Calculus problems: implicit differentiation?

Question

Again, someone posts 5 math questions within a few minutes then deletes them all. In the interest of not letting my work go to waste, here is the question and my answer. Any comments or alternate solutions are welcome.

Question:
Find the point (in the first quadrant) on the lemniscate 2(x^2+y^2)2=25(x^2−y^2) where the tangent is horizontal.

how do you solve for x and y?
______________________________

My answer:

First we find dy/dx by implicit differentiation:
2(x²+y²)² = 25(x²-y²)

4(x² + y²) (2x + 2y dy/dx) = 25 (2x - 2y dy/dx)
2x (4x² + 4y²) + 2y (4x² + 4y²) dy/dx = 50x - 50y dy/dx
2y (4x² + 4y²) dy/dx + 50y dy/dx = 50x - 2x(4x² + 4y²)
2y (4x² + 4y² + 25) dy/dx = 2x (25 - 4x² - 4y²)
dy/dx = x (25 - 4x² - 4y²) / [y (4x² + 4y² + 25)]

Tangent is horizontal when dy/dx = 0

x (25 - 4x² - 4y²) / [y (4x² + 4y² + 25)] = 0
x (25 - 4x² - 4y²) = 0

x = 0
25 - 4x² - 4y² = 0 -----> x² + y² = 25/4

Now we plug these values into equation of lemniscate: 2(x²+y²)² = 25(x²-y²)

x = 0
2(0+y²)² = 25(0-y²)
2y⁴ + 25y² = 0
y² (2y² + 25) = 0
y = 0

But when x = 0 and y = 0, then dy/dx = 0/0 which is indeterminate.
Checking graph of 2(x²+y²)² = 25(x²-y²), we see that tangent is not horizontal at (0,0)
http://www2.wolframalpha.com/input/?i=2%28x%C2%B2%2By%C2%B2%29%C2%B2+%3D+25%28x%C2%B2-y%C2%B2%29

x² + y² = 25/4

2(x²+y²)² = 25(x²-y²)
2(x²+y²)² = 25(x²+y²-2y²)
2(25/4)² = 25(25/4-2y²)
625/8 = 625/4 - 50y²
50y² = 625/4 - 625/8
50y² = 625/8
y² = 25/16
y = 5/4 (we eliminate y = -5/4 ---> not in first quadrant)

x² + y² = 25/4
x² = 25/4 - y²
x² = 25/4 - 25/16
x² = 75/16
x = 5√3/4 (we eliminate x = -5√3/4 ---> not in first quadrant)

Point in first quadrant where tangent is horizontal: (5√3/4, 5/4)


Mαthmφm

Elizabeth M · Accepted Answer

There is a parametric form x=at(1+t^2)/(1+t^4)
y=at(1-t^2)/(1+t^4), where a=5/sqrt(2) and you can find where
dy/dt=0.

Alternatively the polar equation of the curve is r^2=(25/2)(2cos^2(theta)-1) and find
where dy/dtheta=0, giving dr/dtheta=-rcottheta

U V · Answer

Very impressive Mathmom - so I'm going to give you an [Interesting].


Can you solve this then?

Find the equation of a locus of a point on a circle radius 'r' that rolls along the (+) x axis.  The locus starts at the origin.

Anonymous · Answer

ok so there is 3 words right here, the final 2 are undemanding, utilising the chain rule. by-made from x: think of that is x^a million multiply via the skill (a million), then shrink the skill via a million to get: a million*x^0 remember that something to the skill of 0 is basically a million, so the respond is basically: a million by-made from 2y: same technique as above, yet you may get 2 via fact the respond, remember however via fact that is a y term, then you multiply it via component of dy/dx to get: 2dy/dx by-made from xy: it quite is -somewhat- extra sturdy, you are able to desire to apply the product rule the place the (f*g)' = f''(x)g(x) + g'(x)f(x) meaning that the by-made from 2 issues multipled, equals the 1st extensive kind expanded via the by-made from the 2nd - THEN upload THAT - to the by-made from the 1st extensive kind multipled via the 2nd so: = (by-made from x)*y + x*(by-made from y) = a million*y + x*(dy/dx) = x(dy/dx) + y So now you be attentive to all 3 words, basically placed them at the same time to get: x(dy/dx) + y - a million + 2(dy/dx) and that i could rearrange that to: dy/dx(x+2) + y - a million --------------------------- EDIT: and since the unique poster makes a sturdy factor, that is probable xy - x + 2y = 0 wherein case, the answer I even have could = 0 (as 0 is the by-made from 0) So: dy/dx(x+2) + y - a million = 0 dy/dx(x+2) = a million - y dy/dx = a million-y / x+2

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Calculus problems: implicit differentiation?

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