector-valued function π will be the intersection of these three domains.
00:02:42.820 --> 00:02:47.060
So our job is to begin by identifying the domain of each component function.
00:02:47.370 --> 00:02:51.640
Letβs begin by finding the domain of the horizontal component two π‘ squared.
00:02:51.910 --> 00:02:53.450
This is a polynomial.
00:02:53.680 --> 00:02:58.650
Now, we know that the domain of a polynomial is simply the set of all real numbers.
00:02:58.850 --> 00:03:03.390
So the domain of this component function is indeed the set of real numbers.
00:03:03.700 --> 00:03:09.030
And what about the domain of our π component, the domain of the square root of π‘ minus one?
00:03:09.420 --> 00:03:13.400
Well, weβre interested in the values the square root of π‘ minus one can take.
00:03:13.430 --> 00:03:17.470
And we know that the square root of a negative number is not a real number.
00:03:17.690 --> 00:03:24.720
This means we need the expression inside our square root π‘ minus one to be either equal to zero or greater than zero.
00:03:25.140 --> 00:03:30.010
Solving for π‘ by adding one to both sides and we obtain that π‘ must be greater than or equal to one.
00:03:30.440 --> 00:03:33.340
And we found the domain of this component function.
00:03:33.750 --> 00:03:37.950
Our final component function is five over two π‘ plus four.
00:03:38.240 --> 00:03:47.110
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.
00:03:47.480 --> 00:03:49.930
Our numerator and denominator are polynomials.
00:03:50.100 --> 00:03:51.960
So the domain is all real numbers.
00:03:52.340 --> 00:03:56.610
But we must ensure that two π‘ plus four cannot be equal to zero.
00:03:56.950 --> 00:04:01.170
By subtracting four from both sides, we see that two π‘ cannot be equal to negative four.
00:04:01.310 --> 00:04:05.150
And then dividing by two, we find that π‘ cannot be equal to negative two.
00:04:05.640 --> 00:04:09.420
So now, we have the domain of each of our component functions.
00:04:09.690 --> 00:04:14.200
Remember, the domain of our vector-valued function is the intersection of these.
00:04:14.270 --> 00:04:20.210
So itβs real numbers greater than or equal to one, but not including negative two.
00:04:20.570 --> 00:04:23.110
Of course, negative two is less than one.
00:04:23.590 --> 00:04:28.460
So our domain is simply real numbers greater than or equal to one.
00:04:28.810 --> 00:04:31.560
And we can write this using interval notation, as shown.
00:04:32.530 --> 00:04:36.830
In our next example, weβll look how we can sketch the graphs of vector-valued functions.
00:04:37.660 --> 00:04:45.180
Sketch the graph of the vector-valued function π of π‘ equals five cos π‘ π plus five sin π‘ π.
00:04:45.720 --> 00:04:50.330
Letβs begin by thinking about what this vector-valued function actually tells us.
00:04:50.440 --> 00:04:54.880
It takes a real number π‘ and the output is a position vector.
00:04:55.210 --> 00:05:00.690
The horizontal component is five cos π‘ and its vertical component is five sin π‘.
00:05:00.940 --> 00:05:10.470
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‘.
00:05:10.710 --> 00:05:14.450
And then, once weβre armed with this information, we could do one of two things.
00:05:14.660 --> 00:05:21.050
We could construct a table and try inputting various values of π‘ and plotting the π₯- and π¦-coordinates.
00:05:21.290 --> 00:05:27.510
Alternatively, we can try manipulating our equations to eliminate π‘ and see if we get something we recognise.
00:05:27.810 --> 00:05:29.190
Letβs look at that latter method.
00:05:29.480 --> 00:05:34.170
We begin by spotting that we have a function in cos of π‘ and one in sin of π‘.
00:05:34.360 --> 00:05:38.280
Now, we know that cos squared π‘ plus sin squared π‘ equals one.
00:05:38.560 --> 00:05:43.360
So weβre going to square each expression for π₯ and π¦ and find their sum.
00:05:43.840 --> 00:05:49.300
We obtain π₯ squared to be equal to five squared cos squared π‘ or 25 cos squared π‘.
00:05:49.460 --> 00:05:52.430
And similarly, π¦ squared is 25 sin squared π‘.
