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since you're looking for the magnitude of velocity at that point.

what does that s(t) formula tell me in terms of the problem and what does the dy/dx tell me?

im sorry but im trying to understand the formulas

which one excludes time?

ah, but if i divide dy/dt by dx/dt i still have dy/dx in terms of t

The question is just asking, "Where is the object headed at time 0 and at time 1.2s?"

its asking the angle that it left my hand (time 0) and the angle that it hit the ground (time 1.2)

Actually, I should have said, "Where is the object?", not 'headed', since that would be velocity.

\[\frac{dy}{dx}(t)=\tan \theta\]

gradient?

Look at this...

tan (theta) = (dy/dt)/(dx/dt) = dy/dx

that makes sense... what do it do with that?

that makes complete sense, lemmy try that

at the same time?

so i got that it left at approx 137 degrees and landed at 83 degrees which looks right

yes, you're right (not about being stupid!).

alright so to get the greatest speed ill use the dy/dx formula?

what is wrt?

wrt = with respect to

But this is the square of speed, so we have to take the square root of both sides.

\[v=\sqrt{x'(t)^2+y'(t)^2}\]

4am...are you getting up to study, or have you not gone to bed?

not gone to bed

Have you got a test or something?

technically i dont actually take calculus

Is this just for physics or something?

cant take physics either, yet... im a freshman in hs

So what's this for?

Torture?

torture to who?

You.

I wasn't being rude :)

haha no, this is my fun

i got half an hour and about 7 speed questions that i know are wrong now

so i really have no option but taking the root of that big equation? :(

If you want the speed, no.

first thing i want is speed at time 0

dx/dt will be 0
and dy/dt will be 4.864... so i sqare that then root it? which is still 4.864...

what would be another aim? than computation?

dy/dt or y(t) ??

oh

yeah - I'm tired too

lol you on the eat cost too?

No, other side of the world. Sydney. Off sick.

are you a physicist or something or just enjoy math?

that is freaking awesome.

:)

lol, I've worked with people of all ages and abilities. All I care about is the attitude.

You don't want to use an analysis? What do you mean?

logic says that its decreasing till the max height and increasing therein after

i dont want to use any derivative analysiss

oh

One sec...

oh i see my mistake... omg this is pathetically horrible ><

i got 7 this time :DDD

the speed should be fastest when it hits the ground, right?

Sorry, I think I've answered a question you didn't ask.

Yes, the speed should be fastest when it hits the ground. You get that from energy considerations.

yeah i already got everything else

niceee finally i can glue this crap together lol

Good! I can't believe you're up at ~4am and doing this...lol.

Do you want to study math/science after school?

like after i graduate, keep studying?

Yeah

XD

good

( when you can answer) why did you get into math?

a break?

Sabbatical.

Just a formal break from university stuff.

are you a professor or a researcher?

I was a researcher. I'm just teaching for a while without the research.

i can tell that you come on here often haha

Lol, I know...it's a nice distraction from other things I should be doing.

i could probably use a little more help on a related rates problem DX

i'll see what i can do

i can do related rates, i just cant figure out how to set this one up ...

...whaa??

one sec.

and yeah i tried that, but i still cant set up the equation right

with the +6 or -6 or even without it, its the trig that i cant get down ><

Okay, I'm going to draw something.

oh, i'll continue and you can go...

ill prolly be back tonight, i always liked helping people with their math

iight im out- thank you for everything btw

no probs.

From your question, I assume any acceleration on the balloon in the vertical is ignored.

thats precisely what i got! and yes, there is no acceleration in this case

[thumbs up]