anonymous
  • anonymous
curvilinear motion...
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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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anonymous
  • anonymous
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?
anonymous
  • anonymous
since you're looking for the magnitude of velocity at that point.
anonymous
  • anonymous
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?
anonymous
  • anonymous
Do what I wrote for the speed, and do what I wrote for dy/dx...you should get two different answers. Does that help?
anonymous
  • anonymous
what does that s(t) formula tell me in terms of the problem and what does the dy/dx tell me?
anonymous
  • anonymous
im sorry but im trying to understand the formulas
anonymous
  • anonymous
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.
anonymous
  • anonymous
so finding dy/dx pretty much has nothing to do with the speed function and my teacher just put it in there randomly...?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
which one excludes time?
anonymous
  • anonymous
ah, but if i divide dy/dt by dx/dt i still have dy/dx in terms of t
anonymous
  • anonymous
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.
anonymous
  • anonymous
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?"
anonymous
  • anonymous
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*.
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
The question is just asking, "Where is the object headed at time 0 and at time 1.2s?"
anonymous
  • anonymous
its asking the angle that it left my hand (time 0) and the angle that it hit the ground (time 1.2)
anonymous
  • anonymous
Actually, I should have said, "Where is the object?", not 'headed', since that would be velocity.
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[\frac{dy}{dx}(t)=\tan \theta\]
anonymous
  • anonymous
gradient?
anonymous
  • anonymous
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.
anonymous
  • anonymous
Look at this...
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anonymous
  • anonymous
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."
anonymous
  • anonymous
tan (theta) = (dy/dt)/(dx/dt) = dy/dx
anonymous
  • anonymous
that makes sense... what do it do with that?
anonymous
  • anonymous
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.
anonymous
  • anonymous
that makes complete sense, lemmy try that
anonymous
  • anonymous
and so just making sure...: dy/dx (t) gives me both dy vertical height and dx horizontal distance as a function of time?
anonymous
  • anonymous
at the same time?
anonymous
  • anonymous
Nearly. What it's doing is giving you the *rate of change* of y with respect to x (as a function of time).
anonymous
  • anonymous
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).
anonymous
  • anonymous
right, being the derivative, but yeah it does both in relevance to time... jeez im clearly too tired if im being this stupid
anonymous
  • anonymous
so i got that it left at approx 137 degrees and landed at 83 degrees which looks right
anonymous
  • anonymous
yes, you're right (not about being stupid!).
anonymous
  • anonymous
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
anonymous
  • anonymous
alright so to get the greatest speed ill use the dy/dx formula?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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) ?
anonymous
  • anonymous
what is wrt?
anonymous
  • anonymous
wrt = with respect to
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anonymous
  • anonymous
whats the difference between using s(t) and dy/dx, would they get me the same speed?... would dy/dx get me "speed" ?
anonymous
  • anonymous
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\[v^2=v.v=(x'(t),y'(t)).(x'(t),y'(t))=x'(t)^2+y'(t)^2\]
anonymous
  • anonymous
But this is the square of speed, so we have to take the square root of both sides.
anonymous
  • anonymous
\[v=\sqrt{x'(t)^2+y'(t)^2}\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
4am...are you getting up to study, or have you not gone to bed?
anonymous
  • anonymous
not gone to bed
anonymous
  • anonymous
Have you got a test or something?
anonymous
  • anonymous
technically i dont actually take calculus
anonymous
  • anonymous
Is this just for physics or something?
anonymous
  • anonymous
cant take physics either, yet... im a freshman in hs
anonymous
  • anonymous
So what's this for?
anonymous
  • anonymous
Torture?
anonymous
  • anonymous
torture to who?
anonymous
  • anonymous
You.
anonymous
  • anonymous
I wasn't being rude :)
anonymous
  • anonymous
haha no, this is my fun
anonymous
  • anonymous
i got half an hour and about 7 speed questions that i know are wrong now
anonymous
  • anonymous
so i really have no option but taking the root of that big equation? :(
anonymous
  • anonymous
If you want the speed, no.
anonymous
  • anonymous
first thing i want is speed at time 0
anonymous
  • anonymous
dx/dt will be 0 and dy/dt will be 4.864... so i sqare that then root it? which is still 4.864...
anonymous
  • anonymous
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).
anonymous
  • anonymous
what would be another aim? than computation?
