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Thread: Anyone know how to solve this?

  1. #1
    May 2018

    Anyone know how to solve this?

    Really struggling with this one :/ anyone have a clue how to solve this?
    Attached Thumbnails Attached Thumbnails Anyone know how to solve this?-20180510_104127.jpg  
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  2. #2
    MHF Contributor

    Apr 2005

    Re: Anyone know how to solve this?

    Yes, I know how to solve it. The more important question, I suspect, is "do you know how to solve it"? You at given $\displaystyle y= 62.5\left(e^{x/130}+ e^{-x/130}\right)- 1.22$ and are asked to find "length AO" which is just the height at x= 0. Surely you can calculate $\displaystyle y= 62.5\left(e^{0/130}+ e^{-0/130}\right)- 1.22$? You are also asked to find the length of BC which is just the height at x= 70. That is $\displaystyle y= 62.5\left(e^{70/130}+ e^{-70/130}\right)- 1.22$. You are then asked to find the length of ED. You should be able to see that, by symmetry, this is the same as the length of BC.

    Finally, you are asked to find the tangent line to the curve at x= 50. It says that you can do that either "by use of differential calculus or by use of a scale drawing". So the new question is "Do you know differential calculus"? If you don't then you will have to use a scale drawing. Draw the given curve, especially region around x= 50 as accurately as you can, then try to draw the tangent line at x= 50. When I was in secondary school, I learned a cute method to do that: take a small mirror (what women might call a "purse mirror") and hold across the curve at x= 50. Looking at the curve and its reflection in the mirror you will probably see a sharp "corner" at the point the curve appears to enter the mirror. Rotate the mirror (about the point x= 50 on the curve) until the curve appears to go "smoothly" into its image in the mirror. Holding the mirror there, use it edge as a straight-edge to draw the line perpendicular to the curve. Now, do the same thing to draw a second line perpendicular to that line and so tangent to the curve! Extend that line until you can identify the coordinates of two points on the line and use those to determine the coefficients a and b in y= ax+ b.

    If you do know differential calculus, you don't have to do all that! The coefficient b, in y= ax+ b, is just y(50), which you can calculate from the given function, and a is y'(50), the derivative of that function evaluated at x= 50. Do you know how to find the derivative of $\displaystyle e^{x/130}$?
    Last edited by HallsofIvy; May 10th 2018 at 03:47 AM.
    Thanks from topsquark
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