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Help me convert this relationship into a linear equation

  1. Mar 7, 2010 #1
    Is there anyway I can convert the attached equation into a linear relationship so that 'a' is the gradient of the straight line. 't' and 'h' are both variables. It doesn't matter exactly what the x and y values are. I just need 'a' to be the gradient of the straight line. I tried lots of different things but couldn't manage it. I would appreciate it very much if anybody can help me with this.

    Thanks a lot.
     

    Attached Files:

  2. jcsd
  3. Mar 7, 2010 #2
    I typed the formula because I'm not sure if the attachment can be seen. I hope its clear enough

    (at2)/2 = 0.25 - t(2ah)0.5
     
  4. Mar 7, 2010 #3
    You can use the quadratic formula to get t as a function of h, but the relationship is not
    linear.
     
  5. Mar 7, 2010 #4
    I know the equation I wrote is not linear but is it possible to make it linear? For example sometimes it is possible to take logs to make a relationship linear. I'm not saying this is the case here but are there any other such 'tricks' to derive a linear equation from the one I wrote? I just need the gradient to be 'a'. I hope I'm making myself clear.

    The reason I'm trying to do this is that I've got a series of known values for 't' and 'h' and I wanted calculate 'a' from the gradient of a straight line. Obviously if a linear relationship can be derived, the x and y values aren't going to be 't' or 'h' but some other value that can be given in terms of these variables.
     
    Last edited: Mar 7, 2010
  6. Mar 8, 2010 #5
    If what I'm trying to do doesn't make sense to anyone, then what do you think would be the best way of calculating 'a', from the given values of 'h' and 't'?
     
  7. Mar 8, 2010 #6

    Mentallic

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    Homework Helper

    When you took the logarithm of an equation such as y=ex to create a linear equation, you were supposed to be taking the logarithm of BOTH sides, so basically you have ln(y)=x. This means that you can have a linear equation (even though this expression is still the same as y=ex) if you plot the function on axes that are not linear. You can have the usual x axis, but you'll have to change the y-axis to ln(y).

    i.e. at the point y=1, you should instead have ln(y)=0 on your axis. At [itex]y=e\approx 2.718[/itex] you should instead have y=1. At y=e2, y=2 etc.

    OR you can keep the y-axis the same, but change the x-axis to ex so at the usual x=1 point, instead have x=e etc.

    With a bit of ingenuity, you can apply the same concept to your problem.

    However, I'm unsure if this is what you're looking for.
     
  8. Mar 10, 2010 #7
    I am still struggling with this. Can someone at least show me how to solve the equation for 'a'?

    Thanks a lot.
     
  9. Mar 10, 2010 #8

    hotvette

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    Subtract 0.25 from both sides, then square both sides. You should end up with a quadratic equation in a.
     
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