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 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
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)

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

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

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

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

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

calyne
 2 years ago
Best ResponseYou've already chosen the best response.0so 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

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

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

calyne
 2 years ago
Best ResponseYou've already chosen the best response.0i 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

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

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

phi
 2 years ago
Best ResponseYou've already chosen the best response.1You 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'

phi
 2 years ago
Best ResponseYou've already chosen the best response.1you 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.

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

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

phi
 2 years ago
Best ResponseYou've already chosen the best response.1Now 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

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