Mathematics - 2017-04-11 (page 6 of 8)

« first day (2442 days earlier) ← previous day next day → last day (2596 days later) »

7:00 PM

C turn out to be zero?

And $s(0)=0$ tells us that $C=$?

Right.

OK, so the height of the ball at time $t$ is $s(t)=-16t^2+v_0t$. What does the problem mean when it says that the ball "just gets" to Colleen in the balloon?

so now i set both position equations equal to each other?

No ... is the answer to that.

That means, the ball stops moving, and is at the same height as the balloon

OK. So you need to find at what time the ball is at its maximum height and then figure out that maximum height. THEN we need to think about the balloon.

@Dodsy: Can you now finish the problem?

7:03 PM

But the way you're saying makes it sound like there's a possiblity the ball went up higher, then dropped down into her friends palms

No, not at all. I'm asking you to find the moment the ball stops and what the height is right then.

Did you accidentally @ me, Ted?

No, @Dodsy. I purposely did it because you said you were thinking about the problem.

Hi @TedShifrin

Hi Kasmir.

7:04 PM

How are you sir?

I'm leaving in a moment.

you now have the equation for distance given a variable t, if you put in the initial velocity into the v0 then we can find the distance travelled by inputting the time into the equation.

to find the velocity at any given t, we find the first derivative of the function.

At what time is the ball at its maximum height?

Oh haha am not here allways to ask you :D just wanted to ask how you feel =p

I still feel sick, so better not to ask, but thanks. :)

7:05 PM

I hope you get well! :)

when the velocity is equal to 0.

so we find the first derivative.

and equate it to 0

and solve for t.

@TedShifrin get better soon@ thanks a lot for you help! :D

Right. @WillN: So at what time is the ball at its maximum height? And then what is the maximum height?

0 = -32t + 2

Huh?

7:06 PM

nothing.

Uh huh.

:P

Remember that the goal of the problem is to determine the right $v_0$ that makes everything work.

the right initial velocity?

Go reread the problem!

Yeah, I haven't read it.

plugging 0 for $v_0$ gets me an imaginary number for t

7:08 PM

@WillN: Say what?

A hot-air balloon left the ground rising at 3 feet per second. 12 seconds later, Victoria threw a ball straight up to her friend Colleen in the balloon. At what speed did she throw the ball if it just made it to Colle

We're not putting in $v_0=0$. We're putting in $v(t)=0$ and solving for $t$.

Heya @robjohn

@TedShifrin but then i still have an unknown, $v_0$

@WillNjundong I believe that v0 is the initial velocity

ohh rightttt

7:09 PM

Precisely, @WillN. At the end of the whole problem, it is $v_0$ we're trying to find. You better reread the problem, too.

We could turn this into a quadratic and solve for the x's.

NOT YET.

Sorry.

So what is the time at which the ball stops? And what is its height? (This is the fourth time I've asked that.)

Well since we know that our equation for velocity is -16t^2 + v0t and that we are finding when the velocity is 0

right?

Is that our velocity equation or our displacement.

7:12 PM

@Dodsy I have a little thing for you

That's our displacement.

So we have to find the derivative. which would be -32t + v0 = 0

That's our displacement.

Right, so at long last $t=v_0/32$.

What is the height of the ball then?

Do you know Taylor Series, @Dodsy?

the height of the ball can be modelled by -16(v0/32)^2 + v0(v0/32) ?

is that right Ted?

Are you agreeing, @WillN? You're the one who needs to do the problem!

7:15 PM

yes, I was here working with s(t) = 0, thinking it was v(t) -.- My head is a little cloud atm

no wonder it ddint look right

You have to stay super organized and clear. This problem is complicated.

Give me a break

@robjohn sorry to hear that. Hope you're better now.

so plug in $t$ to to $s(t)$

oh so @TedShifrin $x - v0(v0/32) = -16(v0/32)^2$ ?

I should probably talk less and listen more at this point.

Carry on, guys.

Right, @WillN.

You can do that and simplify it a bit.

Now, we need to know that that is the same as the height of the balloon at that moment. What is the height of the balloon at that moment?

7:19 PM

36 feet.

@Givemeabreak Yes, but still very busy these days.

No.

so this is the ball's position when it is stationary: $s(V_0/32)= -V_0/2$

No.

432 feet

7:20 PM

You need to correctly compute $s(v_0/32)$, yes.

hi chat.