00:05:52.740 --> 00:05:58.500
Then, we see that π₯ squared plus π¦ squared is 25 cos squared π‘ plus 25 sin squared π‘.
00:05:58.820 --> 00:06:00.460
We factor by 25.
00:06:00.460 --> 00:06:05.370
And on the right-hand side, our expression becomes 25 times cos squared π‘ plus sin squared π‘.
00:06:05.710 --> 00:06:08.800
Of course, cos squared π‘ plus sin squared π‘ is equal to one.
00:06:09.000 --> 00:06:12.440
So we find that π₯ squared plus π¦ squared equals 25.
00:06:12.870 --> 00:06:15.830
And at this stage, you might recognise this equation.
00:06:16.160 --> 00:06:24.000
Itβs the equation for a circle whose centre is at the origin and whose radius is the square root of 25 or five units.
00:06:24.150 --> 00:06:27.040
And now, we have enough information to be able to sketch our graph.
00:06:27.250 --> 00:06:29.620
Itβs going to look a little something like this.
00:06:29.880 --> 00:06:32.180
Now, we arenβt actually quite finished.
00:06:32.520 --> 00:06:35.510
Notice how we created an equation in π₯ and π¦.
00:06:35.810 --> 00:06:37.810
These are parametric equations.
00:06:37.960 --> 00:06:43.350
And when we plot a parametric graph, we must consider the direction in which the curve is sketched.
00:06:43.420 --> 00:06:45.910
So weβre going to take a couple of values of π‘.
00:06:46.170 --> 00:06:48.900
Letβs take π‘ equals zero and π‘ equals one.
00:06:49.390 --> 00:06:54.780
When π‘ is equal to zero, we know that π₯ is equal to five times cos of zero, which is just five.
00:06:55.160 --> 00:06:58.850
Similarly, π¦ is equal to five sin of zero, which is zero.
00:06:59.070 --> 00:07:01.570
And so, we begin by plotting the coordinate of five, zero.
00:07:01.860 --> 00:07:10.110
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.
00:07:10.420 --> 00:07:14.520
So we move from five, zero to zero, five.
00:07:14.640 --> 00:07:18.910
This tells us weβre moving along this circle in a counterclockwise direction.
00:07:19.160 --> 00:07:20.800
And so, we add the arrows, as shown.
00:07:23.020 --> 00:07:30.910
Letβs have a look at another example, where forming a pair of parametric equations and eliminating the parameter can help us sketch the graph.
00:07:31.910 --> 00:07:38.060
Sketch the graph of the vector-valued function π of π‘ equals π‘ cubed π plus π‘ π.
00:07:38.600 --> 00:07:42.570
Letβs begin by recalling what this vector-valued function actually tells us.
00:07:42.790 --> 00:07:46.500
It takes a real number π‘, and it outputs a position vector.
00:07:46.710 --> 00:07:51.180
Itβs horizontal component is π‘ cubed and its vertical component is π‘.
00:07:51.370 --> 00:07:59.210
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‘.
00:07:59.590 --> 00:08:01.240
And there are two things we could do next.
00:08:01.240 --> 00:08:06.670
We could try forming a table and inputting values of π‘ and plotting the π₯- and π¦-coordinates.
00:08:06.930 --> 00:08:14.330
Alternatively, we can manipulate our equations for π₯ and π¦ to see if we can eliminate our parameter and get something we recognise.
00:08:14.530 --> 00:08:15.880
Letβs look at that latter method.
00:08:16.100 --> 00:08:18.530
We are told that π¦ is equal to π‘.
00:08:18.730 --> 00:08:23.080
We can, therefore, replace π‘ with π¦ in our equation for π₯.
00:08:23.200 --> 00:08:25.770
And we find that π₯ is equal to π¦ cubed.
00:08:25.810 --> 00:08:29.360
We could alternatively say that π¦ is equal to the cube root of π₯.
00:08:29.420 --> 00:08:30.920
So how do we sketch this graph?
00:08:31.040 --> 00:08:34.630
Well, we know how to sketch the graph of π¦ equals π₯ cubed.
00:08:34.920 --> 00:08:41.440
But we also know that π¦ is equal to the cube root of π₯ is the inverse function of π¦ equals π₯ cubed.
