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Equation of the Tangent Line to a Curve at a Given Point

  1. Mar 16, 2015 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    Find an equation of the tangent line to the curve at the given point.

    2. Relevant equations
    y=x¼ Point = (1,1)

    3. The attempt at a solution

    Derivative of y is y' = ¼x
    Plugging in the derivative to the equation for a line: y-1=¼(x-1).
    My book's answer is Y=¼x+¾, but I don't know how they got there. Expanding ¼(x-1) gives me a nasty fraction with a radical that doesn't seem to come out to equal ¼x+¾. I'm not sure what to do next here.
     
  2. jcsd
  3. Mar 16, 2015 #2

    Dick

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    That's not the equation of a line. Where did the exponent -3/4 come from? Review the form for the equation of a line again.
     
  4. Mar 16, 2015 #3

    berkeman

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    You use the y' equation to give you the slope of the line. What is the basic equation for a straight line in terms of its slope and y-intercept?

    Use the slope and the fact that the line goes through the point (1,1) to figure out the y-intercept to complete the equation...
     
  5. Mar 16, 2015 #4
    Last edited: Mar 16, 2015
  6. Mar 16, 2015 #5

    Drakkith

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    Since ¼x is the slope at any point, doesn't the -¾ go there?

    I believe it's: y=mx+b

    Plugging in values for x and y: 1 = ¼(1) + b.
    Simplifying: 1 = ¼ + b, so b = ¾.

    So are they simplifying ¼x to ¼x at the point since x = 1?
     
  7. Mar 16, 2015 #6
    There is a difference between "The Equation of a line" and "The Equation of a Tangent Line".
    Check the links I provided.
     
  8. Mar 16, 2015 #7

    berkeman

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    Not exactly... you find y' = 1/4 at the point (1,1), and then separately substitute that into the equation of a line y = mx + b, and solve for b. Since you know the line passes through (1,1) at its point of tangency, you can use that point to solve for b.

    There are likely many ways to solve this type of problem, but that's the way I prefer to do it.
     
  9. Mar 16, 2015 #8

    Drakkith

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    Okay, so I have a curve given by the first equation. They want the tangent line at a certain point on the curve. The derivative of the original equation gives me another function. At any x-value, this new function gives me the slope of the curve at that x-value, correct? So to find the slope of the curve at (1,1) I'd plug 1 in for X, which gives me 1/4 as the slope. Then, using (1,1), I get y-1 = 1/4(x-1), which simplifies to y= 1/4x + 3/4, which is the equation for the tangent line at that point on the curve.

    How's that look?
     
  10. Mar 16, 2015 #9

    berkeman

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    Looks good! :smile:
     
  11. Mar 16, 2015 #10

    berkeman

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    BTW, I'd probably just plug the point (1,1) into the y = mx + b equation, so 1 = (1/4)*1 + b to solve for b. You know the line goes through (1,1) and has the slope 1/4.
     
  12. Mar 16, 2015 #11

    Drakkith

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    Thanks for the help guys.

    Thanks, Berk. My main problems were that A.) I didn't really understand the overall problem and how everything worked together, and B.) for some reason I was thinking that x-3/4 = 1/3x1/4... Had to look up my exponent laws again to figure out that its one over the fourth root of x cubed, not three times the fourth root of x.
     
  13. Mar 17, 2015 #12
    Any time, sir.
    I hope people understand my way...
    Instead of noticing that I am new here and not everything I say is accurate....People will click the links to notice some other (Experienced human or resources to get all the needed info) and getting the answers themselves....That or I am too lazy to type the answer! :wink:
     
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