Dodsy. You're not thinking.

Give me a break

@robjohn I understand. Find a way to have more time to spend with mathematics. :-)

Hi, @Semiclassic.

Hi semi.

7:20 PM

More physics problems?

A hot-air balloon left the ground rising at 3 feet per second. 12 seconds later, Victoria threw a ball straight up to her friend Colleen in the balloon. At what speed did she throw the ball if it just made it to Collen?

Calc :(

This is calc?

yup

I thought this was physics.

7:21 PM

Sure, this is Calc II.

Ehhh, only just physics.

Calc I at my school

OK, I want to get out of here.

Just as I arrived lmao

So, Will, you're going to have to get faster and pay attention.

7:21 PM

Ted!

Hi Demonark.

Okay, I'll stay out of it, I think I'm making things more complicated.

While it's much faster to do everything if you know calculus, kinematics is basically just two equations.

How's everything going?

@Dodsy Do you know them?

7:22 PM

@Semiclassic: This problem has a slight twist because of a time shift.

@MeowMix Not at all!

(There's a third one, but it's derived from the others and I really count it as a prototype of energy conservation.)

Oh.

I guess you don't really need to know them

Well I might by the end of the summer :P

Sure. But it's still just modifications of said equations. @ted

7:23 PM

Did you read my question to you about the diophantine equations from arithmetica?

Oh sure. I'm just explaining that this problem is a bit harder than "usual."

$16(V_0/32)+V_0^2/32$

Sure. I agree there.

why 12 will?

sorry

7:23 PM

Problems like this usually boil down to some piecewise applications of those formulas, though.

^

Wait, ted.

I think the first v0/32 would be squared

would it not?

Interestingly, @Semiclassic. I've worked the problem another way and am not getting the same answer. I'm puzzled.

it should be

I'll also say, though, that these 'X just barely reaches Y' problems are always kinda annoying to explain the first time.

Yup. I'm trying to get to the end of the damn problem.

7:24 PM

That is interesting.

and ted, shouldn't the second one not be squared?

Okay, i'm sorry.

God damn computer

Oh damn. It's all wrong.

The height is $v_0(v_0/32) - 16(v_0/32)^2$.

Either you have to justify it as "when do you get a double root" or as "if you just barely reach it, then the speeds are also the same at that moment."

Windows is such a terrible operating system

7:25 PM

Oh!

Right, @Semiclassic. And my double root for time is giving something else. I can't figure out why.

Well, it's not bad, but sometimes it annoys me

That's perfect semi.

Huh, weird.

I never thought of that.

7:26 PM

should be $$-16\frac {V_0^2}{32^2}+\frac {V_0^2}{32}$$

You can justify that intuition if you think about what happens as you convert a secant line into a tangent line in a position versus time graph.

@WillNjundong this is the same as ted posed, he just put the second one first and the first one second!

it looks more presentable that way

No, @WillN. There's a $32^2$ in the first term.

oh right

7:27 PM

I'm-a make a mathematica picture of this, in any case.

@Semiclassic: I'm still getting a discrepancy.

Simplify that, @WillN, and you will get (eventually) $v_0^2/64$.

NOW, what about the balloon?

Are we assuming that $v(t)=+3$ for the balloon?

I mean, the balloon also experiences gravity.

Yes.

Mmkay.

We're told it rises at a constant rate.

7:29 PM

Hm.

ehh

"A hot-air balloon left the ground rising at 3 feet per second."

So now is it the time to equate both positions equations to each other ?

That's strictly-speaking only telling you what it's initial velocity is.

But now you have to think about the 12 seconds.

I thought it would be 3(12)

But I was wrong.

:C

7:30 PM

But, I'm also being silly because you'd need to include the bouyant force on the balloon.

Otherwise you'd get the rather silly conclusion that it stops rising in less than a second :P

Right, @Semiclassic. Let's just agree that $v(t)=3$, but $t$ is different.

Right.

It's like those damn ladders sliding down walls, and we all know they come away from the wall at a certain moment.

in any case, easy enough to create a family of plots in mathematica.

mutters about Newton's laws

7:32 PM

@Semiclassic: But when I set the two positions functions equal and look for a double root for $t$, I get a slightly different answer. I'm totally puzzled.

@Dodsy Basically, a Taylor Series is an expansion around a point which provides an approximation for the function using a series of monomials (so like an infinite polynomial)

Well, a picture should at least tell us which is right.

