Solving Trig Substitutions: Getting from 1st Line to 2nd

[KoGs]

Yeah I know this is pretty basic, but I just can't think of it right now.

The slope of the wire is tan(theta) = y' = dy/dx
which yields sin(theta) = y' / sqrt (1 + y'^2)

How do you get from the first line to the second one?



this is what I I have done:
x^2 + y^2 = 1
cos^2(theta) + sin^2(theta) = 1
so cos^2(theta) = 1 - sin^2(theta)
so cos(theta) = sqrt (1 - y^2).

And
tan(theta) = sin(theta) / cos (theta).
So using tan(theta) = y' = dy/dx
sin(theta) / cos (theta) = y'.
sin(theta) = y' cos(theta)
plugging in what we got for cos earlier
sin(theta) = y' [sqrt (1 - y^2)]
now I am stuck.


I am just using this example to try and figure out how to do it. I actually want it in the form cos(theta) = something * x'.

 

[topsquark]

Yeah I know this is pretty basic, but I just can't think of it right now.

The slope of the wire is tan(theta) = y' = dy/dx
which yields sin(theta) = y' / sqrt (1 + y'^2)

How do you get from the first line to the second one?



this is what I I have done:
x^2 + y^2 = 1
cos^2(theta) + sin^2(theta) = 1
so cos^2(theta) = 1 - sin^2(theta)
so cos(theta) = sqrt (1 - y^2).

And
tan(theta) = sin(theta) / cos (theta).
So using tan(theta) = y' = dy/dx
sin(theta) / cos (theta) = y'.
sin(theta) = y' cos(theta)
plugging in what we got for cos earlier
sin(theta) = y' [sqrt (1 - y^2)]
now I am stuck.


I am just using this example to try and figure out how to do it. I actually want it in the form cos(theta) = something * x'.

sin(theta) = y' cos(theta)

Recall that sec(theta)=1/cos(theta). What's the relationship between tan^2 and sec^2?

-Dan

 

[KoGs]

BAM. I think I got it.

1 + tan^2 = sec^2
sec = sqrt (1 + tan^2)
sec = 1/cos
so sin * sqrt (1 + tan^2) = y'
so sin = y' / sqrt (1 + tan^2) , and tan = y'
so sin = y' / sqrt (1 + y'^2)

Thx for your help.