WEBVTT
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All right, we are here to solve this calculus
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problem about throwing a rock upwards on mars. So
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we're looking at the question text here, we've got
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a starting velocity of 10 m per second. We
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have our height in meters given by y equals 10
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t minus 1.8 60 squared. We're asked to find
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the average velocity over these intervals. So in order
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to find the average velocity, if we remember,
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I'll use this notation since it's fairly common in calculus
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textbooks, that's actually going to be given by this
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definition. So this right here is a vector.
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And when we're talking about a change in a vector
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, in this case it's our position, we need
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to take whatever our final position is, again,
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this is our position vector. So this is a
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vector pointing from the origin to our position. Take
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that final minus the initial position vector. Now this
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is where if you're clever with setting up your coordinate
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systems, usually you will set one of these two
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vectors at the origin so that one of these two
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vectors is zero. That's the great thing about these
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vectors you can choose your coordinate system and make it
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so that either X final or X initial is equal
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to zero. So for this case we're throwing a
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rock upwards basically if we find its initial height and
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its final height, find a difference and divide by
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the time it took to get there then we can
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get our average velocities as we were asked. So
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rather than doing this individually every time I like to
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do repeated operations like this by coating them. Um
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I did some command line coding an octave to get
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these calculations. So for part a gonna copy these
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from this other tab. We can see that our
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average velocities are given by this. V average up
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at the top, ignore the code. The numbers
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are what matter? We've got 4.4 to 5.35 6.9
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. So I'm just gonna keep two decimal places on
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these because I don't think we need to be any
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more precise than that. Also, if this is
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for an ap class um maybe I should say we
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should keep three decimal places since I know that's at
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least the convention for ap physics when I took it
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. So our average velocities the shortest time interval for
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the longest time interval. We actually get the lowest
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average velocity. We had 4.42 m per second over
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that interval. So that's for 1 to 2.
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And the reason that's the shortest or the lowest that
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we get is because this thing is being thrown upward
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it's slowing down. And so because it's slowing down
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as it's going the longer we allow it to slow
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down, the lower its average speed is going to
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be. So the shorter and shorter intervals the average
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speed increases and then we have 5.356 point 94 356
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point oh 94 And then for the very very shortest
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time intervals noticed that the speeds are very close together
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. 6.261 and 278 and 6.2 eight. So these
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are the average velocities over these time intervals. This
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one, it was over a time interval of 1.1
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seconds. That's a very very short time interval.
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The reason I point that out is the question be
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asked us to estimate the average velocity of this thing
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. Um Yeah. Over oh no, estimate the
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instantaneous velocity, sorry, when T equals one.
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So this is the average velocity between T equals one
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and equals 1.1 If we look at what these numbers
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are doing, it looks like they're getting closer and
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closer to 6.28 ish 6.266 point 278 If I had
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to guess this thing's actual velocity at T equals one
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is probably around 6.26 and a quarter 6.3. Somewhere
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in there. Well, as it turns out,
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we can actually directly calculate the derivative in this case
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without too much trouble and we can see what our
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actual instantaneous velocity was. This will let me shrink
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it down. So our hide function was given by
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y equals 10 t minus 1.81 point 86 T squared
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. So at this point, if we take our
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time derivative in calculus class, we should find that
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this is equal to 10 minus. We've got to
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multiply by two, we'll get 372 T. So
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if T equals one, it looks an awful lot
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like Y equals 6.28 which is exactly what we had
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right here. So we do in fact, have
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a good estimate by looking at average philosophies over a
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short enough time interval, which is a reasonable thing
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to do. That's your solution. Have a good
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day.