Parametric/calculus??? (1 Viewer)

[Michaelmoo]

Find dy/dx if:

x=2cos(Theta)

y=2sin(Theta)

Im getting dy/dx = +-x/[{root}(4-x^2)]

Doesn't seam too right =\

Please help. Thanks.

 

[adnan91]

Find dy/dx if:

x=2cos(Theta)

y=2sin(Theta)

Im getting dy/dx = +-x/[{root}(4-x^2)]

Doesn't seam too right =\

Please help. Thanks.

chain rule!
dy/dx= cot theta

 

[Arowana21]

dx/d(theta)= -2sin(theta)
dy/d(theta)= 2cos(theta)
therefore
dy/dx= (dy/d(theta)) * (d(theta)/dx)= (2cos(theta)) / (-2sin(theta))

you cancel the 2, and since cos/sin is equal to cot
therefore the final answer is
-cot(theta)

 

[tommykins]

Find dy/dx if:

x=2cos(Theta)

y=2sin(Theta)

Im getting dy/dx = +-x/[{root}(4-x^2)]

Doesn't seam too right =\

Please help. Thanks.

dx/d@ = -2sin@
dy/d@ = 2cos@
dy/dx = 2cos@/-2sin@ = -cot@

 

[gurmies]

My approach:

x = 2cos@

y = 2sin@

x^2 + y^2 = 4sin^2@ + 4cos^2@

x^2 + y^2 = 4(sin^2@ + cos^2@)

x^2 + y^2 = 4

Dx [x^2 + y^2] = Dx [4]

2x + 2y (dy/dx) = 0

dy/dx = -x/y

= -cot@

Probably not the best method, but I haven't learnt to differentiate trigonometry yet =(

 

[Michaelmoo]

Find dy/dx if:

x=2cos(Theta)

y=2sin(Theta)

Im getting dy/dx = +-x/[{root}(4-x^2)]

Doesn't seam too right =\

Please help. Thanks.

OK. But we havnt learnt to differnetiate sine and cos. So basically my question is. Find dy/dx for x^2 + y^2 = 4 (That is after you turn them into a parametric). By the way, does the final derivative have to be with respect to x only or it could be x and y??

 

[gurmies]

Well, I haven't learnt to differentiate trig either, so I used implicit differentiation, and had it in terms of x and y, and seeing as we know what x and y is, it wasn't too hard to find dy/dx...