Video: Finding the Second Derivative of a Polynomial Function

Consider the points 𝐴 = (βˆ’1, 1) and 𝐡 = (4, 2). Parameterize the segment 𝐴𝐡, where 0 ≀ 𝑑 ≀ 1.


Video Transcript

Consider the points 𝐴 equals negative one, one and 𝐡 equals four, two. Parameterize the segment 𝐴𝐡, where 𝑑 is greater than or equal to zero and less than or equal to one.

Let’s begin by simply sketching the diagram, showing our points and our line segment 𝐴𝐡. It looks a little something like this. Now, we know we’re going to go from left to right generally. So we’ll begin by letting 𝑑 be equal to zero at the first point, at negative one, one. And that means we need to let 𝑑 be equal to one at the other end of our interval at the point four, two. Now, we’re going to use the vector form for the equation of a line. It’s π‘Ÿ equals π‘₯ nought, 𝑦 nought plus 𝑑 times π‘Žπ‘.

Now, whilst this is an equation of a line in two dimensions, this process can be extended to working with a line in three dimensions. We’ll next say that the vector π‘Ÿ is of the form π‘₯, 𝑦. Then, we’ll combine the vectors on the right to π‘₯ nought plus π‘‘π‘Ž, 𝑦 nought plus 𝑑𝑏. And we see that the only way for the vector on the left to be equal to the vector on the right is if its component parts are equal. That is, if π‘₯ is equal to π‘₯ nought plus π‘‘π‘Ž and 𝑦 is equal to 𝑦 naught plus 𝑑𝑏. In fact, this pair of equations is called the parametric Form of the equation of a line. Now, we can use these along with the information in our question to parameterize the line segment from π‘Ž to 𝑏.

Let’s go back to what we said earlier. We said 𝑑 is equal to zero at negative one, one. When 𝑑 is equal to zero, this gives us the values for π‘₯ nought, 𝑦 nought. So π‘₯ nought must be equal to negative one and 𝑦 nought must be equal to one. So we see the π‘₯ is equal to negative one plus π‘‘π‘Ž and 𝑦 is equal to one plus 𝑑𝑏. We’ll now use the fact that when 𝑑 is equal to one, π‘₯ is equal to four and 𝑦 is equal to two. So our first equation becomes four equals negative one plus one π‘Ž. And our second equation becomes two equals one plus one 𝑏. By adding one to both sides of our first equation, we find π‘Ž to be equal to five. And by subtracting one from both sides of our second, we find 𝑏 is equal to one. And we have our parametric equations to describe the line segment 𝐴 to 𝐡 for values of 𝑑 between zero and one. They are π‘₯ equals five 𝑑 minus one and 𝑦 equals 𝑑 plus one.

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