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# Extra: completar la demostración

Transcripción del video
(clunking) - Thanks for hanging in there. I know this discussion is getting a little bit technical, but we finally have all the tools we need to complete the derivation of a formula for the touching point on a parabola. But, before we continue, let's back up just little bit, and remind ourselves why we're doing this again. Well, we need that touching point formula so that shots like this in Brave can be computed really efficiently. Right, because that touching point will allow us to write computer programs to draw each blade of grass, without having to draw all the individual string on lines. So to turn this into formulas, let's again label things. So this light magenta line is controlled by the parameter t, So I'm going to label, as before, this point Q and this point R. The dark magenta line is controlled by the parameter S, so I'm gonna call this point Q prime, and this point R prime. Now, let's start writing down a few things we know. Well, we know that Q is a fraction t along the line segment AB, which means I can write Q as one minus t times A plus t times B. Similarly, R is a fraction t along the line segment BC, so it can be written as one minus T times B plus t times C. Similarly, Q prime is s along the way from A to B, so I can write Q prime as one minus s times A, plus s times B, and I can write R prime, finally, as one minus s times B plus s times C. Okay, now this intersection point here that we're looking for, P, is somewhere on the line segment QR. But where on the line segment is it? Well, I'm gonna prove in a second that it's at fraction s. That is, I claim that P can be written as one minus s, times Q, plus s times R. Now, if that's true, something nice happens, because as s approaches t, this expression here approaches one minus t, times Q, plus t times R. And that's the thing I ultimately want to prove. So the only thing left to show is that the intersection point can be written this way. Why should that be the case? What I'm going to do, is I'm gonna substitute this expression for Q in here, this expression for R in here, and if I do that and rearrange, I'll leave that rearrangement to you, but the result is that P can be written as one minus s, times one minus t, times A, plus s times one minus t, plus t times one minus s, times B, plus s, t, times C. And now, if I rewrite this, using these expressions for Q prime, R prime, I see I can write P as one minus t, times Q prime, plus t times R prime. Well, this expression says that P is somewhere on the line segment Q prime R prime, and this expression says that P is somewhere on the line segment QR. And the only point that can be on both line segments is the intersection point. So our proof is complete. (bow stretching) (arrow thumps) Bullseye!