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If a rock is thrown upward on the planet Mars with a velocity of $ 10 m/s $, its height in meters $ t $ seconds after is given by $ y = 10t - 1.86t^2 $.
(a) Find the average velocity over the given time intervals: (i) $ [1, 2] $ (ii) $ [1, 1.5] $ (iii) $ [1, 1.1] $ (iv) $ [1, 1.01] $ (v) $ [1, 1.001] $(b) Estimate the instantaneous velocity when $ t = 1 $.
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Calculus 1 / AB
Limits and Derivatives
The Tangent and Velocity Problems
September 30, 2020
Oregon State University
Idaho State University
In mathematics, the limit of a function is the value that the function gets very close to as the input approaches some value. Thus, it is referred to as the function value or output value.
In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.
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All right, we are here to solve this calculus problem about throwing a rock upwards on mars. So we're looking at the question text here, we've got a starting velocity of 10 m per second. We have our height in meters given by y equals 10 t minus 1.8 60 squared. We're asked to find the average velocity over these intervals. So in order to find the average velocity, if we remember, I'll use this notation since it's fairly common in calculus textbooks, that's actually going to be given by this definition. So this right here is a vector. And when we're talking about a change in a vector, in this case it's our position, we need to take whatever our final position is, again, this is our position vector. So this is a vector pointing from the origin to our position. Take that final minus the initial position vector. Now this is where if you're clever with setting up your coordinate systems, usually you will set one of these two vectors at the origin so that one of these two vectors is zero. That's the great thing about these vectors you can choose your coordinate system and make it so that either X final or X initial is equal to zero. So for this case we're throwing a rock upwards basically if we find its initial height and its final height, find a difference and divide by the time it took to get there then we can get our average velocities as we were asked. So rather than doing this individually every time I like to do repeated operations like this by coating them. Um I did some command line coding an octave to get these calculations. So for part a gonna copy these from this other tab. We can see that our average velocities are given by this. V average up at the top, ignore the code. The numbers are what matter? We've got 4.4 to 5.35 6.9. So I'm just gonna keep two decimal places on these because I don't think we need to be any more precise than that. Also, if this is for an ap class um maybe I should say we should keep three decimal places since I know that's at least the convention for ap physics when I took it. So our average velocities the shortest time interval for the longest time interval. We actually get the lowest average velocity. We had 4.42 m per second over that interval. So that's for 1 to 2. And the reason that's the shortest or the lowest that we get is because this thing is being thrown upward it's slowing down. And so because it's slowing down as it's going the longer we allow it to slow down, the lower its average speed is going to be. So the shorter and shorter intervals the average speed increases and then we have 5.356 point 94 356 point oh 94 And then for the very very shortest time intervals noticed that the speeds are very close together. 6.261 and 278 and 6.2 eight. So these are the average velocities over these time intervals. This one, it was over a time interval of 1.1 seconds. That's a very very short time interval. The reason I point that out is the question be asked us to estimate the average velocity of this thing. Um Yeah. Over oh no, estimate the instantaneous velocity, sorry, when T equals one. So this is the average velocity between T equals one and equals 1.1 If we look at what these numbers are doing, it looks like they're getting closer and closer to 6.28 ish 6.266 point 278 If I had to guess this thing's actual velocity at T equals one is probably around 6.26 and a quarter 6.3. Somewhere in there. Well, as it turns out, we can actually directly calculate the derivative in this case without too much trouble and we can see what our actual instantaneous velocity was. This will let me shrink it down. So our hide function was given by y equals 10 t minus 1.81 point 86 T squared. So at this point, if we take our time derivative in calculus class, we should find that this is equal to 10 minus. We've got to multiply by two, we'll get 372 T. So if T equals one, it looks an awful lot like Y equals 6.28 which is exactly what we had right here. So we do in fact, have a good estimate by looking at average philosophies over a short enough time interval, which is a reasonable thing to do. That's your solution. Have a good day.
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