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Find the points on the lemniscate 2(x^2 + y^2)^2 = 25(x^2  y^2) where the tangent is horizontal. The derivative using implicit differentiation is (4x^3  4xy^2 + 25x)/(4x^3 + 4x^2 y + 25y)
 2 years ago
 2 years ago
Find the points on the lemniscate 2(x^2 + y^2)^2 = 25(x^2  y^2) where the tangent is horizontal. The derivative using implicit differentiation is (4x^3  4xy^2 + 25x)/(4x^3 + 4x^2 y + 25y)
 2 years ago
 2 years ago

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calyneBest ResponseYou've already chosen the best response.0
leminscate is just the name of that particular type of curve but makes no difference
 2 years ago

Mani_JhaBest ResponseYou've already chosen the best response.0
Where the tangent is horizontal, dy/dx=0 Just differentiate it and set the result equal to zero.
 2 years ago

RohangrrBest ResponseYou've already chosen the best response.0
follow this !! cims.nyu.edu/~kiryl/Calculus/.../Smithimplicitdifferentiation.pdf
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
damn it i'd love to because my textbook doesn't actually explain it just provides examples but i've done lright so far anyway
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
but what do i do after i set it equal to zero i solve for x or y??
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
so ok what 4(x^2 + y^2)(2x  2yy') = 25(2x  2yy') and then y'=0 so 4(x^2 + y^2)(2x) = 25(2x) NOW WHAT
 2 years ago

RohangrrBest ResponseYou've already chosen the best response.0
it will give u a better explanation cims.nyu.edu/~kiryl/Calculus/.../Smithimplicitdifferentiation.pdf
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
yeah no i can't access it i'm now allowed oh well
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
i don't like go to nyu maybe if you wanna download it and upload it here go for it but i can't check that url
 2 years ago

Mani_JhaBest ResponseYou've already chosen the best response.0
Find y' from that result. Set that equal to zero Then solve for x and y.
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
right well that's what i did at first and either way i was stuck there too that's why i tried this
 2 years ago

phiBest ResponseYou've already chosen the best response.1
You got this 4(x^2 + y^2)(2x  2yy') = 25(2x  2yy') I think the 2nd factor on the lefthand side should be (2x+2y y'). So it should be 4(x^2 + y^2)(2x + 2yy') = 25(2x  2yy') Now solve for y'
 2 years ago

phiBest ResponseYou've already chosen the best response.1
you get \[ y' = \frac{4x^34xy^2+25x}{25y+4y(x^2+y^2)}=0 \] It looks like we can solve the numerator for the x values that make it zero.
 2 years ago

phiBest ResponseYou've already chosen the best response.1
Are you asking how to solve for x in the expression \[ 4x^34xy^2+25x =0 \]
 2 years ago

phiBest ResponseYou've already chosen the best response.1
x=0 is one solution now solve \[ 4x^2+4y^225=0 \] moving the \( 4y^225 \) to the righthand side, and dividing by 4: \[x^2= \frac{244y^2}{4} \]
 2 years ago

phiBest ResponseYou've already chosen the best response.1
Now try each of these x values in the original equation. Notice that x=0 results in an imaginary value for y, so we can discard that point. with x^2 substituted into the original equation, solve for y, and then find x. you should get y=±5/4
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
oh no i'd gotten 2x+2yy' for the second term where you said i got  btw
 2 years ago
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