Question

Find points on the curve x^2 - xy + y^2 = 9 that are closest to and furthest from the origin.

Answer 1

Go to polar coordinates!

Answer 2

you familiar wid lagrange ?

Answer 3

okay distance function is simply : x^2+y^2

Answer 4

Going that way, you need to minimize $x^2+y^2$ subject to the constraint that $x^2-xy+y^2-9=0$

Answer 5

I have differentiated implicitly and gy/dx= 2x - y / x - 2y. The next step is to write down a function that represents the distance of a point from the origin. I am stuck.

Answer 6

define your function to be the distance function and subtract the constraint multiplied by the lagrangian multiplier. then diffeentiate with respect to x, y and the multiplier.

Answer 7

.....easier, though is to go to polar coordinates. Then you can avoid constraints altogether.

Answer 8

we have not yet covered the above methods so there must be a simpler way.
step 1 is to implicitly differentiate which I have done
step 2 is to write down a function that represents the distance of a point from the origin and to differentiate D(x)
step 3 is to solve dy/dx=0, d'(x)=0 and x^2 - xy + y^2 = 9 together to determine the required points of the curve.

I am stuck on step 2

Answer 9

What are you differentiating and with respect to which variable?

Answer 10

you are differentiating the distance function. I am not sure what the distance function is.

Answer 11

So you need to start with step 2

Answer 12

distance between (0, 0 ) and any point on curve (x,y ) is : sqrt(x^2+y^2)

you may as well optimize its square : x^2+y^2

Answer 13

ok thanks I will try work it out and come back with my solution to see if im on the right track

Answer 14

still stuck on this ?

Answer 15

sqrt(x^2+y^2) and x^2+y^2 has extrema for the same values of x and y. The latter is easier to differentiate with than the former.

So in other words, we need to find extrema of x^2+y^2. One approach is to solve the constraint for y and eliminate y from this expression. Then you get an expression that only has x in it, and you can differentiate to find minima.

Easier is to substitute x=r cos(theta) and y=r sin(theta) into the equation for the curve. This results in a very simple equation for r^2 that you can easily minimize.

Answer 16

Ok I am lost. I am using the steps provided see above. So step 2 I have the distance function and to take it's derivative. If I use the method of taking it's square I get d'=-x/y
So now I am up to step 3. Making sense?

Answer 17

There are several approaches:

(1) in the equation for the curve, substitute x = r cos(theta), y=r sin(theta) and solve for r^2 and then look for max/min of r^2 as a function of theta. This is the easiest approach.
(2) Solve the equation for the curve for y in terms of x and substitute this into x^2 + y^2. This will give you an expression for the distance function in terms of x only. Differentiate it to find its max/min.
(3) use lagragian multipliers. (This is out since you do't know this method.)

Answer 18

x^2 - xy + y^2 = 9 ----- (1)

To find the maximum/minimum distance of a point (x,y) on the curve from the origin
we find the max/min of x^2 + y^2 (distance squared).

Let D = x^2 + y^2
Differentiate with respect to x:
D' = 2x + 2yy' = 0
2x = -2yy'
x = - yy' ----- (2)

Find y' by implictly differentiating the original equation: x^2 - xy + y^2 = 9
2x - xy' - y + 2yy' = 0
2x - y + y'(2y - x) = 0
y' = -(2x-y)/(2y-x) ---- (3)

Put y' from (3) into (2):
x = -y * -(2x-y)/(2y-x) = y(2x-y)/(2y-x)
x(2y-x) = y(2x-y)
2xy - x^2 = 2xy - y^2
x^2 = y^2
y = x or -x

Put y = x in (1): x^2 - x^2 + x^2 = 9
x^2 = 9
x = -3 or +3
y = -3 or +3

The two points are: (-3,3) and (3,3)

Put y = -x in (1): x^2 + x^2 + x^2 = 9
3x^2 = 9
x^2 = 3
x = -sqrt(3) or sqrt(3)
y = +sqrt(3) oe -sqrt(3)


The two points are: (-sqrt(3),sqrt(3)) and (sqrt(3),-sqrt(3))

Substitute the points (-3,3) and (3,3) into the distance-squared equation:
x^2 + y^2 = 9 + 9 = 18
distance = sqrt(18) = sqrt(2*9) = 3*sqrt(2)
Both points give the same distance of 3*sqrt(2)

Substitute the points (-sqrt(3),sqrt(3)) and (sqrt(3),-sqrt(3))into the distance-squared equation:
x^2 + y^2 = 3 + 3 = 6
distance = sqrt(6)
Both points give the same distance of sqrt(6)

Closest points are: (-sqrt(3),sqrt(3)) and (sqrt(3),-sqrt(3))
Farthest points are: (-3,3) and (3,3)

Minimum distance = sqrt(6)
Maximum distance = 3*sqrt(2)

Answer 19

The solution in polar coordinates: let x =r cos(t), y = r sin(t). Plugging into the curve: x^2 - xy + y^2 = 9 we get
r^2 - r^2 sin(t) cos(t) = 9
r^2 (1 - 0.5 sin(2t)) = 9
r^2=9/(1 - 0.5 sin(2t)) =
Hence r^2 has max when sin(2t) = 1 i.e. when t = pi/4 or 5pi/4 that is r^2 = 18 and x = y = 3 or x = y = -3.

r^2 has min when sin(2t) = -1 i.e. when t = -pi/4 or -5pi/4 that is r^2 = 6 and x = -y = sqrt(3) or x = -y = -sqrt(3).

Answer 20

Thank you so much for your help
I looked for similar examples but it was hard to find.

Answer 21

You are welcome.

Answer 22

Thanks ljensen for the polar coordinates example. A lot simpler but we have not yet covered that topic so I can't use it :(