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What is this question asking?

  1. Apr 16, 2016 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the curve for which y'' = 8 if the curve is tangent to the line y = 11x at (2, 22).

    2. Relevant equations

    ?

    3. The attempt at a solution

    y' = 8x + c1

    y = 4x2 + c1x + c2

    What exactly is the question asking me to do, especially with the y = 11x bit? ∫ydx?
     
  2. jcsd
  3. Apr 16, 2016 #2

    LCKurtz

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    It wants your curve to be tangent at (2,22). That means it must pass through that point and have the same slope there as the line.
     
  4. Apr 16, 2016 #3
    But doesn't tangent mean derivative in this case? My curve should have a slope of 11 at (2,22), fine; how does that apply to y'' and y' and y as I said? My curve would be y, right, because y'' = 8 is the double derivative of that? How to solve it with two different constants, then?
     
  5. Apr 16, 2016 #4

    LCKurtz

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    You have two constants with which you can make the curve agree with point and slope.
     
  6. Apr 16, 2016 #5
    So y = 4x2 + c1x + c2 = 11? Without knowing the slope for y', how can we solve for this?

    I mean, so far, I've got -4x2 - c1x + 11 = c2, but now what?
     
  7. Apr 17, 2016 #6

    LCKurtz

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    You have ##y = 4x^2 + c_1x + c_2## with two unknown constants. What do the constants have to be so that ##y(2) = 22## and ##y'(2) = 11##?
     
  8. Apr 17, 2016 #7
    Oh, y' = 11 as well? Because y = 11x is a line?
     
  9. Apr 17, 2016 #8

    LCKurtz

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    There's really nothing more to tell you without working it for you. Just do it.
     
  10. Apr 17, 2016 #9
    Got it. WileyPlus accepted my answer.

    So basically, being tangent to a line means y' = 11 at that x-value, so it's solvable. If someone else asks a similar question, answer like that, and it might just make a lot more sense to them.
     
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