Question

parametric equations...please help me with last 3...i am totally confused...explain them

Answer 1

http://assets.openstudy.com/updates/attachments/517d6401e4b0be6b54ab3130-best_mathematician-1367172224273-1.png

Answer 2

arclength is just\[\int ds=\int_{a}^{b}\sqrt{(x')^2+(y')^2}~dt\]

Answer 3

i know tht but how to place the functions

Answer 4

v in terms of s? i assume that they are refering to s(t) as a position equation maybe? then ds = v(t)

Answer 5

replace v by ds/dt

\[\frac12m(\frac{ds}{dt})^2+mgy=k\]

Answer 6

solve for dt

Answer 7

\[dt = \frac{ ds }{ \sqrt{1+2g(1-y)} }\]

Answer 8

k = mg
\[\frac12m(\frac{ds}{dt})^2+mgy=mg\]
\[\frac12(\frac{ds}{dt})^2+gy=g\]
\[(\frac{ds}{dt})^2=2(g-gy)\]
\[\frac{ds}{\sqrt{2g(1-y)}}=dt\]

Answer 9

now how to express the arc length

Answer 10

it says to express it in terms of f' and g' wrt u

\[\int_{}^{}\sqrt{(\frac{df}{du}\frac{du}{dt})^2+(\frac{dg}{du}\frac{du}{dt})^2}~dt\]
\[\int_{}^{}\sqrt{(\frac{du}{dt})^2\left((\frac{df}{du})^2+(\frac{dg}{du})^2\right)}~dt\]
\[\int_{}^{}\sqrt{(\frac{df}{du})^2+(\frac{dg}{du})^2}~\frac{du}{dt}dt\]
\[\int_{0}^{\pi}\sqrt{(f')^2+(g')^2}~du\]

Answer 11

pfft, f and h, but samesame

Answer 12

the nest thing is a straight line equation between 2 points .....

Answer 13

how did u do that lol

Answer 14

*next

Answer 15

so that's the answer for f)

Answer 16

thats what i get for "f" yes

Answer 17

but i used a g instead of an h :/

Answer 18

g stands for gravity so should i just replace g by h?

Answer 19

since y = h(u), yes .... i simply misread, mistyped, it is all

Answer 20

haha thanks...how to do g?

Answer 21

if i asked you to find the equation of a line between say: (0,1) and (5,0) what would you do?

Answer 22

y-y0 = m(x-x0)

Answer 23

yep, thats all g is asking for ... and also to put it into parametric form

Answer 24

so y=-0.2x+1 ????

Answer 25

first determine the cartesian, xy plane, y=mx+b format

Answer 26

good, and with that we can see that: x=x
y=mx+b

Answer 27

just use the points they gave you instead

Answer 28

\[y = -\frac{ 2 }{ \pi }x+1\]

Answer 29

what next ?

Answer 30

thats the cartesian form

the parametric form is: x = x
y = -2/pi x + 1

but replace x on the right by t

Answer 31

ok lemme try that

Answer 32

oh just replace x by t???

Answer 33

yeah

Answer 34

y = -2/pi t + 1???

Answer 35

x = t

y = -2t/pi + 1

this gives us parametric equations in terms of "t", but i have to be leery as to if this is "correct" to do or not

Answer 36

i thought parametric equations is something like x^2 + y^2 = r^2 where r is radius...and this is cartesian equation

Answer 37

nah, thats polar

Answer 38

parametric simply define x and y seperately

Answer 39

oh then what is cartesian

Answer 40

y = f(x) is cartesian ...

parametric define x and y seperately by parameters

x = f(t)
y = g(t)

Answer 41

ohk...so cartesian would be f(t) = -2t/pi + 1

Answer 42

we can define a vector setup in terms of:\[\vec r(t)=(x(t),y(t))\]

Answer 43

no, cartesian is the usual xy plane .. y = f(x) = mx+b type stuff, defining y in terms of x

Answer 44

f(x) = -(2/pi)x + 1 ??

Answer 45

yes

Answer 46

and for time just replace y by zero and solve for t...right?

Answer 47

at t=0, x=0
at t=0, y=1

at t=?, x=pi/2
at t=?, y=0

Answer 48

im just not sure if its a good or proper idea to equate x=t .... the speed determines the time it takes to travel the slope, not the shape of the slope

Answer 49

x=t so at x = pi/2, t=pi/2
at y=0, t= pi/2 again

Answer 50

it says at the start that x and y are functions of u

so lets use u instead of t

Answer 51

oh man where does this u come from

Answer 52

3rd sentence

Answer 53

when u=0, x=0
when u=pi, x=pi/2

let x = u/2

Answer 54

\[x = x\\
y = -\frac{2}{\pi}x+1
\]
\[x = f(u) = \frac u2\\
y = h(u)=-\frac{2}{\pi}\frac u2+1=-\frac u{\pi}+1
\]

Answer 55

\[f' = \frac12\\h' = -\frac{1}{\pi}\]
\[\int_{u=0}^{u=pi}~\sqrt{(f')^2+(h')^2}du\]
\[\int_{u=0}^{u=pi}~\sqrt{\frac14+\frac{1}{\pi^2}}~du\]

Answer 56

1.86

Answer 57

\[\int_{u=0}^{u=pi}~\frac{\sqrt{\pi^2+4}}{2\pi}~du\]

Answer 58

1.86 yes, but not sure how that would relate to time

Answer 59

18.38

Answer 60

wait y do we got two different equations

Answer 61

we dont ... one is a simplification is all

Answer 62

i got different answer for second one

Answer 63

sqrt(pi^2+4)/2

Answer 64

5.85

Answer 65

wait without integral it is 1.86

Answer 66

http://www.wolframalpha.com/input/?i=sqrt%28pi%5E2%2B4%29%2F2

Answer 67

ya i integrate it...sry abt that

Answer 68

lemme check our answers with @electrokid

Answer 69

how does this relate u to t tho?

Answer 70

im thinking that line "e" is important

Answer 71

@electrokid already helped me solve that...we went upto there then i got confused and couldnt solve anything...thx for ur help...i just need him to verify answers

Answer 72

k