Quick P3 differentation questionWatch

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Thread starter 14 years ago
#1
Given that y = 18x - 16sinx + sin2x, prove that dy/dx > 0 for all values of x.

This is what I have done so far:

dy/dx = 18 - 16cosx + 2cos2x

now how do i prove that dy/dx is > 0 for all values of x? 0
14 years ago
#2
(Original post by devesh254)
Given that y = 18x - 16sinx + sin2x, prove that dy/dx > 0 for all values of x.

This is what I have done so far:

dy/dx = 18 - 16cosx + 2cos2x

now how do i prove that dy/dx is > 0 for all values of x? y=18x - 16sinx + sin2x
dy/dx = 18 - 16cosx + 2cos2x
dy/dx = 2cos2x - 16cosx + 18
dy/dx = 2[2cos^2(x) - 1] - 17cosx + 18
dy/dx = 4(cosx)^2 - 17cosx + 16
dy/dx = 4c^2 - 17c + 16
dy/dx = 4[c^2 - 17c/4 + 4]
dy/dx = 4[ (c - 17/8)^2 - (17/8)^2 + 4]
dy/dx = 4[ (c-17/8)^2 + 4 - (17/8)^2]

Simplify the expression by multiplying back by the 4 and tidying up the constants. The part outside the square should make a positive constant, proving the expression is greater than 0 for all x. Replace the c with cosx too. This is just use of completing the square.
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Thread starter 14 years ago
#3
daaaaamn thx...

just wondering u sure there isnt an easier way? lol otherwise i will have to seriously brush up my completing the square skills....
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Thread starter 14 years ago
#4
(Original post by Gaz031)
y=18x - 16sinx + sin2x
dy/dx = 18 - 16cosx + 2cos2x
dy/dx = 2cos2x - 16cosx + 18
dy/dx = 2[2cos^2(x) - 1] - 17cosx + 18
dy/dx = 4(cosx)^2 - 17cosx + 16
dy/dx = 4c^2 - 17c + 16
dy/dx = 4[c^2 - 17c/4 + 4]
dy/dx = 4[ (c - 17/8)^2 - (17/8)^2 + 4]
dy/dx = 4[ (c-17/8)^2 + 4 - (17/8)^2]

Simplify the expression by multiplying back by the 4 and tidying up the constants. The part outside the square should make a positive constant, proving the expression is greater than 0 for all x. Replace the c with cosx too. This is just use of completing the square.
hold on i fink u misread it...lol....see the red bit:

y=18x - 16sinx + sin2x
dy/dx = 18 - 16cosx + 2cos2x
dy/dx = 2cos2x - 16cosx + 18
dy/dx = 2[2cos^2(x) - 1] - 17cosx + 18
dy/dx = 4(cosx)^2 - 17cosx + 16
dy/dx = 4c^2 - 17c + 16
dy/dx = 4[c^2 - 17c/4 + 4]
dy/dx = 4[ (c - 17/8)^2 - (17/8)^2 + 4]
dy/dx = 4[ (c-17/8)^2 + 4 - (17/8)^2]

shouldnt it be 16cosx? then it would factorise easily:

dy/dx = 2[2cos^2(x) - 1] - 16cosx + 18
dy/dx = 4cos^2(x) - 16cosx + 16
dy/dx = 4[cos^2(x) - 4cosx + 4]
dy/dx = 4(cosx - 2)(cosx - 2)
dy/dx = 4(cosx - 2)^2

Therefore dy/dx>0 for all values of x as (cosx - 2)^2 will always be positive.

is that right?
0
14 years ago
#5
Yes. That's right. Sorry about that.
That's precisely why i hate doing mathematics on computers Pen and paper is superior in every way.
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Thread starter 14 years ago
#6
lol yeh temme bout it...takes so long to type out equations...lol...
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14 years ago
#7
(Original post by devesh254)
just wondering u sure there isnt an easier way? lol otherwise i will have to seriously brush up my completing the square skills....
dy/dx = 18 - 16cosx + 2cos2x = 18 - 16cosx + 2sinxcosx

Try to find the smallest value of dy/dx. This would occur when 16cosx is maximum and 2sinxcosx is minumum.
max(16cosx) = 16 (since cosx<1)
min(2sinxcosx) = -0.5 (let d(2sinxcosx)/dx = 0)

So min(dy/dx) = min(18 - 16cosx + 2cos2x) = 18 - 16 - 0.5 = 1.5 > 0.

I don't think this is a particulary good method, and I doubt you'd get any marks if you used it in an exam.
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