Hey everyone!

@WillN. When the ball has been in the air $t$ seconds, how long as the balloon been rising?

I'm using g=32 ft/s^2 for specificity.

7:33 PM

Yeah, we are too.

Well, -.

Eh, I tend to take g to always be positive and insert the minus sign appropriately.

Since otherwise talking about accelerations of 10g wouldn't make much sense :)

The Taylor series expansion for $\sin(x)$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$

womp womp

What a mess, Zach.

Oh @Ted Difftop today was real fun, we finished the proof of smooth BFT, derived continuous BFT, then we did homotopy for the rest of class

Hey @Mike!

7:34 PM

@TedShifrin the balloons position is given by 3t, and i dont know how to preserve the t

@Dodsy Given that, can you find the Taylor series expansion for $\cos(x)$?

Nevermind. I definitely misread that.

But it's not $3t$. When the clock starts, how high is the balloon already?

0 feet

What is t=0 here?

7:35 PM

The clock started when the ball is thrown.

Mmkay.

so t=0, v= 0 s =0 for the balloon

NO

You aren't listening to me.

oh right

my mistake gimme a moment

When Victotria throws the ball, things will be rather boring if Colleen is still on the ground.

7:36 PM

Yup.

balloon has been rising for 12 seconds

Semiclassic, fix my mistake

So where is it?

Trying.

36 feet up

So what should we use for the height of the balloon?

7:37 PM

36 feet

s= 36 + 3t

for the balloon

Right.

user image

Semi, your graphs are always so beautiful, I must say.

So I get 51 using the double root and $3(1+\sqrt{257})$ doing it the "right" way we're going through. WTF, Semiclassic?

The initial velocities shown are 48, 51, and 54 ft/s respectively.

7:38 PM

Mathematica draws beautiful graphs.

Yeah, it's mathematica. All credit to Wolfram for that.

I see now why the time is different, the ball was thrown after the balloon had already been in the air for some seconds.

Right.

Oh, @Semiclassic. I have it.

There's another version of this problem where you have a train accelerating and a person who starts to run after it.

7:40 PM

The one you just drew is having the ball have the same instantaneous speed as the balloon, not just stop.

It was thrown when the balloon was at 36 feet.

Right. Hence what I said earlier :)

so i plug in what i got for t

I'm pretty sure they intended it to be that Colleen grabs the ball just as it stops, but that's not really the right question. I doubt a Calc I class expects students to do what you and I just did.

into the balloon's position equation

7:41 PM

@Dodsy Do you see how to get cosine from that?

Right, @WillN, and then you will get a quadratic to solve for $v_0$. I really am heading out now.

I disagree. "Stops" in what reference frame?

well, the answer to this question was 51.

From colleen's perspectives, it won't have stopped unless the instantaneous speeds match.

So thanks a lot

7:41 PM

Agreed, @Semiclassic, but this is a Calc I question.

I just wouldnt let it go

Ehhh.

@MeowMix hmmmmmm well cos(x) would be d/dx(sin(x))

I dunno. But I'm outta here.

If you do it the double root way then you get that regardless.

7:42 PM

Haha bye ted.

later @ted

Thanks for helping me understand the question.

See you @Ted!

thanks alot!

also, the picture looks even better if you excise the first 12 seconds:

user image

7:43 PM

@Dodsy Yep.

So, what would you do to that series to get cosine?

(Hint, you just said it)

WTH i got 36 :(

it would be $1 - 3x^2 + 5x^4 + 7x^6....$ ?

is that it?

But there's denominators!!

You're close.

v0=36 doesn't even get you up to the balloon, alas.

where did i go wrong again

7:45 PM

Would it be over the natural log?

Noooooo

I'll write it again

$V_0^2/64 - V_0/32 -36 = 0$

$x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$

Go term by term.

What is the derivative of $x$?

1

Good.

7:47 PM

That doesn't look right, @will.

Now, what about $-\frac{x^3}{3!}$

1 - 3x^2 / 3!(ln3!) ?

If you're putting ln into these, you're making things waaay harder than it needs to be.

Where are you getting a logarithm?

$3!$ is a constant, remember

hm.

7:47 PM

The correct series for cosine will only have rational coefficients.

Think of it as just $3 \cdot 2 \cdot 1$

What is the derivative of $-\frac{x^3}{3 \cdot 2 \cdot 1}$?

