Find the average velocity over the given time intervals | Mathematics | Page 1

rich furnace Aug 31, 2024, 7:12 PM

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hasty jettyBOT Aug 31, 2024, 7:12 PM

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rich furnace

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rich furnace Aug 31, 2024, 7:12 PM

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So I'm a lil confused

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Bc I plugged in the numbers where I thought they were supposed to go but apparently that's incorrect

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This was an example problem I did earlier

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So I used this for reference when completing it

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I think if I can figure out what I did wrong in the first one I'll have no trouble solving the other ones

burnt fiber Aug 31, 2024, 7:17 PM

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rich furnace So I'm a lil confused

Can I know how you computed the average velocity over said intervals?

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the first one, for instance

rich furnace Aug 31, 2024, 7:19 PM

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burnt fiber Can I know how you computed the average velocity over said intervals?

Like this

burnt fiber Aug 31, 2024, 7:20 PM

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That doesn't sound right

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it should be (y(2) - y(1))/1

rich furnace Aug 31, 2024, 7:20 PM

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.

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ohhhhhhhhhhhh........

burnt fiber Aug 31, 2024, 7:20 PM

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with y(t) = 14t - 1.86t²

rich furnace Aug 31, 2024, 7:23 PM

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Wait wait
So should I just switch the places of the 2 numbers?

burnt fiber Aug 31, 2024, 7:26 PM

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No

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What is y(2) ?

rich furnace Aug 31, 2024, 7:28 PM

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I.... Am not sure

burnt fiber Aug 31, 2024, 7:32 PM

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What is y(t) ?

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for any t

rich furnace Aug 31, 2024, 7:33 PM

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1?

burnt fiber Aug 31, 2024, 7:42 PM

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y(t) is 14t - 1.86t²

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as indicated by the question

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now, y(2) is just that, but you replace t by 2

rich furnace Aug 31, 2024, 7:43 PM

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So for question 1

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I would just replace both with 2

burnt fiber Aug 31, 2024, 7:43 PM

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that is to compute y(2)

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you also need to compute y(1)

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and then compute their difference

rich furnace Aug 31, 2024, 7:47 PM

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And get the answer 12.14? Or am I still squaring the second number to get 12.40 instead?

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I think I did some math wrong actually

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Ohhhhhhhhh

burnt fiber Aug 31, 2024, 7:53 PM

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So what in the end?

rich furnace Aug 31, 2024, 7:57 PM

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So the answer ends up being 8.42

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But if I use the same method for the others it doesn't work?

burnt fiber Aug 31, 2024, 7:59 PM

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you also need to divide by the length of the interval

rich furnace Aug 31, 2024, 7:59 PM

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Ohhhhh

burnt fiber Aug 31, 2024, 7:59 PM

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in the first case, the interval is [1, 2] so the length of this interval is 1

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in the second case, the interval is [1, 1.5]

rich furnace Aug 31, 2024, 8:01 PM

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So the answer would be 21.49?

burnt fiber Aug 31, 2024, 8:05 PM

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I got something different for the 2nd one

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what did you do?

rich furnace Aug 31, 2024, 8:10 PM

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burnt fiber Aug 31, 2024, 8:12 PM

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No

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The first line is indeed y(1.5)

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but you seem to be computing y(1.5)/0.5 - y(1)

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which is not what you should be computing

rich furnace Aug 31, 2024, 8:13 PM

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Because I divided by 0.5?

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I thought that was the length?

burnt fiber Aug 31, 2024, 8:14 PM

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you should be computing $\frac{y(1.5) - y(1)}{0.5}$

viscid zephyrBOT Aug 31, 2024, 8:14 PM

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Rion

burnt fiber Aug 31, 2024, 8:15 PM

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not $\frac{y(1.5)}{0.5} - y(1)$

viscid zephyrBOT Aug 31, 2024, 8:15 PM

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Rion

rich furnace Aug 31, 2024, 8:15 PM

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.

