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LogicalApple

  • one year ago

Challenge (calculus)

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  1. LogicalApple
    • one year ago
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  2. LogicalApple
    • one year ago
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    For what value of b is the line y = 10x tangent to the curve y = e^(bx)?

  3. N00bstyle
    • one year ago
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    @hartnn, what are your thoughts about it?

  4. hartnn
    • one year ago
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    10 = be^(bx) and maybe to find point of intersection, 10x = e^(bx) still thinking further steps...

  5. LogicalApple
    • one year ago
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    Medal is awaiting .. . :)

  6. N00bstyle
    • one year ago
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    Haha, screw the medal :P

  7. LogicalApple
    • one year ago
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    Oh wait I wrote the wrong thing down. I meant to write down what was in the attachment... one sec.

  8. LogicalApple
    • one year ago
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    For what value of b is the line y = 5x tangent to the curve y = e^(bx)?

  9. LogicalApple
    • one year ago
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    Same logic though.

  10. N00bstyle
    • one year ago
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    yeah, idd, phew :P

  11. mathmate
    • one year ago
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    At the required point, e^(bx) has a slope of 10 => be^(bx) = 10. We can solve for b using newton's method.

  12. hartnn
    • one year ago
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    5 =be^(bx) 5x = e^bx ---> bx =1 ---->5x = e^1 x= e/5 b = 5/e

  13. LogicalApple
    • one year ago
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    You have demonstrated your ability. Medal earned!

  14. hartnn
    • one year ago
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    lol thanks :)

  15. LogicalApple
    • one year ago
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    Although I like the idea of using Newton's method.. Sometimes it doesn't work though.

  16. mathmate
    • one year ago
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    I agree. There are known strict conditions of convergence. We need to get a close starting point. In any case, we can solve for b=5/e here, but still you need to calculate x one way or another.

  17. N00bstyle
    • one year ago
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    Good question btw, I like. Now back to drinking beer.

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