Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

bettyboop8904

  • 2 years ago

Can anyone help me figure this problem out pretty please? Find an equation of the tangent line to the curve xe^(y)+ye^(x)=1 at the point (0,1) The equation is rewritten in the post with the equation converter helper = )

  • This Question is Closed
  1. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[xe ^{y}+ye ^{x}=1\] at (0,1)

  2. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    anybody??

  3. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @zepdrix

  4. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    any idea how to do this one? = )

  5. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Bahhhhh gimme few minutes XD eating some food.

  6. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    kk take your time = )

  7. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\huge xe^y+ye^x=1\]Were you able to take a derivative of this function? Or are we having some trouble on that part? :)

  8. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Or confused about the directions? heh

  9. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I was a little confused, since my teacher added implicit differentiation to our course last semester I always get confused. This is what I got when I took the derivative but I'm not sure if its correct: \[ye ^{x}=1-xe ^{y}\] \[y=\frac{ 1-xe ^{y} }{ e ^{x} }\] \[y'=\frac{ e ^{x}(y'xe ^{y})-(1-xe ^{y})(e ^{x}) }{ (e ^{x})^{2} }\]

  10. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ew ew ew don't try to turn it into an explicit function in \(x\). It's actually impossible in this case. So just leave it alone, and take the derivative from where it is at the start D:

  11. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hahaha ok so show me = )

  12. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Implicit isn't too bad, just remember that whenever you take the derivative of a `y` term, a y' will pop out. Ok ok ok :)

  13. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so is it everytime the derivative of y will be y*y' ?

  14. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Everything you take the derivative of y, you'll apply the power rule just like you would for x. Giving you 1 (not y), and then you attach a y' to it. So the derivative of y will be 1*y'.

  15. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\huge \color{salmon}{xe^y}+ye^x=1\]We'll take the derivative of the pink term first. Applying the `Product Rule` will give us,\[\large \color{salmon}{(x)'e^y+x(e^y)'}\]Which simplifies to,\[\large \color{salmon}{e^y+x(e^y)y'}\]

  16. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Understand how we did the first term? :)

  17. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok that's right I remember now = )

  18. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that was for above hold on let me read what u posted lol

  19. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hehe yah I understand :3

  20. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    On the second part, we had to apply the chain rule. e^y gave us e^y. But then we also have to multiply by the derivative of the exponent.

  21. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok so the next part would be (and altogether): \[(e ^{y}+xy'e ^{y})+(y'e ^{x}+ye ^{x})\]

  22. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    =1 oops lol

  23. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes good c: `Except` we took the derivative of `both sides`, so what should the right side produce?

  24. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    0

  25. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\huge e^y+xe ^y\color{orangered}{y'}+e^x\color{orangered}{y'}+ye^x=0\]Our goal from here is to solve for y'

  26. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, good 0* blah that disappeared somehow lol

  27. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    haha its ok i was trying to figure out what to do next lol i was going to say plug 0 in for x but solve for y' that makes sense lol

  28. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Understand how to do that? It's not too bad, it just requires a bit of nasty algebra :)

  29. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah give me a sec i love algebra = )

  30. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    haha nice XD

  31. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh my goodness betty... -_- are you using the equation tool? zzzzzzz lol

  32. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok I do love algebra but I could be wrong with this one but I think I might be right at the same time lol \[xy'e ^{y}+y'e ^{x}=-e ^{y}-e ^{x}y\] \[y'(xe ^{y}+e ^{x})=-e ^{y}-e ^{x}y\] \[y'=\frac{ -e ^{y}-e ^{x}y }{ xe ^{y}+e ^{x} }\]

  33. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    haha yes i was

  34. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    looks good c:

  35. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok so now I plug 0 in for x?

  36. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what is that lol jk

  37. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1359347099057:dw|

  38. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This isn't necessarily what the function looks like. It's just to give us an idea of what they're asking.

  39. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  40. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    They want us to form an equation for the `straight line` that is tangent to the function at x=0.

  41. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Remember the form of a straight line? :)

  42. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That's right! Your good ole buddy \(y=mx+b\). Remember him?

  43. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh yeah ok i wasn't sure what u were asking lol

  44. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Our slope \(m\) is going to be equal to the slope of the tangent line at x=0. Meaning \(y'(0)=m\).

  45. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I was actually helping a person out today a lot with that form and the slope form with two points

  46. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh cool XD

  47. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I guess I should have written it like this,\[y'(0,1)=m\]We're going to plug in the given x and y value to get our m.

  48. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok i was wondering what i would do with the y-values

  49. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i got \[\frac{ -e-1 }{ 1 }\]

  50. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yep looks good.

  51. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large y=mx+b \qquad \rightarrow \qquad y=-(e+1)x+b\] And now we only need to find the y-intercept `b`. To do so, we'll again use the point that ws given (0,1).

  52. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    kk let me work it out = )

  53. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok so this is what i got: \[y=-(e+1)x+1\] Now would you multiply it out like this: \[y= -ex-x+1\rightarrow -x(e+1)+1\]

  54. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No don't multiply it out c: looks good! Hopefully we didn't make any silly mistakes in there. Looks correct though :D

  55. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and then thats it? = )

  56. zepdrix
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, we found an equation for the line tangent to our curve at x=0, y=1.

  57. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    awesome = ) ty

  58. bettyboop8904
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I got a couple others = ) stay posted lol

  59. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.