Question Video: Vector-Valued Functions Mathematics • Higher Education

Sketch the graph of the vector-valued function π‘Ÿ(𝑑) = (5 cos 𝑑) 𝑖 + (5 sin 𝑑) 𝑗.


Video Transcript

Sketch the graph of the vector-valued function π‘Ÿ of 𝑑 equals five cos 𝑑 𝑖 plus five sin 𝑑 𝑗.

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

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

And at this stage, you might recognise this equation. It’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. It’s going to look a little something like this. Now, we aren’t actually quite finished. Notice how we created an equation in π‘₯ and 𝑦. These are parametric equations. And when we plot a parametric graph, we must consider the direction in which the curve is sketched. So we’re going to take a couple of values of 𝑑. Let’s take 𝑑 equals zero and 𝑑 equals one.

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

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