curvilinear motion...

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curvilinear motion...

Mathematics
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parametric equations: y(t)=-4.9t^2 +4.864t+1.2192 and x(t)=5.08t... i found the equations for velocity individually etc etc. i got s(t)=-1.9291t+.9574 for the speed of the particle at any time, t. and the next question is asking to find dy/dx?
You can find dy/dx using the chain rule. That is,\[\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}\]All you need to do is take the derivatives of your equations with respect to time and multiply them appropriately.
that is what i did to get the speed function s(t)... if i found dy/dx as that, why is it asking for it again, or is it in reference to another derivative of something else?

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I'm thinking the speed is given by,\[s(t)=\sqrt{\left( \frac{dy}{dt} \right)^2+\left( \frac{dx}{dt} \right)^2}\]Is that how you found it?
since you're looking for the magnitude of velocity at that point.
nope but that does make sense, now that i reread the question, it just wants that formula, which then means that the next question wants me to take the derivative of that?
Do what I wrote for the speed, and do what I wrote for dy/dx...you should get two different answers. Does that help?
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
s(t) is the speed - the magnitude of the velocity. dy/dx just gives you the rate of change of the y co-ordinate as x varies - so it's just looking at how vertical displacement varies with horizontal...nothing special.
so finding dy/dx pretty much has nothing to do with the speed function and my teacher just put it in there randomly...?
Sometimes it's useful to have motion problems in terms of displacements only, excluding time. It's easier for understanding the geometrical properties of the path taken.
Well, speed is defined as the *time* rate of change of distance. dy/dx makes no mention of time. I'm not sure what your teacher intends.
which one excludes time?
ah, but if i divide dy/dt by dx/dt i still have dy/dx in terms of t
Well, sometimes we solve one coordinate for time and sub. it into the other coordinate to combine the coordinates into a relationship between each other, that's all.
Yes, you may have it in terms of t, and that's fine. That would just allow you to answer the question, "What is the path rate of change in y versus x at time t?"
Or you could eliminate t using one of the equations and sub. it in and you'd have dy/dx in terms of x and y only...you can do anything! Okay, not *anything*.
ok, how about the trajectory of the particle at time 0 and time 1.2? i dont think i even heard about what that means
The trajectory is the path a moving object follows through space as a function of time. You have your displacement vector in terms of t, namely, \[(x(t),y(t))\]All you'd do is sub. in the times to get a displacement vector for the object at that time.
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.
Oh - so that's why you needed dy/dx in terms of t...remember what dy/dx means geometrically? It's the gradient of the function. dy/dx is telling you where the particle is headed.
\[\frac{dy}{dx}(t)=\tan \theta\]
gradient?
I'm assuming from what you've said you're throwing something in this question. Draw a right-angled triangle where the hypotenuse is sloping up. Let theta be the angle...wait, I'll draw something.
Look at this...
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so basically its the tennisball project in two directions. up and out. i threw the ball, from 1.2192m and it took 1.2s to land 6.096m away. i got all the equations above then it asked for the formula for speed of the particle at any time. and right after that said "step 6: find dy/dx."
tan (theta) = (dy/dt)/(dx/dt) = dy/dx
that makes sense... what do it do with that?
Well, from that setup (picture), dy/dt is always vertical and dx/dt is always horizontal. Their sizes will change depending on time, and therefore, so will dy/dx. All that picture is doing is to show the relationship between the angle with the horizontal and the velocity vector. You're looking for the angle at a certain time. You have dy/dx as a function of t, and that picture shows you how dy/dx relates to that angle (i.e. tan(theta) = dy/dx). So, find dy/dx at your times and take the inverse tangent to find the angles. Check to see if the answers make sense.
that makes complete sense, lemmy try that
and so just making sure...: dy/dx (t) gives me both dy vertical height and dx horizontal distance as a function of time?
at the same time?
Nearly. What it's doing is giving you the *rate of change* of y with respect to x (as a function of time).
So you're getting the rate of change in y wrt x given the time...you don't have to know anything about x (or y) explicitly (like you do when you're asked more elementary questions like, y=x^2 -> dy/dx = 2x. To fing dy/dx here, we need to know x).
right, being the derivative, but yeah it does both in relevance to time... jeez im clearly too tired if im being this stupid
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!).
nah man im definitely being stupid right now... idk where you are but here on the east coast its 4am and i got school first thing tomorrow
alright so to get the greatest speed ill use the dy/dx formula?
83 degrees...might be (180-83) degrees if you think about the fact the coordinate system and how the ball lands. I'll draw something.
