1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculus - Derivative Help?

  1. Nov 27, 2007 #1
    [SOLVED] Calculus - Derivative Help??

    1. The problem statement, all variables and given/known data
    Consider the curve given by the equation y^3 + 3x^(2)y + 13 = 0. Find the minimum y - coordinate of any point on the curve. Justify your answer.


    2. Relevant equations

    I solved parts a&b:
    dy/dx = -6xy / (3y^2 + 3x^2)
    Equation for line tangent to curve at (2,-1): y + 1 = 4/5 (x-2)

    3. The attempt at a solution
    This is the other problem I cannot get. I tried to make up my own minimum to find a relative minimum, which my instructor said we could find, but I still couldn't get critical points. I also tried using the second derivative ... but that got too messy and I didn't know what to do.

    Any help would be fantastic! Thank you for your time and explanations!!
     
  2. jcsd
  3. Nov 27, 2007 #2

    rock.freak667

    User Avatar
    Homework Helper

    at the minimum point...the gradient of the tangent at that point is zero...i.e. gradient function=0 => dy/dx=0
     
  4. Nov 27, 2007 #3
    sorry, we haven't learned gradient functions yet. can you explain this concept to me? or is there another way to do it?
     
  5. Nov 27, 2007 #4

    rock.freak667

    User Avatar
    Homework Helper

    Well the gradient at any point (x,y) on a curve is given by dy/dx...this is called the gradient function. So to find the gradient at any point you just sub the x value and y value(if there is y in your expression fot dy/dx) and you'll get the gradient of the tangent at that point.

    At a min point, the tangent drawn to that point is a horizontal line i.e. the line has a gradient at zero. So to find this point you equate dy/dx to zero and solve...

    dy/dx = -6xy / (3y^2 + 3x^2)=0

    for this to be equal to zero, the numerator must be zero...so solve for x and y in here and you will get a stationary point...find the 2nd derivative and sub the min point you get and check the sign, If [itex]\frac{d^2y}{d^2x}[/itex] is positive,then it is a minimum point, if it is negative, it is a maximum point.
    Is this clear? If not I shall try to explain better
     
  6. Nov 27, 2007 #5
    ok, one more question. so i got (0,0) from the first deriv, now i plug that in to the second to find if it is a max or min? i believe it must be a max anyway since it is higher than the point (2,-1) which the problem gives.
     
  7. Nov 28, 2007 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Neither (0,0) nor (2, -1) is even on the curve y^3 + 3x^(2)y + 13 = 0!

    In order that the derivative be 0, either x= 0 or y= 0. If x= 0, what is y? If y= 0, what is x?
     
  8. Nov 28, 2007 #7
    ahhhh, i understand now. i was being dense! but the sad thing is i already handed my problem in. oh, well, at least i will get it right next time!

    thank you all for the wonderful attention and help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Calculus - Derivative Help?
  1. Calculus Help! (Replies: 6)

  2. Calculus help (Replies: 4)

  3. Help with calculus (Replies: 3)

  4. Calculus Help (Replies: 4)

  5. Calculus Help (Replies: 5)

Loading...