Question

curvilinear motion...

Answer 1

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?

Answer 2

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.

Answer 3

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?

Answer 4

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?

Answer 5

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

Answer 6

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?

Answer 7

Do what I wrote for the speed, and do what I wrote for dy/dx...you should get two different answers. Does that help?

Answer 8

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

Answer 9

im sorry but im trying to understand the formulas

Answer 10

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.

Answer 11

so finding dy/dx pretty much has nothing to do with the speed function and my teacher just put it in there randomly...?

Answer 12

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.

Answer 13

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.

Answer 14

which one excludes time?

Answer 15

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

Answer 16

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.

Answer 17

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?"

Answer 18

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*.

Answer 19

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

Answer 20

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.

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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.

Answer 25

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

Answer 26

gradient?

Answer 27

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.

Answer 28

Look at this...

Answer 29

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."

Answer 30

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

Answer 31

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

Answer 32

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.

Answer 33

that makes complete sense, lemmy try that

Answer 34

and so just making sure...: dy/dx (t) gives me both dy vertical height and dx horizontal distance as a function of time?

Answer 35

at the same time?

Answer 36

Nearly. What it's doing is giving you the *rate of change* of y with respect to x (as a function of time).

Answer 37

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).

Answer 38

right, being the derivative, but yeah it does both in relevance to time... jeez im clearly too tired if im being this stupid

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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.

Answer 44

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.

Answer 45

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) ?

Answer 46

what is wrt?

Answer 47

wrt = with respect to

Answer 48

whats the difference between using s(t) and dy/dx, would they get me the same speed?... would dy/dx get me "speed" ?

Answer 49

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\[v^2=v.v=(x'(t),y'(t)).(x'(t),y'(t))=x'(t)^2+y'(t)^2\]

Answer 50

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

Answer 51

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

Answer 52

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.

Answer 53

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

Answer 54

not gone to bed

Answer 55

Have you got a test or something?

Answer 56

technically i dont actually take calculus

Answer 57

Is this just for physics or something?

Answer 58

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

Answer 59

So what's this for?

Answer 60

Torture?

Answer 61

torture to who?

Answer 62

You.

Answer 63

I wasn't being rude :)

Answer 64

haha no, this is my fun

Answer 65

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

Answer 66

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

Answer 67

If you want the speed, no.

Answer 68

first thing i want is speed at time 0

Answer 69

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

Answer 70

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).

Answer 71

what would be another aim? than computation?

Answer 72

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.

Answer 73

\[y(t)=-4.9t^2+4.864t+1.2192 \rightarrow y(0)=1.2192\]\[x(t)=5.08t \rightarrow x(0)=0\]So,\[s(0)=\sqrt{1.2192^2+0^2}=1.2192\]

Answer 74

dy/dt or y(t) ??

Answer 75

oh

Answer 76

yeah - I'm tired too

Answer 77

lol you on the eat cost too?

Answer 78

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

Answer 79

are you a physicist or something or just enjoy math?

Answer 80

I'm trained in pure maths and physics. I'm a UK citizen, worked with people in physics from Oxford, Cambridge, Yale.

Answer 81

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?

Answer 82

that is freaking awesome.

Answer 83

:)

Answer 84

that makes me wonder, how many teens have you worked with that your first answer is "torture" to math :P

Answer 85

and i need help with this one: on what intervals is the speed increasing and decreasing? (i dont want to use an analysis -.- )

Answer 86

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

Answer 87

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

Answer 88

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

Answer 89

i dont want to use any derivative analysiss

Answer 90

oh

Answer 91

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

Answer 92

One sec...

Answer 93

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

Answer 94

i got 7 this time :DDD

Answer 95

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

Answer 96

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).

Answer 97

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