Well, to get the greatest speed, I would use s(t) and take the derivative wrt to t to find the time at which the speed was greatest. Then you can use that time to find position.
yeah it would be about -263 or +97 degrees technically but same thing in this perspective... and i can understand why dy/dx is a speed formula but whats the whole root of dy/dt sqrd + dx/dt sqrd) ?
what is wrt?
wrt = with respect to
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whats the difference between using s(t) and dy/dx, would they get me the same speed?... would dy/dx get me "speed" ?
The whole 'root' thing is because of the following. The velocity vector is\[v=(x'(t),y'(t))\]and since the speed is the magnitude of the velocity vector, you have\[v^2=v.v=(x'(t),y'(t)).(x'(t),y'(t))=x'(t)^2+y'(t)^2\]
But this is the square of speed, so we have to take the square root of both sides.
\[v=\sqrt{x'(t)^2+y'(t)^2}\]
Re. using s(t) and dy/dx...speed is the time rate of change, which is what s(t) is. dy/dx is *not* a time rate of change, it's a displacement rate of change; it measures the change in the y-coordinate as a function of the change in x-coordinate. It's different. It gives you information about the geometry of what's going on - as in the case where you have to find angles for where things fall.
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...
I would just take the derivative of x(t) and y(t) and sub. the times in to find numerical values, square each, add, take square root of sum (if your aim is computation only).
what would be another aim? than computation?
Well, if you're asked to find the actual formula...if you're not asked to, and only asked to find values (and you're in an exam), it would take up unnecessary time.
\[y(t)=-4.9t^2+4.864t+1.2192 \rightarrow y(0)=1.2192\]\[x(t)=5.08t \rightarrow x(0)=0\]So,\[s(0)=\sqrt{1.2192^2+0^2}=1.2192\]
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?
I'm trained in pure maths and physics. I'm a UK citizen, worked with people in physics from Oxford, Cambridge, Yale.
wait a sec... it asked for the speed as the ball approached the max height.. if the max height is when y' is 0 then ill end up with dy/dt is 0 and dx/dt is 5.08?
that is freaking awesome.
:)
that makes me wonder, how many teens have you worked with that your first answer is "torture" to math :P
and i need help with this one: on what intervals is the speed increasing and decreasing? (i dont want to use an analysis -.- )
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
that makes no sense, my calculations say that it had a speed of approx, 3 when i let go... 8 when it hit the ground, and 5.08 at the max height
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?
Well, you could find the maximum height by understanding that the vertical component of velocity must be zero there...so \[\frac{dy}{dt}=0\]should give you the time, which you can use to find the height, y. I know you said no calculus, but I figured you already had y'(t).
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
i mean, i intend to get a Ph.D in both mathematics and probably EE... ill be doing this till the day i die
XD
good
( when you can answer) why did you get into math?
School...I accelerated my studies. After school, I had offers for medicine and law, but stuck to mathematics and physics...to my father's grief. I've worked in and biophysics with a chemist from Ox. and nuclear physicist (turned biophysicist) from Yale. Now, I'm taking a break.
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
whether baloon is launched on a windless day, goes straight up... rises at a rate of 1000ft/minutes, observer stands 10ft away, his eyes are 6 ft off the ground at what rate is the angle of elevation changing when the baloon is 10000 ft high
i can do related rates, i just cant figure out how to set this one up ...
well, you won't have, by definition, an angle of elevation until the balloon hits 6 ft. So I would consider finding the vertical velocity and other 'initial' conditions at the y=6ft mark and work from there...
...whaa??
one sec.
no i get what youre saying... nvrmnd that "whaa?" but we are taking the calculations mostly at 10000ft where that 6ft will have been very much covered
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.
actually, given that it is 6 o clock now.. its probably best if you dont... i need to get my 15 minutes of sleep XDDD
oh, i'll continue and you can go...
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ill prolly be back tonight, i always liked helping people with their math
iight im out- thank you for everything btw
no probs.
Had a break...so, you need to find \[\frac{d \theta }{dt}\] as some function of height. From the pic. I sent, you should see that\[\tan \theta = \frac{h}{10} \rightarrow \frac{d}{dt}\tan \theta = \frac{d}{dt}\frac{h}{10} \rightarrow \sec^2 \theta \frac{d \theta}{dt}=\frac{1}{10}\frac{dh}{dt}\]That is,\[\frac{d \theta }{dt}=\frac{dh/dt}{10\sec^2 \theta}=\frac{dh/dt}{10(1+\tan^2 \theta)}=\frac{dh/dt}{10(1+h^2/100)}\]\[\frac{d \theta}{dt}=\frac{10(dh/dt)}{100+ h^2}\]You're given\[\frac{dh}{dt}=1000ft/\min\]and your height at this point is (10,000 - 6) ft since our calculations for angle of elevation begin at 6ft. Then\[\frac{d \theta }{dt }|_{h=9994}=\frac{10 \times 1000}{100+(10000-6)^2}\approx 1 \times 10^{-4} radians/\min\]
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]

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