Consider the curve defied by the equation y+cosy = x +1 for 0< or to equal y < or to equal 2pi

- anonymous

Consider the curve defied by the equation y+cosy = x +1 for 0< or to equal y < or to equal 2pi

- schrodinger

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- anonymous

@Goten77

- Goten77

hmm

- anonymous

Okay, i considered it, now what?

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## More answers

- anonymous

a.) Find dy/dx in terms of y.
b.) Write an equation for each vertical tangent to the curve
c.) Find d^2y/dx^2 in terms of y.

- anonymous

wouldn't be like this? |dw:1358131334593:dw|

- anonymous

sorry about my bad handwriting my hands are cold :P

- Goten77

no it wont be like that cause you with with respect to Y

- anonymous

ok so part a.) is just 1-siny=x'

- dumbcow

they are looking for dy/dx or y'
derivative wrt x BUT in terms of y

- Goten77

oh derp.... i did it wrong

- anonymous

lol

- anonymous

so was my technique correct? :D

- Goten77

|dw:1358139066897:dw|

- Goten77

k so y did it not show both images.... the 1st part wwas expanded 2nd part is simplified

- anonymous

wait how did u get that?

- Goten77

|dw:1358139147928:dw|

- Goten77

this program some times deletes my images or it doesnt appear for whatever reason

- anonymous

o.o

- Goten77

|dw:1358139329704:dw|

- Goten77

get it now?

- anonymous

how does x cancel from the other side?

- Goten77

|dw:1358139425312:dw|

- Goten77

when/where did i cancel a x?

- anonymous

from the right original side of equation (x+1)

- dumbcow

btw here is graph of curve
http://www.wolframalpha.com/input/?i=y+%2B+cos%28y%29+%3D+x%2B1+%2C+from+x%3D0+to+2pi%2C+from+y%3D0+to+2pi

- anonymous

oh ok i think i see why u canceled the variables on the right side of the equation :)

- anonymous

so 1/(1-siny) is part a.)?

- Goten77

the x didnt cancel
the derivative of x+1 = just 1 since its wrt x

- Goten77

and yes thats your part a

- anonymous

ok how do i do part b.)

- Goten77

hmm

- Goten77

so vertical asymptotes is what makes the bottom of a fraction = 0

- Goten77

|dw:1358140100389:dw|

- Goten77

but we need the vertical asymptote so we needa find the x value so we put the equation in the form of x = #

- Goten77

so now we use the original equation and use that y value to solve for the x

- anonymous

how did u get 1=siny

- Goten77

|dw:1358140410824:dw|

- Goten77

so the vertical graph is
x= (pi - 2)/ 2

- Goten77

i got 1=siny because it was 1st
1-siny= 0
i just moved it over

- Goten77

|dw:1358140649853:dw| sorry im bad at spelling

- Goten77

ok so part c is finding the 2nd derivative.... this will be fun

- Goten77

|dw:1358140714426:dw|

- anonymous

so part b.) is (pi -2)/2

- Goten77

|dw:1358140838951:dw|

- Goten77

|dw:1358140982676:dw|

- Goten77

this is basically the quotient rule + implicint differentation + production rule + chain rule THEN use that negative .... you dont have to simplify unless ur told to.. which i dont think u can

- anonymous

so cosy/(1-siny)^2 is the answer? :)

- Goten77

yee

- anonymous

k, ty ^^

- anonymous

can you help me with 1 more pls? :)

- Goten77

sure

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