Video: Vector-Valued Functions

Sketch the graph of the vector-valued function π‘Ÿ(𝑑) = (𝑑³) 𝑖 + (𝑑) 𝑗.

02:08

Video Transcript

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

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

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

Now, of course, we aren’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 let’s take a couple of values. Let’s consider 𝑑 equals zero and 𝑑 equals one. When 𝑑 is equal to zero, π‘₯ is equal to zero cubed or zero and 𝑦 is equal to zero. Similarly, when 𝑑 is equal to one, π‘₯ is equal to one cubed, which is, of course, one and 𝑦 is also equal to one. So we start at the point zero, zero and we move up to the point one, one. This means we’re moving on this curve from left to right. And now, we’ve finished. We’ve sketched the graph of the vector-valued function π‘Ÿ of 𝑑 equals 𝑑 cubed 𝑖 plus 𝑑 𝑗.

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