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MathEd-PII

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8
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of
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I
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X I
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plus V of XI + 1.
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over 2
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Delta T
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T
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And you just have to consider each step.
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volunteer
right? So there's
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one
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23456 steps.
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volunteer
Does that make sense
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I know I went a little fast
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Um, if you need me to explain anything.
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a little bit more, I can try to give more examples. I can
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um,
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I can explain to you how the trapezoid rule is effective.
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you know, what's the purpose of it? How does it actually work?
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student
It makes sense I just cant see the numbers on your bottow drawing good
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volunteer
Oh, you need me to expand my formula?
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student
yes please
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volunteer
Is that better
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And also, I, um
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volunteer
I see I got this, this, this X shouldn't be here. It's supposed to be a T.
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volunteer
And let me just rewrite this.
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volunteer
So it should be
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um T of I and your TI + 1, right? Basically, so it's your current, whichever um
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instance you're starting from, right, whichever, uh tea you're starting from, and then plus the next increment.
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student
ok ok
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Yeah
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If you want
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student
and this is all ddor part a so fart
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volunteer
I can also describe, that's a little too big.
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student
far
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volunteer
and this is all for
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parts so far. Yeah. And that's just for A. Yeah, this is just for A.
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volunteer
Um.
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but if you've seen how
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the trapezoid rule works.
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student
ok im just making sure bc im writing this on my paper
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volunteer
um, as opposed to like a right rhyming sums or left.
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Are you sure we can
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Yeah.
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we can use
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say
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X
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I
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all the way to
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XI + 2.
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XI + 1.
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So as you know, um, or you may know that Riemann sums and trapezoid rule are all just approximations.
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right? They're just attempts to
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uh, figure out the actual value below the curve, right? The area below the curve, which is very similar to what you do with derivatives.
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Right? Derivatives we had to take the limits, right, to approximate
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uh
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well, we took uh
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steps to figure out what the
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um I don't know if you remember, if you know the air function.
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So for the dri a derivative, you would have your function and then you would
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find that you'd find a point
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you would take the tangent out of the function at that point. And then if you wanted to, based on that tangent, try to and not knowing what the actual value of what the actual function is. You just have a curve and try to approximate the next value
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on the curve
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you would have um
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uh let's say you have the function
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of X2, right, the point X1, you have X1 is your original point and you have X2 is the point you want to know.
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You could say that F of X2 is equal to F of X1.
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Um, let me get some space
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By the way, are you following along? I don't wanna just keep running and running without, you know, you kind of being with me.
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It's not really helpful to either of us.
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student
no yeah im following
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right?
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So, um
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we have this approximation. What we're saying is that if we have some tangent line Y with some function F.
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right, that we could say why one
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is equal to F of X1, right? So we're saying that the tangent to the point.
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X1, F of X1.
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right?
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is, you know, that Y1 is the same at that point, right? We're saying that the tangent value is the same.
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But if we know the tangent function
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right, we say Y is equal to
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um the slope, which you can just say is M
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uh
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B +
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X
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right? So this would be the line, the tangent line.
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Actually, hold on, that's wrong.
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I forgot you got a
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you gotta take your vertices X minus X1 plus.
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Y1
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right? So this would be the tangent line
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at the point X1Y1. And this is derived from the um point form, which is Y minus Y1 is equal to MX minus X1, right? All they did was just move Y1 over.
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OK
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And if we
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you know, use this funk, use this value, we would say that Y is equal to M
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X minus X1 plus F of X1.
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OK? And the next step,
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will be to try to figure out what is this M? What does that mean?
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right? So we know
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M is
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well, you know what the derivative is, but it's just the same what Delta is just the change in why.
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