CPhill

Nice one, rosala!!!

Here's (b)

 q =  [P₁P₂ + 2P₁ ] / [P₁P₂ - 2P₂ ].......we can use the quotient rule here.....

∂q/∂P₁ =  [ [ (P₂ + 2)(P₁P₂ - 2P₂) ] - [ (P₁P₂ + 2P₁) (P₂) ] ] /  [P₁P₂ - 2P₂ ]²   =

[ P₁P₂² + 2P₁P₂ - 2P₂² - 4P₂  - P₁P₂² - 2P₁P₂ ] / [P₁P₂ - 2P₂ ]² =

- [ 2P₂² + 4P₂ ] /  [P₁P₂ - 2P₂ ]² =

-2P₂ [ P₂ + 2 ] / [P₁P₂ - 2P₂ ]²

-2P₂ [ P₂ + 2 ] / [P₂ (P₁ - 2)]² =

-2 [ P₂ + 2 ] / [P₂ (P₁ - 2)² ]

Happy New Year, Rosala and Andre.....!!!

Here's (b)

x^2 + xy - y^2  = 1    using implicit differentiation, we have

2x + y + xy' - 2yy'  = 0

xy' - 2yy'  = -2x - y       multiply through by - 1 on both sides

2yy' - xy'  = 2x + y

y'(2y - x) = 2x + y

y' =  [2x + y ] /  [2y - x ]      and the slope of the tangent line at (2.3) is given by

y' = [2(2) + 3] / [2(3) - 2 ]   =   [7] / [ 6 - 2]  =  7 / 4  

∫√(4 - x^2) dx       

Let x = 2sinΘ       dx = 2cosΘ dΘ  .....    when x =0, Θ = 0  and when x = 2, Θ = pi/2 ... so we have

∫√(4 - 4sin^2 Θ) 2cosΘ dΘ =

2∫√[(4(1 - sin^2  Θ) ] cos Θ  dΘ =

2 ∫2 √[ 1 - sin^2  Θ) ] cos Θ  dΘ =

4 ∫ cos Θ * cos Θ  dΘ =

4 ∫ cos ^2 Θ  dΘ          and using  cos^2 Θ = (1/2)(1 + cos2 Θ)  , we have

2 ∫ [1 + cos2 Θ]  dΘ  =

2Θ   + sin 2Θ  ....and substituting in the limits of integration, we have

2 [ pi/2 - 0 ] + [sin 2 (pi/2) - sin 2(0)] =

pi + [ sin (pi) - sin(0)] =

[pi ] +  [0]  =

pi =  about 3.1416

We have

n/3 - 50  = n/4       multiply both sides by 12

4n - 600 = 3n          simplify

n = 600

And that's the number.....!!!

That's a "gouda" question......one thing's for certain....you can't "edam" all at once.......!!!

f(x)=-2x^2-4x

Without using Calculus, we can find the x coordinate of the vertex by ... x = -b/2a......where, in this case, b= -4 and a = -2.....so we have

x = -b/2a  = -(-4) / 2(-2) = 4 / -4  = -1

Now, to find the "y" coordinate of the vertex, just put -1 back into the function for each "x" and evalute the result......

So you will have.... -2(-1)^2  - 4(-1)   =  'y"

Can you take it from here???

Sorry, rosala...I should have explained more....Phi is the irrational number (1 + √5) / 2   ≈ 1.61803398874989485

If you have regular pentagon of side length = 1, then the length of one of its diagonals turns out to be "Phi"

Also....if you're familiar with the Fibonacci Series.......as the series grows larger, the ratio between a successive term and its preceding term tends towards Phi....!!

1.61803398874989485^4 = 6.85410196624968457504

We could have also found this by summing ..... 1.61803398874989485^2 +1.61803398874989485^3........since Phi has the interesting property that Phi^n + Phi^(n + 1)  = Phi^(n + 2)......!!!!