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Derivatives and Tangent Help

  1. Mar 8, 2006 #1
    I am stuck on two question and have no idea how to solve them:

    1) Find the equation of the tangent line to the graph of y=-3x^2 + 4x +18 at the point where x = -1

    2) At what points on the curve y=x^3 - 9x^2 +29x +5 is the tangent line perpendicular to x+2y-7=0
     
    Last edited: Mar 8, 2006
  2. jcsd
  3. Mar 8, 2006 #2

    Hootenanny

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    (1)start by finding [itex]\frac{dy}{dx}[/itex] at that point.
     
    Last edited: Mar 8, 2006
  4. Mar 8, 2006 #3
    for 1) I have y'=-6x+4

    and for 2) I have y'=3x^2 -18x +29
     
  5. Mar 8, 2006 #4
    b) Remember that the slope of the line perpendicular is the negative multiplicative inverse of the original line. Hope this helps...

    Beau
     
  6. Mar 8, 2006 #5
    Your derivatives are correct!
     
  7. Mar 8, 2006 #6
    I can get the derivatives but I don't know what to do afterwards...
     
  8. Mar 8, 2006 #7

    Hootenanny

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    (1) And you know that at this point x =1, so you can obtain the gradient. Then sub x = 1 into the origonal equation to obtain the y coordinate. You know have the gradient of the tangent and a point on it. You can then use:
    [tex]y - y_1 = m(x - x_1 )[/tex]
    to obtain an equation for your tangent.
     
  9. Mar 8, 2006 #8
    (Meant to say x=-1 but the idea is the same. Working on it now)
     
  10. Mar 8, 2006 #9
    okay I did y-y1=m(x-x1)

    2-6x+4=m(-1-x)

    x(-6+6) / x= -m
     
  11. Mar 8, 2006 #10
    That doesn't seem right though...
     
  12. Mar 8, 2006 #11

    VietDao29

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    The tangent line equation to the graph of f(x) at the point (x0, f(x0)), is:
    y - f(x0) = f'(x0) (x - x0)
    Now that you have f'(x), what you have to do is to find f'(x0).
    In (1), if x0 = -1, then what's f(x0)? What's f'(x0)?
    Can you go from here? :)
     
  13. Mar 8, 2006 #12

    Hootenanny

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    No, you substituted the values in wrong. [itex]x_1[/itex] and [itex]y_1[/itex] are the coordinated of the point. m is the gradient. You should end up with an eqaution in the form;
    [tex]ay = bx + c[/tex]
     
  14. Mar 8, 2006 #13

    Would f'(xo) be the derivative? or -6x+4, so I would substitute

    [-6(1)+4]-[6x+4] / x - 1 ?
     
  15. Mar 8, 2006 #14

    VietDao29

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    No, if f'(x) = -6x + 4, then you can find f'(-1) by simply replacing every x in f'(x) by -1, like this:
    f'(-1) = -6(-1) + 4 = 6 + 4 = 10.
    I think you should skim through your book again to understand the main concept before trying to solve problems. And also try to understand some examples the book gives you.
    Can you go from here? :)
     
  16. Mar 8, 2006 #15
    The book didn't explain it very well which is why I am so confused >_<. I usually understand something through an example question but there weren't any in the book and there weren't any on the web >_<. I'll keep trying though.
     
  17. Mar 8, 2006 #16

    Hootenanny

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    Here's my working to see if it helps:
    [tex]\frac{dy}{dx} -3x^2 + 4x + 18 \;\; dx = 4 - 6x[/tex]
    At the point x = -1 the gradient of the tangent to the give is given by;
    [tex] \frac{dy}{dx} = 4 - 6 \times -1 = 10[/tex]
    At x = -1, finding y:
    [tex]-3\times -1^2 + 4 \times - 1 + 18 = 11[/tex]
    Therefore the co-ordinates of the point of intersection of the tangent and curve is (-1,11). Using;
    [tex]y - y_1 = m(x - x_1 )[/tex]
    gives;
    [tex]y - 11 = 10(x + 1) \Rightarrow y = 10x + 22[/tex]

    Hope this helps.

    p.s. I'll apologise for any stupid errors in advance.
     
    Last edited: Mar 8, 2006
  18. Mar 8, 2006 #17
    Ooooooo I see what I did wrong! Thank you!
    Also, any help for the 2nd one?
     
  19. Mar 8, 2006 #18
    I just wish to analyze the solution. Makes it easier to understand ._.
     
  20. Mar 8, 2006 #19

    Hootenanny

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    (2) Okay, the product of the gradients of two lines which are perpendicular is always -1. i.e:
    [tex]m_1 \cdot m_2 = -1[/tex]
    So you need to find the point on the line where the gradient of the tangent times the gradient of the line equals -1. You will need to re-arrange the equation of the line into the form [itex]y = mx + c[/itex] where [itex]m[/itex] is the gradient.

    Hope this helps :smile:
     
  21. Mar 9, 2006 #20

    VietDao29

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    What do you mean by this????
    Hootenanny gives you an example by solving the problem (1). Now, it's your turn to do the problem (2), which can be done in the same way as problem (1).
    We do not give out COMPLETE SOLUTIONS here, at PhysicsForums. We do, however give you examples, and explain things to you!!! :grumpy: :grumpy: :grumpy:
     
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