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## calyne Group Title 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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1. calyne Group Title

leminscate is just the name of that particular type of curve but makes no difference

2. Mani_Jha Group Title

Where the tangent is horizontal, dy/dx=0 Just differentiate it and set the result equal to zero.

3. Rohangrr Group Title

follow this !! cims.nyu.edu/~kiryl/Calculus/.../Smith-implicit-differentiation.pdf

4. calyne Group Title

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

5. calyne Group Title

but what do i do after i set it equal to zero i solve for x or y??

6. calyne Group Title

PLEASE HELP ME

7. calyne Group Title

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

8. Rohangrr Group Title

it will give u a better explanation cims.nyu.edu/~kiryl/Calculus/.../Smith-implicit-differentiation.pdf

9. calyne Group Title

yeah no i can't access it i'm now allowed oh well

10. calyne Group Title

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

11. Mani_Jha Group Title

Find y' from that result. Set that equal to zero Then solve for x and y.

12. calyne Group Title

right well that's what i did at first and either way i was stuck there too that's why i tried this

13. phi Group Title

You got this 4(x^2 + y^2)(2x - 2yy') = 25(2x - 2yy') I think the 2nd factor on the left-hand side should be (2x+2y y'). So it should be 4(x^2 + y^2)(2x + 2yy') = 25(2x - 2yy') Now solve for y'

14. phi Group Title

you get $y' = \frac{-4x^3-4xy^2+25x}{25y+4y(x^2+y^2)}=0$ It looks like we can solve the numerator for the x values that make it zero.

15. calyne Group Title

show me pleasee

16. phi Group Title

Are you asking how to solve for x in the expression $-4x^3-4xy^2+25x =0$

17. phi Group Title

x=0 is one solution now solve $4x^2+4y^2-25=0$ moving the $$4y^2-25$$ to the right-hand side, and dividing by 4: $x^2= \frac{24-4y^2}{4}$

18. phi Group Title

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

19. calyne Group Title

oh no i'd gotten 2x+2yy' for the second term where you said i got - btw