00:08:41.580 --> 00:08:48.220
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₯.
00:08:48.450 --> 00:08:53.790
And we obtain the graph of π¦ is equal to the cube root of π₯ or π₯ is equal to π¦ cubed, as shown.
00:08:54.280 --> 00:08:57.190
Now, of course, we arenβt actually quite finished.
00:08:57.440 --> 00:09:00.840
Remember, we created a pair of parametric equations.
00:09:01.120 --> 00:09:06.300
And we know that when we plot a parametric graph, we must consider the direction in which the curve is sketched.
00:09:06.440 --> 00:09:08.260
So letβs take a couple of values.
00:09:08.260 --> 00:09:11.160
Letβs consider π‘ equals zero and π‘ equals one.
00:09:11.350 --> 00:09:16.680
When π‘ is equal to zero, π₯ is equal to zero cubed or zero and π¦ is equal to zero.
00:09:17.070 --> 00:09:23.340
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.
00:09:23.700 --> 00:09:28.190
So we start at the point zero, zero and we move up to the point one, one.
00:09:28.410 --> 00:09:32.240
This means weβre moving on this curve from left to right.
00:09:32.480 --> 00:09:33.290
And now, weβve finished.
00:09:33.290 --> 00:09:38.550
Weβve sketched the graph of the vector-valued function π of π‘ equals π‘ cubed π plus π‘ π.
00:09:39.920 --> 00:09:46.470
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.
00:09:47.780 --> 00:09:55.500
Sketch the curve whose vector-valued function is given by π of π‘ equals sin π‘ π plus cos π‘ π plus π‘ π.
00:09:55.980 --> 00:09:58.710
Letβs begin by setting up some parametric equations.
00:09:58.990 --> 00:10:00.960
This time weβre working in three dimensions.
00:10:01.100 --> 00:10:06.420
So we set these as π₯, π¦, and π§, where π₯ is sin π‘, π¦ is cos π‘, and π§ is π‘.
00:10:06.780 --> 00:10:09.900
We could go about just substituting some values of π‘ in.
00:10:10.340 --> 00:10:11.590
Letβs see what that looks like.
00:10:11.880 --> 00:10:16.550
Letβs try values of π‘ at zero, π by four, π by two, three π by four, and π.
00:10:16.990 --> 00:10:23.760
When π‘ is zero, π₯ is sin of zero, which is zero, π¦ is cos of zero, which is one, and π§ is zero.
00:10:24.290 --> 00:10:31.160
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.
00:10:31.510 --> 00:10:38.030
Correct to three significant figures at 0.707, π§ is equal to π by four, which is 0.785.
00:10:38.530 --> 00:10:45.070
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.
00:10:45.820 --> 00:10:51.010
So we can see that the values of π§ increase at the same rate as the values of π‘.
00:10:51.430 --> 00:10:55.230
Itβs a little bit more difficult to spot whatβs happening in the π₯π¦-plane though.
00:10:55.450 --> 00:11:00.230
We do, however, know that cos squared π‘ plus sin squared π‘ is equal to one.
00:11:00.470 --> 00:11:04.930
So we can say that π₯ squared plus π¦ squared must be equal to one.
00:11:05.250 --> 00:11:10.900
This means in this plane we have a circle centred at the origin with a radius of one unit.
00:11:11.060 --> 00:11:12.780
We can also look at the values in our table.
00:11:12.780 --> 00:11:18.530
And we see that this starts at the point with coordinates zero, one and then moves to the point with coordinates one, zero.
00:11:18.890 --> 00:11:22.000
In this plane, it must be moving in a clockwise direction.
00:11:22.270 --> 00:11:29.600
As it does, according to our π§-coordinates, it spirals upwards, almost as if itβs moving around a cylinder.
00:11:29.820 --> 00:11:33.250
We can sketch the graph for π of π‘ then, as shown.
00:11:35.160 --> 00:11:38.790
In this video, weβve learned how to work with vector-valued functions.
00:11:38.980 --> 00:11:44.520
These are of the form π of π‘ equals π of π‘ π plus π of π‘ π plus β of π‘ π.
00:11:44.740 --> 00:11:52.220
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.
00:11:52.280 --> 00:12:12.760
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.