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Okay, i considered it, now what?
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.
wouldn't be like this? |dw:1358131334593:dw|
sorry about my bad handwriting my hands are cold :P
no it wont be like that cause you with with respect to Y
ok so part a.) is just 1-siny=x'
they are looking for dy/dx or y' derivative wrt x BUT in terms of y
oh derp.... i did it wrong
so was my technique correct? :D
k so y did it not show both images.... the 1st part wwas expanded 2nd part is simplified
wait how did u get that?
this program some times deletes my images or it doesnt appear for whatever reason
get it now?
how does x cancel from the other side?
when/where did i cancel a x?
from the right original side of equation (x+1)
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
oh ok i think i see why u canceled the variables on the right side of the equation :)
so 1/(1-siny) is part a.)?
the x didnt cancel the derivative of x+1 = just 1 since its wrt x
and yes thats your part a
ok how do i do part b.)
so vertical asymptotes is what makes the bottom of a fraction = 0
but we need the vertical asymptote so we needa find the x value so we put the equation in the form of x = #
so now we use the original equation and use that y value to solve for the x
how did u get 1=siny
so the vertical graph is x= (pi - 2)/ 2
i got 1=siny because it was 1st 1-siny= 0 i just moved it over
|dw:1358140649853:dw| sorry im bad at spelling
ok so part c is finding the 2nd derivative.... this will be fun
so part b.) is (pi -2)/2
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
so cosy/(1-siny)^2 is the answer? :)
k, ty ^^
can you help me with 1 more pls? :)