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Find a tangent line?

  1. Sep 1, 2010 #1
    Say I have a curve is called C: y=1287*x^-1.5

    Find a tangent line to the C, and the tangent line has to have a intercept of 150.

    This is not a homework, not at all.
  2. jcsd
  3. Sep 1, 2010 #2


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    What do you mean by has an intercept of 150, which axis?

    Students usually do so many of the same kinds of questions that they get accustomed to just shortening things like as you've done, because saying x-intercept all the time just gets repetitive. We don't do your homework though so we have no clue :tongue:
  4. Sep 2, 2010 #3
    the line is always defined by Y=aX+b right? b is the intercept. This is a problem my boss gave to me. I just don't know the calculus, please help.
  5. Sep 2, 2010 #4


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    Oh so you're looking for the y-axis intercept then. I don't understand why your boss would give you a question you cannot even start to answer :confused:

    Anyway, for a function y=axn where a and n are constants, the derivative is y'=anxn-1

    You will have to use the point-gradient formula for a line, which is
    y-y0=m(x-x0) where (x0,y0) is a point on the line and m is the gradient of the line.

    For one line we know the x and y intercept because it's what is required of us, mainly (0,150) and we don't know the gradient, so we will leave it as m.

    For the next equation we will use the other info we can determine which is the point of contact between the line and the curve, we don't know this point but since it lies on the curve we can say it is (x1,ax1n) where x1 is the x-coordinate we are looking for.

    Now from the derivative equation, it is telling us what the gradient of the tangent is at an x-coordinate on the curve, so we will substitute x1 for x since it is the same point, so we have the gradient of the curve is anx1n-1

    Setting up our second line equation, we have


    So you have two line equations, and you have a whole bunch of constants and stuff that you can throw in later (I prefer to keep any large ugly numbers out of equations so it's easier to see). You need to find out what x1 is because then you can find out the point of intercept between the line and curve, and then the gradient, which is finally what you need. Do you know how to solve equations simultaneously?
  6. Sep 2, 2010 #5
    Ok, I get it now, thanks. all the best
  7. Sep 2, 2010 #6


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    No problem. By the way, why has your boss asked you to solve this?
  8. Sep 2, 2010 #7


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    No, b is the y-intercept. The x-intercept is -b/a.

    Interesting! What kind of job do you have that your boss gives you such interesting problems?

    Not that peculiar. A boss has a problem he can't solve so he hands it to a lackey. Probably the boss's son's or daughter's homework!
  9. Sep 2, 2010 #8


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    Hahaha :approve:
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