Question

parametric equations

Answer 1

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

Answer 2

Uh, which one? Lol.

Answer 3

last 3, i know rest of the stuff

Answer 4

you have to make the effort. I did the rest. All you have to do is combine the steps.

Answer 5

http://openstudy.com/users/best_mathematician#/updates/517d6401e4b0be6b54ab3130

Answer 6

ok i will try...what's f ' and h '

Answer 7

for arc length, am i suppose to integrate like this \[ds = \sqrt{1+(\sqrt{2g(1-y)})^2} dt\]

Answer 8

for g? what is cartesian and parametric equation

Answer 9

i mean i know for parametric equation, \[x^2 + y^2 = r^2..\] r is radius here...so should i insert y=1 and x=\[\frac{ \pi }{ 2 }\] and solve it

Answer 10

yes.
g = acceleration due to gravity.

Answer 11

we are still in cartesian co-ordinates

Answer 12

i know parametric equation but what is cartesian equation

Answer 13

Cartesian system is also called as rectangular co-ordinate system

Answer 14

oh so is it simple equations like y = mx + b

Answer 15

yes

Answer 16

my arc length step is correct right?

Answer 17

for part "f" of the questiom,
they asked for "differential element" not complete length

Answer 18

and then integrate with "u"

Answer 19

i m confused i know replace 9.8 for g, then what am i suppose to do

Answer 20

what does question "f" say

Answer 21

Express ds......

Answer 22

so integrate \[\int\limits_{}^{} ds = \int\limits_{}^{} \sqrt{1+\sqrt{2*9.8(1-y)^2}}\]
like this

Answer 23

the question asks for ds interms of "f" and "h"

Answer 24

oh ya what's f and h

Answer 25

then you have to read the question

Answer 26

x and y?

Answer 27

yes. then read our solution from last time

Answer 28

omg this is so confusing ... how to replce it with f and h

Answer 29

what was the "ds" from previous steps?

Answer 30

the sqrt of 1+2g(1-y)^2

Answer 31

times dt

Answer 32

oh ya sry abt tht

Answer 33

\[
\text{we had:}\\
{ds\over st}=\sqrt{1+2g(1-y)}\\
\frac{ds}{\sqrt{1+2g(1-y)}}=dt\\
{\rm but}\quad y=h(u)\quad ds=\sqrt{f'^2+y'^2}du
\]plug in

Answer 34

what is st?

Answer 35

where do you see "st"?

Answer 36

oh, thats a typo. it should be "dt"

Answer 37

so is it y = \[\sqrt{(1+ 2g(1-y)^2)} du\]

Answer 38

ugh nm its not true

Answer 39

no we are concerned with "ds" and "dt"we have to change all "ds" terms to "du"
all x,y to their parametric form

Answer 40

would you please do it...

Answer 41

wait i think i got it

Answer 42

the main relation is \[{ds\over\sqrt{1+2g(1-y)}}=dt\]
in this, plug in the relation for "ds" and "y"

Answer 43

lemme try first

Answer 44

\[ds = \sqrt{1^2+19.6(1-y)^2} du\]

Answer 45

its parametric right

Answer 46

For Solving this Question, I would like to ask you that, do you have the detailed conception of : Paramatric Equations

Answer 47

@goformit100 no i don't. I am studying on my own. That's the reason I am not understanding this concept