Horizontal tangent points (implicit differentiation)

[TsAmE]

Homework Statement

Find the points on the lemniscate: 2( x^2 + y^2 )^2 = 25( x^2 - y^2 ) where the tangent is horizontal

Homework Equations

Horizontal tangent: y' = 0

The Attempt at a Solution

I got the correct gradient of y' = [ 50x - 8x^3 - 8y^2 ] / [ (8x^2)y + 50y + 8y^3 ], and then solved for y = + ( -8x^3 + 50x)^1/2 ), after which I subbed this y into the original equation and tried to solve for x, whereby I reached a dead end. The last line I got to was complicated when solving for x: 128x^6 - 32x^5 + 2x^4 - 1800x^23 + 175x^2 + 1250x = 0.

 

[LCKurtz]

To tell the truth, I didn't work through your stuff, but here's another idea that will work. First, notice that your equation has all symmetries so if you find an HTL in the first quadrant you will have three more by symmetry.

Put your equation into polar form r = f(θ), then calculate y' by using the polar formula:

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}<br /> <br /> = \frac{r(\theta)\cos\theta+r&#039;(\theta)sin\theta}{-r(\theta)\sin\theta+r&#039;(\theta)cos\theta}

You might want to get r' implicitly when you get there. To get a slope of 0 you just need to think about getting the numerator 0. With a bit of work you should be able to find a θ and then (x,y) that works.

 

[TsAmE]

I haven't learned polar form, in a test we are expected to do it the hard way

 

[LCKurtz]

Well, looking at your problem in rectangular coordinates, I get:

\frac{dy}{dx}= \frac {8(x^2+y^2)x-50x} {8(x^2+y^2)y +50y}

You can set the numerator to zero, factor out x and solve the other factor for x in terms of y. Plug that back into your original equation and you get a 4th degree equation in x, but no odd powers, so it is a quadratic equation in x² which you can solve for x², then for x. Seems a little long for an exam, but you can work through it.

 

[TsAmE]

Sorry but I don't understand. I tried it again and found a mistake in my previous calculation and got:

0 = [ 50x - 8x^3 - 8y^2 ] / [ (8x^2)y + 50y + 8y^3 ]
0 = 2( 25x - 4x^3 - 4y^2 ) then solved for y^2
y^2 = (25 / 4)x - x^3 subbed that back into my original equation:
2[ x^(2) + (25 / 4)x - x^3]^2 = 25[ x^(2) - (25 / 4)x + x^3 ]

which seems right to me, but now I am not sure how to simplify it

 

[LCKurtz]

Sorry but I don't understand. I tried it again and found a mistake in my previous calculation and got:

0 = [ 50x - 8x^3 - 8y^2 ] / [ (8x^2)y + 50y + 8y^3 ]

By my calculations, you are missing an x in the last term in the numerator. It should be -8y²x, so you get:

x(50-8x²-8y²)

in the numerator.

 

[TsAmE]

Oh ok I got y^(2) = -2x^(2) + (25x / 2) then subbed it back into the equation to get

2[x^(2) - 2x^(2) + (25x / 2)] = 25[x^(2) + 2x^(2) - (25x / 2)]
2x^2 - 4x^(2) + (25x / 2) = 25x^(2) + 50x^(2) - (625x / 2)

and I can't see that getting factorised anytime soon

 

[LCKurtz]

Oh ok I got y^(2) = -2x^(2) + (25x / 2)

Once again, the first thing you write down is wrong. You are apparently in a calculus class and you ought to be able to do simple algebra correctly. Once you correctly solve for y² you should get a very easy equation to solve for x² then x.

Also, please learn to use the X² button to make your equations more readable.