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OHHHHH

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So for the very last part when it asks for the instantaneous velocity when t=1, how come it isn't 12.14?

burnt fiber Aug 31, 2024, 8:20 PM

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The instantaneous velocity is not computed the same way as the average velocity over a period

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it's given by the derivative

rich furnace Aug 31, 2024, 8:31 PM

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I'm a little confused

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So

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I learned about derivative yesterday when finding the slope of a secant

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I thought that was what a derivative was

void oracle Aug 31, 2024, 8:32 PM

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rich furnace I thought that was what a derivative was

Derivative is the slope of tangent, not secant.

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Besides, there are infinitely many secants at a given point, but at most one tangent.

rich furnace Aug 31, 2024, 8:33 PM

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Oh

void oracle Aug 31, 2024, 8:34 PM

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rich furnace

Better to first write the general formula for ∆y(1, 1 + ∆t)/∆t, simplify it, and then substitute various values of ∆t.

rich furnace Aug 31, 2024, 8:35 PM

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Sorry- What are the triangles?

void oracle Aug 31, 2024, 8:36 PM

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What?

rich furnace Aug 31, 2024, 8:36 PM

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∆

void oracle Aug 31, 2024, 8:36 PM

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Oh, delta.

rich furnace Aug 31, 2024, 8:36 PM

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OH

void oracle Aug 31, 2024, 8:37 PM

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That is used to show change in something. So, for example, ∆y(t1, t2) = y(t2) - y(t1).

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Anyway, try calculating y(1) and y(1 + ∆t) first.

burnt fiber Aug 31, 2024, 8:40 PM

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rich furnace I thought that was what a derivative was

As Darpinger said, the derivative at 1 would be to find the average speed on an interval [1, 1 + ∆t] for an infinitely small ∆t

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that would be the instantaneous speed

void oracle Aug 31, 2024, 8:42 PM

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Yes, but also doing it generally first will allow you to more easily find average velocities, too.

burnt fiber Aug 31, 2024, 8:42 PM

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The question here asks you to "estimate" this instantaneous speed based on your previous answers

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in question (a)

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which should converge to a certain value if you did it right

void oracle Aug 31, 2024, 8:44 PM

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And you can calculate it exactly, too.

burnt fiber Aug 31, 2024, 8:44 PM

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void oracle And you can calculate it exactly, too.

(Given more context, I think the OP does not know what a derivative is, this exercise seems to be an introduction to it)

void oracle Aug 31, 2024, 8:45 PM

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burnt fiber (Given more context, I think the OP does not know what a derivative is, this exe...

Well, for such a simple function you can use a difference quotient and let ∆t -> 0, as I proposed above.

rich furnace Aug 31, 2024, 8:50 PM

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Wait so

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Calculating y(1) would be 12.14 like earlier right? So what am I doing with y(1 + ∆t) if ∆ is to show a change in something?

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What's changing if I'm just using y(1)?

burnt fiber Aug 31, 2024, 8:52 PM

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∆t is a small quantity

void oracle Aug 31, 2024, 8:52 PM

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Well, for a secant it isn't necessarily small.

burnt fiber Aug 31, 2024, 8:52 PM

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Let's try to reformulate the problem in a different way

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We need to (algebraically, not numerically) compute y(t+h)

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for any t and h

void oracle Aug 31, 2024, 8:53 PM

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I think I'll move to my PC, I'll be back in a sec.

burnt fiber Aug 31, 2024, 8:53 PM

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So here, the task is not that complex

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you take the expression of y(t), which is: y(t) = 14t - 1.86t²

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and now, to compute y(t+h), you take the expression above and replace every occurrence of the symbol "t" with "(t+h)"

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what does that give?

rich furnace Aug 31, 2024, 8:55 PM

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14(t+h) - 1.86(t+h)²

burnt fiber Aug 31, 2024, 8:55 PM

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correct

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now, please expand the new expression

void oracle Aug 31, 2024, 8:55 PM

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Oh, a more general form, ok.

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That's even better.

burnt fiber Aug 31, 2024, 8:56 PM

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what would an expansion give? @rich furnace

rich furnace Aug 31, 2024, 8:58 PM

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14t + 14h - 1.86t² - 1.86h² - 3.72th

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?

burnt fiber Aug 31, 2024, 8:58 PM

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Perfect

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now, sneakily in there, you see y(t)

void oracle Aug 31, 2024, 8:59 PM

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Wait, should we expand it, even?

burnt fiber Aug 31, 2024, 8:59 PM

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if you group the 14t and -1.86t²

void oracle Aug 31, 2024, 8:59 PM

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I think it's better to find y(t + h) - y(t) without expanding everything.