@WillNjundong I didn't try to follow what @ted said above. What are your position functions?

hmmmmm

-1/2 ?

is that right?

That's the coefficient...

balloon = 36 +3t
ball $= s(t) = (\frac {-16V_0}{32})^2+\frac {(V_0)^2}{32} = \frac{(V_0)^2}{64}$
@Semiclassical

7:50 PM

@Dodsy Start with just $x^3$.

3x^2

but how would you find the derivative of a factorial?

Now, divide by $3\cdot2\cdot1$

It's a constant

isn't the derivative of 3 nothing?

It is, but it's multiplied by something.

So far i know the ball stops moving at $t = V_0/32$

7:51 PM

$\frac{-1}{3!}$ is a constant

And then we multiply it by $x^3$

yes that's right @WillNjundong

@WillNjundong The s(t) you wrote out isn't a function of time. So with that s(t) you'd just have the ball standing still forever.

He got this by using v(t)

and equating v(t) to 0

and solving for t.

That's what ted did. It's also what he concluded was ultimately -wrong-.

s(t) = -16t^2 + v0t

oh!

7:52 PM

So, let's not do that.

Sorry I am not on the same page anymore.

@Dodsy this was correct right?

I'd listen to semi

Do you agree that $\frac{-1}{3!}$ is a constant?

Understandable. Ted and I understood the issue, but that was at a higher level.

7:53 PM

he's a physicist.

Let's start with a simple thought experiment.

@MeowMix yes.

Suppose that you hurl the ball as hard as you can, and it passes Colleen in midair.

And what do we do when we have a constant in the power rule?

In that case, Colleen has two opportunities to catch it: As it passes Colleen, and as it falls back down.

7:54 PM

we would put it as the natural log I thought?

Okay, lets go step by step. I need to master this: Here is the same thing with new numbers: A hot-air balloon left the ground rising at 9 feet per second. 19 seconds later, Victoria threw a ball straight up to her friend Colleen in the balloon. At what speed did she throw the ball if it just made it to Colleen?

No!!!

The power rule is $\frac{d}{dx}ax^n = nax^{n-1}$

@WillNjundong Thought experiment first :)

ok

but if it's a constant

like 1

then it dissapears

7:55 PM

Not when it's multiplied by something...

So what we conclude from that is that, if we throw the ball 'too hard', then it'll pass Colleen with some speed.

oh so how do you do it?

from the question, colleen catches it as its coming up

Sure.

Well, We have $\frac{x^3}{3!}$

7:56 PM

Hi everyone

Now, suppose I throw it just a -little- slower at first.

That guy's derivative is $\frac{3x^2}{3!}$

What do you notice we can do now?

we can find the 3! ?

Then it'll still pass Colleen on the way up, but it'll take a little longer to reach Colleen and the speed with which is passes Colleen will be smaller as well. Right?

so it'd be 3x^2 / 6

7:56 PM

No, we can cancel out the 3's

yes

oh

To get $\frac{x^2}{2!}$

so then it's 2!

Yep!

7:57 PM

ah!

Time for the next one. $\frac{x^5}{5!}$

so then the series would be all the even numbers.

right?

Now, if you keep on with that logic, you'll eventually find the following scenario: The ball reaches Colleen, but is moving with the same speed as Colleen at that instant.

Yeah, you basically just get $5x^4/5!$ which is $x^4/4!$

cos(x) = 1 - x^2/2!

7:57 PM

And so on.

Oh cool!

That's awesome.

If it was moving any faster, it'd pass Colleen with some speed. If it was moving any slower, Colleen would outrun it.

So, our cosine Taylor series is now $1-\frac{x^2}{2!}+\frac{x^4}{4!} -\dots$

Now time for some even cooler stuff

Does that make sense?

I think there is some discussion going on..I will come later

7:58 PM

Do you know Euler's equation?

$e^{ix} = \cos(x) + i \sin(x)$

@Semiclassical thats some beautiful logic right there :)

makes perfect sense

Heh. I've done versions of this problem a lot :)

Right. So we need two things to happen.

One: y1(t)=y2(t) at some common time t

Two: v1(t)=v2(t) at that same time.

« first day (2442 days earlier) ← previous day next day → last day (2596 days later) »

Transcript for

Apr '1711

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$

site design / logo © 2024 Stack Exchange Inc; legal

mobile