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[y(t)=-4.9t^2+4.864t+1.2192 \rightarrow y(0)=1.2192\]\[x(t)=5.08t \rightarrow x(0)=0\]So,\[s(0)=\sqrt{1.2192^2+0^2}=1.2192\]
anonymous
  • anonymous
dy/dt or y(t) ??
anonymous
  • anonymous
oh
anonymous
  • anonymous
yeah - I'm tired too
anonymous
  • anonymous
lol you on the eat cost too?
anonymous
  • anonymous
No, other side of the world. Sydney. Off sick.
anonymous
  • anonymous
are you a physicist or something or just enjoy math?
anonymous
  • anonymous
I'm trained in pure maths and physics. I'm a UK citizen, worked with people in physics from Oxford, Cambridge, Yale.
anonymous
  • anonymous
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?
anonymous
  • anonymous
that is freaking awesome.
anonymous
  • anonymous
:)
anonymous
  • anonymous
that makes me wonder, how many teens have you worked with that your first answer is "torture" to math :P
anonymous
  • anonymous
and i need help with this one: on what intervals is the speed increasing and decreasing? (i dont want to use an analysis -.- )
anonymous
  • anonymous
lol, I've worked with people of all ages and abilities. All I care about is the attitude.
anonymous
  • anonymous
You don't want to use an analysis? What do you mean?
anonymous
  • anonymous
logic says that its decreasing till the max height and increasing therein after
anonymous
  • anonymous
i dont want to use any derivative analysiss
anonymous
  • anonymous
oh
anonymous
  • anonymous
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
anonymous
  • anonymous
One sec...
anonymous
  • anonymous
oh i see my mistake... omg this is pathetically horrible ><
anonymous
  • anonymous
i got 7 this time :DDD
anonymous
  • anonymous
the speed should be fastest when it hits the ground, right?
anonymous
  • anonymous
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).
anonymous
  • anonymous
Sorry, I think I've answered a question you didn't ask.
anonymous
  • anonymous
Yes, the speed should be fastest when it hits the ground. You get that from energy considerations.
anonymous
  • anonymous
yeah i already got everything else
anonymous
  • anonymous
niceee finally i can glue this crap together lol
anonymous
  • anonymous
Good! I can't believe you're up at ~4am and doing this...lol.
anonymous
  • anonymous
Do you want to study math/science after school?
anonymous
  • anonymous
?
anonymous
  • anonymous
like after i graduate, keep studying?
anonymous
  • anonymous
Yeah
anonymous
  • anonymous
i mean, i intend to get a Ph.D in both mathematics and probably EE... ill be doing this till the day i die
anonymous
  • anonymous
XD
anonymous
  • anonymous
good
anonymous
  • anonymous
( when you can answer) why did you get into math?
anonymous
  • anonymous
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.
anonymous
  • anonymous
a break?
anonymous
  • anonymous
Sabbatical.
anonymous
  • anonymous
?
anonymous
  • anonymous
Just a formal break from university stuff.
anonymous
  • anonymous
are you a professor or a researcher?
anonymous
  • anonymous
I was a researcher. I'm just teaching for a while without the research.
anonymous
  • anonymous
i can tell that you come on here often haha
anonymous
  • anonymous
Lol, I know...it's a nice distraction from other things I should be doing.
anonymous
  • anonymous
i could probably use a little more help on a related rates problem DX
anonymous
  • anonymous
i'll see what i can do
anonymous
  • anonymous
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
anonymous
  • anonymous
i can do related rates, i just cant figure out how to set this one up ...
anonymous
  • anonymous
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...
anonymous
  • anonymous
...whaa??
anonymous
  • anonymous
one sec.
anonymous
  • anonymous
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
anonymous
  • anonymous
and yeah i tried that, but i still cant set up the equation right
anonymous
  • anonymous
with the +6 or -6 or even without it, its the trig that i cant get down ><
anonymous
  • anonymous
Okay, I'm going to draw something.
anonymous
  • anonymous
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
anonymous
  • anonymous
oh, i'll continue and you can go...
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anonymous
  • anonymous
ill prolly be back tonight, i always liked helping people with their math
anonymous
  • anonymous
iight im out- thank you for everything btw
anonymous
  • anonymous
no probs.
anonymous
  • anonymous
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\[\frac{d \theta }{dt }|_{h=9994}=\frac{10 \times 1000}{100+(10000-6)^2}\approx 1 \times 10^{-4} radians/\min\]
anonymous
  • anonymous
From your question, I assume any acceleration on the balloon in the vertical is ignored.
anonymous
  • anonymous
thats precisely what i got! and yes, there is no acceleration in this case
anonymous
  • anonymous
[thumbs up]

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