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A bit less clutter.

burnt fiber Aug 31, 2024, 8:59 PM

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void oracle I think it's better to find y(t + h) - y(t) without expanding everything.

We'll get there

burnt fiber Aug 31, 2024, 9:00 PM

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rich furnace 14t + 14h - 1.86t² - 1.86h² - 3.72th

In other words: y(t+h) = y(t) + 14h - 1.86h² - 3.72th

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So that gives you the expression of y(t+h) - y(t) for free

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y(t+h) - y(t) = 14h - 1.86h² - 3.72th

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Now, all that is left to compute the average speed in [t, t+h] is to divide by the length of this interval

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What is the length of this interval?

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@rich furnace

rich furnace Aug 31, 2024, 9:08 PM

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burnt fiber What is the length of this interval?

The length would be the distance between t and t+h right? But how am I supposed to find the distance between them?

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For a second there I thought I was just supposed to subtract all the ts

burnt fiber Aug 31, 2024, 9:08 PM

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algebraically, by substracting them

rich furnace Aug 31, 2024, 9:08 PM

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Should've gone with my first thought

burnt fiber Aug 31, 2024, 9:08 PM

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I'm not trying to trip you here, don't worry

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Just trying to steer you in the right direction

rich furnace Aug 31, 2024, 9:10 PM

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So
Is 14h - 1.86h² - 3.72th the difference?

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Since I would need to divide to get rid of that other t

burnt fiber Aug 31, 2024, 9:11 PM

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My question was more so about the length of the interval [t, t+h]

rich furnace Aug 31, 2024, 9:11 PM

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Is the length not the difference?

burnt fiber Aug 31, 2024, 9:11 PM

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Which is?

rich furnace Aug 31, 2024, 9:12 PM

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14h - 1.86h² - 3.72th

burnt fiber Aug 31, 2024, 9:12 PM

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no I'm not asking about y(t+h) - y(t)

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I'm asking about the thing we need to divide this with

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the length of the interval

rich furnace Aug 31, 2024, 9:13 PM

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So..... h??

burnt fiber Aug 31, 2024, 9:13 PM

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Yes

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That was it

rich furnace Aug 31, 2024, 9:13 PM

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ohhhhhh.....

burnt fiber Aug 31, 2024, 9:13 PM

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So the average speed is given by $\frac{y(t+h) - y(t)}{h}$

viscid zephyrBOT Aug 31, 2024, 9:13 PM

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Rion

burnt fiber Aug 31, 2024, 9:13 PM

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but since you know what y(t+h) - y(t) is, you can plug that in

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And you will get the average speed in [t, t+h]

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So what does that yield?

rich furnace Aug 31, 2024, 9:15 PM

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Plug in with the actual numbers?

burnt fiber Aug 31, 2024, 9:15 PM

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rich furnace 14h - 1.86h² - 3.72th

This

rich furnace Aug 31, 2024, 9:16 PM

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OH
14 - 1.86h - 3.72t

burnt fiber Aug 31, 2024, 9:16 PM

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Very good

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now, you can see that as the interval length (h here) becomes very very small

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the term 1.86h will go closer and closer to 0

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so you can throw it away (it's so small that we'll ignore it)

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what remains?

rich furnace Aug 31, 2024, 9:18 PM

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14 - 3.72t

burnt fiber Aug 31, 2024, 9:18 PM

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that's your instantaneous speed at time t 🙂

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and it just happens that in the case of t = 1, you get the quantity you've been trying to approach all this time in question (a)

rich furnace Aug 31, 2024, 9:21 PM

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WA

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Wow

void oracle Aug 31, 2024, 9:22 PM

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Yeah, and in general the rate of change of f(t) is the limit of Δf(t, t + Δt)/Δt as Δt -> 0 (if that limit exists).

rich furnace Aug 31, 2024, 9:22 PM

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Thank you so much for your patience tytytyty sobsquietly

burnt fiber Aug 31, 2024, 9:23 PM

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You are welcome

rich furnace Aug 31, 2024, 9:40 PM

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+close

#Find the average velocity over the given time intervals