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anonymous
 5 years ago
curvilinear motion...
anonymous
 5 years ago
curvilinear motion...

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0parametric 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
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0that 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.0since you're looking for the magnitude of velocity at that point.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nope 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
 5 years ago
Best ResponseYou've already chosen the best response.0Do what I wrote for the speed, and do what I wrote for dy/dx...you should get two different answers. Does that help?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what does that s(t) formula tell me in terms of the problem and what does the dy/dx tell me?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im sorry but im trying to understand the formulas

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0s(t) is the speed  the magnitude of the velocity. dy/dx just gives you the rate of change of the y coordinate as x varies  so it's just looking at how vertical displacement varies with horizontal...nothing special.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so finding dy/dx pretty much has nothing to do with the speed function and my teacher just put it in there randomly...?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sometimes 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
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0which one excludes time?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah, but if i divide dy/dt by dx/dt i still have dy/dx in terms of t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Or 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok, 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
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0The question is just asking, "Where is the object headed at time 0 and at time 1.2s?"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its asking the angle that it left my hand (time 0) and the angle that it hit the ground (time 1.2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, I should have said, "Where is the object?", not 'headed', since that would be velocity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh  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
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dy}{dx}(t)=\tan \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm assuming from what you've said you're throwing something in this question. Draw a rightangled triangle where the hypotenuse is sloping up. Let theta be the angle...wait, I'll draw something.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0tan (theta) = (dy/dt)/(dx/dt) = dy/dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that makes sense... what do it do with that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0that makes complete sense, lemmy try that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and so just making sure...: dy/dx (t) gives me both dy vertical height and dx horizontal distance as a function of time?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Nearly. What it's doing is giving you the *rate of change* of y with respect to x (as a function of time).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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
 5 years ago
Best ResponseYou've already chosen the best response.0right, being the derivative, but yeah it does both in relevance to time... jeez im clearly too tired if im being this stupid

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i got that it left at approx 137 degrees and landed at 83 degrees which looks right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, you're right (not about being stupid!).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nah 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
 5 years ago
Best ResponseYou've already chosen the best response.0alright so to get the greatest speed ill use the dy/dx formula?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.083 degrees...might be (18083) degrees if you think about the fact the coordinate system and how the ball lands. I'll draw something.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0yeah 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
 5 years ago
Best ResponseYou've already chosen the best response.0wrt = with respect to

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whats the difference between using s(t) and dy/dx, would they get me the same speed?... would dy/dx get me "speed" ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0But this is the square of speed, so we have to take the square root of both sides.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[v=\sqrt{x'(t)^2+y'(t)^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Re. 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 ycoordinate as a function of the change in xcoordinate. 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
 5 years ago
Best ResponseYou've already chosen the best response.04am...are you getting up to study, or have you not gone to bed?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have you got a test or something?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0technically i dont actually take calculus

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this just for physics or something?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cant take physics either, yet... im a freshman in hs

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I wasn't being rude :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha no, this is my fun

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got half an hour and about 7 speed questions that i know are wrong now

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i really have no option but taking the root of that big equation? :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you want the speed, no.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first thing i want is speed at time 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dx/dt will be 0 and dy/dt will be 4.864... so i sqare that then root it? which is still 4.864...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0what would be another aim? than computation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[y(t)=4.9t^2+4.864t+1.2192 \rightarrow y(0)=1.2192\]\[x(t)=5.08t \rightarrow x(0)=0\]So,\[s(0)=\sqrt{1.2192^2+0^2}=1.2192\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol you on the eat cost too?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, other side of the world. Sydney. Off sick.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you a physicist or something or just enjoy math?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm trained in pure maths and physics. I'm a UK citizen, worked with people in physics from Oxford, Cambridge, Yale.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait 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
 5 years ago
Best ResponseYou've already chosen the best response.0that is freaking awesome.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that makes me wonder, how many teens have you worked with that your first answer is "torture" to math :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and i need help with this one: on what intervals is the speed increasing and decreasing? (i dont want to use an analysis . )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, I've worked with people of all ages and abilities. All I care about is the attitude.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You don't want to use an analysis? What do you mean?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0logic says that its decreasing till the max height and increasing therein after

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont want to use any derivative analysiss

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that 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
 5 years ago
Best ResponseYou've already chosen the best response.0oh i see my mistake... omg this is pathetically horrible ><

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got 7 this time :DDD

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the speed should be fastest when it hits the ground, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, I think I've answered a question you didn't ask.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, the speed should be fastest when it hits the ground. You get that from energy considerations.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah i already got everything else

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0niceee finally i can glue this crap together lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good! I can't believe you're up at ~4am and doing this...lol.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you want to study math/science after school?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like after i graduate, keep studying?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i mean, i intend to get a Ph.D in both mathematics and probably EE... ill be doing this till the day i die

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0( when you can answer) why did you get into math?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0School...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
 5 years ago
Best ResponseYou've already chosen the best response.0Just a formal break from university stuff.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you a professor or a researcher?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I was a researcher. I'm just teaching for a while without the research.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i can tell that you come on here often haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lol, I know...it's a nice distraction from other things I should be doing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i could probably use a little more help on a related rates problem DX

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'll see what i can do

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whether 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
 5 years ago
Best ResponseYou've already chosen the best response.0i can do related rates, i just cant figure out how to set this one up ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, 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
 5 years ago
Best ResponseYou've already chosen the best response.0no 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
 5 years ago
Best ResponseYou've already chosen the best response.0and yeah i tried that, but i still cant set up the equation right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0with the +6 or 6 or even without it, its the trig that i cant get down ><

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, I'm going to draw something.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually, 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
 5 years ago
Best ResponseYou've already chosen the best response.0oh, i'll continue and you can go...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ill prolly be back tonight, i always liked helping people with their math

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0iight im out thank you for everything btw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Had 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\[\frac{d \theta }{dt }_{h=9994}=\frac{10 \times 1000}{100+(100006)^2}\approx 1 \times 10^{4} radians/\min\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From your question, I assume any acceleration on the balloon in the vertical is ignored.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats precisely what i got! and yes, there is no acceleration in this case
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