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Math Help - Proving that a distance is always equal to 1

  1. #1
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    Proving that a distance is always equal to 1

    Hi guys, this is the second problem I have here in the math forums (and should be my last for a while, hopefully).

    I am given a question that says I need to prove that the distance between a and c is always equal to one. The question also says that line "l" is the tangent line to the graph of the function y = e^x (of course, at the point where (a,b). C is when that line touches the x-axis.

    The image below is what I made to try and make things clear for you.



    So, I think I have an idea of how to prove it. Should I be trying to do something with the pythagorean theorem where the vertical line is the dotted line from point (a,b) to point (a) on the x axis and the other leg of the triangle is from a to c. The a - to - c leg is what I'm looking for to equal one.

    x^2 + y^2 = c^2

    Where x is a-to-c. Thus ...

    x^2 = 1, so how do I prove this?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Lord Darkin View Post
    Hi guys, this is the second problem I have here in the math forums (and should be my last for a while, hopefully).

    I am given a question that says I need to prove that the distance between a and c is always equal to one. The question also says that line "l" is the tangent line to the graph of the function y = e^x (of course, at the point where (a,b). C is when that line touches the x-axis.

    The image below is what I made to try and make things clear for you.



    So, I think I have an idea of how to prove it. Should I be trying to do something with the pythagorean theorem where the vertical line is the dotted line from point (a,b) to point (a) on the x axis and the other leg of the triangle is from a to c. The a - to - c leg is what I'm looking for to equal one.

    x^2 + y^2 = c^2

    Where x is a-to-c. Thus ...

    x^2 = 1, so how do I prove this?
    I would approach it like this. The derivative of e^x at the point x=a is e^a. So the line tangent to e^x at (a,e^a) is given by e^a=e^a\cdot a+b, where b is the y-intercept. Solving the equation for b gives b=e^a(1-a).

    So the equation of your tangent is y=e^a x+e^a(1-a); in other words, y=e^a(x+1-a).

    From this equation, it's easy to see that it is equal to 0 only when x=a-1, which is exactly what you were trying to prove. \square
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  3. #3
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    Quote Originally Posted by redsoxfan325 View Post
    I would approach it like this. The derivative of e^x at the point x=a is e^a. So the line tangent to e^x at (a,e^a) is given by e^a=e^a\cdot a+b, where b is the y-intercept. Solving the equation for b gives b=e^a(1-a).

    So the equation of your tangent is y=e^a x+e^a(1-a); in other words, y=e^a(x+1-a).

    From this equation, it's easy to see that it is equal to 0 only when x=a-1, which is exactly what you were trying to prove. \square
    Thanks. I understand now. This site is awesome, hope I can donate once I get a credit card.
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  4. #4
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    By the way, did you use the point slope formula to get

    <br />
e^a=e^a\cdot a+b

    If not, how did you get that?
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  5. #5
    Super Member redsoxfan325's Avatar
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    It was worded sort of poorly by me. I was using the slope intercept equation for a line: y=mx+b. I plugged in e^a for y, e^a for m, and a for x, and solved for b.
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