derivative trig question (1 Viewer)

[icyeyes]

hi can somebody help me with this
find dy/dx for
sinx + cosy = 5

thanx!

 

[KFunk]

This is outside the 3U course. In order to do it you want to use implicit differentiation.

sinx + cosy = 5

d/dx(sinx + cosy) = 0

cosx - dy/dx.siny = 0

dy/dx.siny = cosx

dy/dx = cosx/siny

you'll probably want an explanation of implicit differentiation to grasp that fully. I'll put up something about it tomorow if you want. Otherwise if anyone feels the urge... go for it.

 

[icyeyes]

thanx so much!! that helped me a lot

 

[wrong_turn]

fuck i should have knwon how to do it....aww so simple...:|

 

[Trev]

Why is this extension 1 work?
I thought implicit differentiation was only part of the extension 2 syllabus?

 

[grendel]

how about:

cosy=5-sinx
y=cos^-1(5-sinx)

let u=5-sinx

tf y=cos^-1(u)
dy/du=-1/sqrt(1-u^2)
dy/du=-1/sqrt(1-(5-sinx)^2)

now, from u=5-sinx
du/dx=-cosx

using the chain rule

dy/dx=dy/du * du/dx
dy/dx=-1/sqrt(1-(5-sinx)^2) * -cosx
dy/dx=cosx/sqrt(1-(5-sinx)^2)

this is all 3 unit theory

 

[grendel]

This is outside the 3U course. In order to do it you want to use implicit differentiation.

sinx + cosy = 5

d/dx(sinx + cosy) = 0

cosx - dy/dx.siny = 0

dy/dx.siny = cosx

dy/dx = cosx/siny

you'll probably want an explanation of implicit differentiation to grasp that fully. I'll put up something about it tomorow if you want. Otherwise if anyone feels the urge... go for it.

dy/dx must be a function of x.

so you must continue as follows:

cosy = 5 - sinx

draw a right angle triangle and label one angle y.

the adjacent side = 5-sin x

the hypotenuse = 1

using pythagoras, the opposite side = sqrt(1-(5-sinx)^2)

you can now express sin y as sqrt(1-(5-sinx)^2).

this then must be substituted for sin y in the last line of your solution.

 

[KFunk]

hi can somebody help me with this
find dy/dx for
sinx + cosy = 5

thanx!

btw, where did you get that equation from? I don't actually think you can graph it.
Given that -1≤sinx≤1 for all x ∈ R , and -1≤cosy≤1 for all y ∈ R.
hmm...

 

[grendel]

btw, where did you get that equation from? I don't actually think you can graph it.
Given that -1≤sinx≤1 for all x ∈ R , and -1≤cosy≤1 for all y ∈ R.
hmm...

hey i didn't notice that!

probably some teacher trying to write a clever question.

 

[acmilan]

Thats right!

sinx + cosy = 0 would give a continuous triangular pattern above the x-axis and sinx + cosy = 1 would give a semi-circular pattern above the x-axis (similarly for sinx + cosy = -1). Anything else for the integer on the right hand side wont give anything.