# What do I do next (trig)Watch

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#1
A curve has the equation: xsiny +cos2y=5/2
for x>/ 0 and 0</y<2 pi

Determine the exact coordinates of each point on the curve at which the tangent to the curve is parallel to the y axis?

So far I used implicit differentiation:
xcosy dy/dx +siny -2sin2y dy/dx=0
dy/dx= siny/2sin2y - xcosy

What do I need to do next?
0
2 months ago
#2
Can you rewrite sin2y in terms of siny and cosy, and factorise the denominator?

Also, what value of dy/dx are you looking for if the curve is parallel to the y-axis?
0
#3
(Original post by ThomH97)
Can you rewrite sin2y in terms of siny and cosy, and factorise the denominator?

Also, what value of dy/dx are you looking for if the curve is parallel to the y-axis?
I simplified the dy/dx by doing:
siny/ 2(2sinycosy)-xcosy

siny/4sinycosy-xcosy

siny/cosy(4siny-x)

If it's parallel to the y axis am I meant to look at the normal to dy/dx?
0
2 months ago
#4
(Original post by Anonymous1502)
I simplified the dy/dx by doing:
siny/ 2(2sinycosy)-xcosy

siny/4sinycosy-xcosy

siny/cosy(4siny-x)
Looks good.

If it's parallel to the y axis am I meant to look at the normal to dy/dx?
What is the gradient of a line parallel to the y axis?
0
2 months ago
#5
(Original post by Anonymous1502)
A curve has the equation: xsiny +cos2y=5/2
for x>/ 0 and 0</y<2 pi

Determine the exact coordinates of each point on the curve at which the tangent to the curve is parallel to the y axis?

So far I used implicit differentiation:
xcosy dy/dx +siny -2sin2y dy/dx=0
dy/dx= siny/2sin2y - xcosy

What do I need to do next?
You've simplified it wrong, you missed a negative sign on the siny numerator. After this you set dy/dx to equal to 1 because the gradient parallel to the y axis is 1. Then you work out the coordinates somehow, idk.
Last edited by JJJJJAAAAMES; 2 months ago
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