[bigbadb1319]

Ex. 4B
q). 4
Find the coordinates of hte points of zero gradient on the curve with implicit equations: x² + 4y² - 6x - 16y + 21
can any1 help me wid da procedure...i know u gotta differentiate, and then do you differentiate again? bu thten how do you find out the co-ordinates...n-e help will be much appreciated!
thanx....

[Vazzyb]

Ok, starting from,

y = x^2 + 4y^2 - 6x - 16y + 21
dy/dx = 2x + 8y(dy/dx) - 6 - 16(dy/dx)
therefore,

17(dy/dx) - 8y(dy/dx) = 2x - 6
(dy/dx)(17 - 8y) = 2x - 6

and therefore,

dy/dx = (2x - 6)/(17 - 8y)

and therefore,

the stationary point happens when that equals 0, ie. x = 3?

Not sure though!

[<A-S-H-I-Q-U-E>]

Originally Posted by bigbadb1319

Ex. 4B
q). 4
Find the coordinates of hte points of zero gradient on the curve with implicit equations: x² + 4y² - 6x - 16y + 21
can any1 help me wid da procedure...i know u gotta differentiate, and then do you differentiate again? bu thten how do you find out the co-ordinates...n-e help will be much appreciated!
thanx....

"x² + 4y² - 6x - 16y + 21" - is this an equation?

[Nima]

Originally Posted by bigbadb1319

Ex. 4B
q). 4
Find the coordinates of hte points of zero gradient on the curve with implicit equations: x² + 4y² - 6x - 16y + 21 = 0

2x + 8y.dy/dx - 6 - 16.dy/dx = 0
=> 8.dy/dx.(y - 2) = 6 - 2x = 2(3 - x)
=> dy/dx = (3 - x)/[4(y - 2)]

At the points of 0 gradient:
=> 0 = 3 - x
=> x = 3

When x = 3:
=> 9 + 4y² - 18 - 16y + 21 = 0
=> 4y² - 16y + 12 = 0
=> y² - 4y + 3 = 0
=> (y - 3)(y - 1) = 0
=> y = 3, y = 1

:. the points of zero gradient are: (3, 1) and (3, 3)

--------------

Originally Posted by <A-S-H-I-Q-U-E>

"x² + 4y² - 6x - 16y + 21" - is this an equation?

He means = 0

[Dekota]

Originally Posted by bigbadb1319

Ex. 4B
q). 4
Find the coordinates of hte points of zero gradient on the curve with implicit equations: x² + 4y² - 6x - 16y + 21

d[x² + 4y² - 6x - 16y + 21]/dx
=> 2x + 8y.dy/dx - 6 - 16.dy/dx
=> 2x + dy/dx(8y - 16) - 6 = 0
=> dy/dx(8y - 16) = 6 - 2x
=> dy/dx = (6 - 2x)/(8y - 16)
=> dy/dx = (3 - x)/(4y - 4)

sub dy/dx = 0...

=> 0 = (3 - x)/(4y - 4)
=> 0 = (3 - x)

x = 3

sub (x = 3) into x² + 4y² - 6x - 16y + 21 = 0
=> 3² + 4y² - 6(3) - 16y + 21 = 0
=> 4(y² - 4y) = -12
=> y² - 4y + 3 = 0
=> (y - 1)(y - 3) = 0

y = 1 or 3

coordinates: P(3,1) and Q(3,3)

[Nima]

Be careful now with your y values! And the poster means = 0, but that doesn't affect the answer.

[Vazzyb]

oh he meant 0!