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Equation of tangent line w/ sqrt.

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the equation of the tangent line to y=sqrt(25-x^2) at the point (4,3)


    2. Relevant equations



    3. The attempt at a solution
    (25-x^2)^1/2
    1/2(25-x^2)^-1/2
    (2x)1/2(25-x^2)^-1/2
    x(25-x^2)^-1/2

    I have no idea what to do or where to go after this...
    Please Help!
     
  2. jcsd
  3. Nov 7, 2009 #2

    lanedance

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    so is that chain rule differntition you tried? very hard to follow...

    if you have a slope & point, you should be able to come up with a an equation of a line no worries
     
  4. Nov 7, 2009 #3
    yes I tried the chain rule...and I cant seem to find the slope
     
  5. Nov 7, 2009 #4

    lanedance

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    you were given f(x), using chain rule you have found f'(x) which is the slope at any x, try substituting in the x valuue of the point you are given
     
  6. Nov 7, 2009 #5
    the point is (4,3)

    f'(x)= x(25-x^2)^-1/2
    f'(4)= 4(25-4^2)^-1/2
    f'(4)= 4(25+16)^-1/2
    f'(4)= 4(41)^-1/2
     
    Last edited: Nov 7, 2009
  7. Nov 7, 2009 #6

    lanedance

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    lost a negative
     
  8. Nov 7, 2009 #7
    Im checking over my work thus far and i cannot see where I lost a neg.
    at (25-4^2)
    -4^2=+16


    Please explain
     
  9. Nov 7, 2009 #8

    lanedance

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    it think it should be (25-(4^2))
     
  10. Nov 7, 2009 #9

    Mentallic

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    [tex](-4)^2 = (-1*4)^2 = (-1)^2*(4)^2 = 16[/tex]

    [tex]-(4^2) = (-1)*4^2 = -16[/tex]
     
  11. Nov 7, 2009 #10
    so..
    f'(4)=4(25-(4^2))^-1/2
    f'(4)=4(25-16)^-1/2
    f'(4)=4(9)^-1/2
    f'(4)=4(1/3)
    f'(4)=4/3

    ?
     
  12. Nov 7, 2009 #11

    Mentallic

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    You made a slight error with the derivative.

    [tex]y=(25-x^2)^{1/2}[/tex]

    [tex]\frac{dy}{dx}=\frac{1}{2}(25-x^2)^{-1/2}*(-2x)[/tex]

    [tex]\frac{dy}{dx}=-x(25-x^2)^{-1/2}[/tex]

    Can you take it from here?
     
  13. Nov 7, 2009 #12
    put the x value into you equation for y', to get the slope, then just find the line with that slope and the point you initially had
     
  14. Nov 7, 2009 #13
    Find tangent line to y= sqrt(25-x^2) at the point (4,3)

    f'(x)= (25-x^2)^1/2
    f'(x)= 1/2(25-x^2)^-1/2
    f'(x)= (-2x)1/2(25-x^2)^-1/2
    f'(x)= -x(25-x^2)^-1/2

    f'(4)=-4(25-4^2)^-1/2
    f'(4)=-4(25-(4^2))^-1/2
    f'(4)=-4(25-16)^-1/2
    f'(4)=-4(9)^-1/2
    f'(4)=-4(1/3)
    f'(4)=-4/3

    3=-4/3(4)+b
    3=-16/3+b
    0=(-16/3)/3+b
    0=-16/9+b
    -b=-16/9
    b=16/9

    y= -4/3x+16/9

    please tell me im done with this problem
     
  15. Nov 7, 2009 #14
    The second line above is wrong. What you should do is add 16/3 to both sides so you get b by itself. After that, the equation of the tangent line will be correct.
    You can check if you have the correct tangent line on a graphing calculator, or if you don't have one, look for one online.
     
  16. Nov 7, 2009 #15

    Mentallic

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    Just follow Bohrok's advice and you are done with this problem. However, judging by these few lines of working, it is evident that you are VERY innefficient in algebra manipulation and equation solving. Before you started calc you should've known how to solve equations properly, since you'll be doing so a lot throughout the course and it will be impossible for you to pass if you don't already know these prerequisites inside out.

    Sorry, but it's just how it is. You need to go back and learn algebra and equation solving (even more-so).

    For e.g.

    You've attempted to divide both sides by 3. Instead, you only divide one term on the right-hand side and ignore the b (probably because it made things look easier for you). Also, when you divided the left-hand side by 3, you resulted in the number 0? 3/3=1, 10/10=1, a/a=1 if a[itex]\neq[/itex]0.

    Good luck!
     
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