## anonymous one year ago what is the slope of the tangent line to the curve (x^3 + y)^2 -2a^2xy = b^2 at the point P(0,1) when a=(2)^(1/2) and b = 1

1. anonymous

@ganeshie8 @Robert136 @nincompoop

2. anonymous

I'm getting DNE but its wrong

3. anonymous

For this problem can you input the values for a and b, then find the derivative? Or does the inputting of values happen after differentiating..

4. anonymous

It looks really ugly if you differentiate it along with a and b :\

5. anonymous

I would use implicit differentiation to find the derivative.

6. anonymous

i really have no idea , iv imputed the values for a and be and then solved for y and derived that , but that seems to be wrong

7. anonymous

Yeah it is hahahahaha

8. anonymous

Because that's basically stating the tangent line is found at a point before even being found.

9. anonymous

you only need to find f'(x,y)

10. anonymous

but before u start it says when a=(2)^(1/2) and b = 1 , a and b seems like small values and you can sub them why it mention them in ur question ? seems like there is a formula in ur book or something

11. IrishBoy123

$\vec n = <2(x^3+y)3x^2 - 2a^2y, 2(x^3+y)-2a^2x>$ $= <6x^2(x^3+y) - 2a^2y, 2(x^3+y)-2a^2x>$ x = 0 $= <-2a^2y,2>$ $y = 1, a = \sqrt{2}$ $\vec n = <-4,2,>$

12. freckles

don't solve for y just find the derivative of both sides w.r.t. x a and b are constants you can input those values in later or not once you have differentiated both sides you can input the values for a and b if you haven't already and then also input 0 for x and 1 for y then solve for y' to find the slope.

13. anonymous

see thats what i'm talking about :D like @irishboy123

14. IrishBoy123

|dw:1441778557798:dw|

15. anonymous

just noticed the name of the chapter is implicit differention

16. freckles

why is there y'=that mess=0?

17. freckles

wait I just had a problem with y'= part

18. anonymous

well the question said (tangent line) so it's line equation u can do it in both ways with irish normal vector or with slope and a point

19. freckles

$2(x^3+y)(3x^2+y')-(2a^2xy'+2a^2y)=0 \text{ was right }$

20. anonymous

Thanks.

21. freckles

I just didn't agree with the y'=that stuff=0

22. freckles

because saying y'=0 is saying you already know the slope to be 0

23. freckles

$2(x^3+y)(3x^2+y')-(2a^2xy'+2a^2y)=0 \text{ was right }$ so copying this @Jdosio just enter in 0 for x and 1 for y simplify and solve for y'

24. anonymous

okay

25. anonymous

a^2 ?

26. freckles

yes and a was?

27. anonymous

a = 2^(1/2) so final answer is 2

28. freckles

right but hey listen do you understand how she got that equation?

29. anonymous

not a single clue ^_^

30. freckles

she used chain rule and product rule (there are some other small potato rules she used)

31. freckles

$\frac{d}{dx}(x^3+y)^2$ for this you should use chain rule

32. anonymous

$(x^3 + y)^2 -2a^2xy = b^2$$2(x^3+y)(3x^2+y')-(2a^2xy'+2a^2y)=0$$6x^5+2x^3y'+6x^3y+2yy'-2a^2xy'-2a^2y=0$$y'(2x^3+2y-2a^2x)=-6x^5 -6x^3y+2a^2y$$y'=...$

33. freckles

$\frac{d}{dx}(x^3+y)^2 =2(x^3+y) \cdot (x^3+y)'$ derivative of outside times the derivative of inside

34. freckles

$\frac{d}{dx}(x^3+y)=3x^2+y' \\ \text{ so } \frac{d}{dx}(x^3+y)^2=2(x^3+y) \cdot (3x^2+y')$

35. anonymous

chain rule: $$f(g(x)) = f'(x) \cdot g'(x)$$

36. anonymous

oh okay i see i see

37. anonymous

i was just rather lost on the second part of the subtraction where you used the product rule , im familiar with it i just wasn't entirely sure how to go about the a but i see what you did

38. freckles

the other term you can use constant multiple rule and product rule $\frac{d}{dx}-2a^2xy \\ -2a^2 \frac{d}{dx}(xy) \text{ by constant multiple rule } \\ \text{ do you think you can use product rule \to find } \frac{d}{dx}(xy)$

39. anonymous

yea x' y + x y'

40. anonymous

You're trying to keep in mind that taking the derivative of $$y$$ will give you $$\dfrac{dy}{dx}$$ and that you are taking the derivative of $$x$$ as well.

41. freckles

that is correct but x'=?

42. freckles

or dx/dx=?

43. anonymous

wait , are you asking about the example or the actual question?

44. freckles

well you said $\frac{d}{dx}(xy)=\frac{dx}{dx} y+x \frac{dy}{dx}$ and I was asking you to simplify dx/dx

45. anonymous

well 1 right?

46. freckles

yep

47. freckles

\frac{d}{dx}-2a^2xy \\ -2a^2 \frac{d}{dx}(xy) \text{ by constant multiple rule } \\ \text{ do you think you can use product rule \to find } \frac{d}{dx}(xy) right $\frac{d}{dx}(xy)=y+xy' \\ \text{ so } \frac{d}{dx}-2a^2xy=-2a^2(y+xy') =-2a^2y-2a^2xy'$

48. freckles

and of course the other side of the equation is b^2 so when differentiating a constant you get 0

49. freckles

@Jdosio do you have any questions on this?

50. anonymous

not on this question specifically , if i come up with anything else i missed ill ask you guys if your still around. thanks for all your help once again.

51. anonymous

No problem.