Identifying vertical extremum of parametric planar motion functions
When completing the square, the narrator factors -4t^2 + 16t as -4(t^2 + 4t) instead of -4(t^2 - 4t), so the vertex comes out at t = -2 instead of t = 2. The maximum y value of 13 is still correct. Notes in the interventions flag the sign error.
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The video finds the vertical extremum of a parametric function g(t) whose y-part is y(t) = -4t^2 + 16t - 3. Because the t^2 coefficient is negative, y(t) is a downward parabola with a maximum. Completing the square gives the vertex form, showing the highest y value is 13.
You'll be able to
- explain that a vertical extremum of a parametric function is the highest or lowest y value it reaches.
- tell whether a quadratic y(t) has a maximum or minimum from the sign of its t^2 coefficient.
- complete the square to find the vertex and the maximum value of a quadratic.
Key words
- parametric function
- A function where x and y are each given as functions of a third variable t.
- vertical extremum
- The highest or lowest y value a graph reaches.
- vertex
- The turning point of a parabola, where it reaches its maximum or minimum.
- completing the square
- Rewriting a quadratic as a(t - h)^2 + k so the vertex (h, k) can be read off.
Transcript
- [Instructor] We're told let g of t be a parametric function defined by g of t is equal to. So the x of t as a function of t is right over here, three to the t minus three, and then y of t as a function of t is right over there. Find the vertical extremum of g of t.
So pause this video, have a go at that before we do this together. Okay, when we're talking about vertical extremum, that's a very fancy way of saying how high do we get in terms of our y value? Vertical, that's the Y coordinate. And extremum is how far do we get there?
Now we don't know if this is gonna be a maximum or a minimum, but let's look at how y of t is defined. We could write y of t is equal to negative 4t squared plus 16t minus three. So given that the coefficient on the second degree term right over here, this is going to be some type of a parabola.
And since it's a negative coefficient, it's gonna be a downward facing parabola. So as a function of time, we're going to have some type of a parabola like this where you're going to have some maximum point. Well, how do we figure that out? Well, we just have to figure out the y value of the vertex.
And there's a bunch of ways that we could do that. I'm gonna do that by completing the square over here. So if we want to complete the square, what I could do is, this is... Let me write this over here. Y of t is equal to, lemme just factor the negative four out of this part right over here.
So actually let me do use the fact that I can use different colors. So if I factor out a negative four out of this, I get t squared plus 4t. I'm going to leave some space over here, and then I have a minus three. Let me write that in the other original blue color.
Minus three. No,w how can I turn what I have in the parentheses into a perfect square? So this into a perfect square. Well, if I take half of this coefficient and if I were to square it, and if I were to add it here, so I'll put add a four there. This makes this into a perfect square, but I can't just do it there.
I can't just add to one side. I either have to add the same thing to the other side or subtract it. And I'm not just adding four, I add the four here, but it's gonna be multiplied by negative four. So really, I'm adding a negative 16 to this side of this equation.
And so, what I wanna do, if I want to just make it net out, I could also add a 16. Just a reminder, I'm just completing the square here. I put a four here. But since that's going to be multiplied by this negative four out here, I actually added a negative 16 to the right hand side.
So to cancel that out, I'm now going to add a positive 16 right over there. Now what does this become? Well, this now becomes, let me write it this way. This is equal to negative four. And then I will write it like this, times x plus two squared. Oh no, yx plus two, t plus two.
So used to dealing with xs. T plus two squared. And then last but not least, these two parts right over here, I'm going to have a plus 13. So the question is, is when do we hit the maximum value? Well, we hit a maximum value because this part right over here is always going to be positive, but it's gonna be multiplied by a negative value.
So there's always going... This whole expression is always going to be non-positive. So it's always going to be either zero or take away from this 13. So what we want to think about is, all right, when this is equal to zero and that's going to be equal to zero, when t is equal to negative two, when t is equal to negative two, you have zero squared times zero, and then you would have y is equal to 13.
So we could say when t is equal to negative two, y is equal to 13. And this is going to be the maximum y that you can get because for any other t, this squared thing is going to start having positive value, but you multiply it by negative four, it's gonna have negative value.
It's gonna take away from that 13. So the vertical extreme room of g of t how high do we get in the vertical direction, in the y direction? It is our maximum value for y, which is 13.
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