Recent content by Howard Fox

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    Using hyperbolic substitution to solve an integral

    It was dx but then we substituted it to du, no?
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    Using hyperbolic substitution to solve an integral

    Okay, went back to my textbook. This should be the way to simplify the denominator, right?
  3. H

    Using hyperbolic substitution to solve an integral

    / Just take the radical of the values like this?
  4. H

    Using hyperbolic substitution to solve an integral

    Raising both denominator and numerator to the power of two?
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    Using hyperbolic substitution to solve an integral

    So the final integral should look something like this? This does seem very complicated to solve though!
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    Using hyperbolic substitution to solve an integral

    I really don't know. I have trouble reconciling the two parts. I guess we could substitute x with asinh?
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    Using hyperbolic substitution to solve an integral

    Okay, I tried to correct the previous mistakes, Hope this one is good? Thank you
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    Using hyperbolic substitution to solve an integral

    Homework Statement Homework Equations So the question is asking to solve an integral and to use the answer of that integral to find an additional integral. With part a, I don't have much problem, but then I don't know how to apply the answer from it to part b. I know I should subsitute all...
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    Equation for sum of torques on a ladder and minimum angle

    Homework Statement [/B] Homework Equations Drawing a diagram for the forces is the easy part. I am not sure I am doing the equation of the sum of the torques well. The Attempt at a Solution This is my attempt for the forces[/B] And this for the torques:
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    Initial and final power supplied by a person pushing a stone

    No, my calculator gave me the number 1.3 because it wasn't set to linear calculations. It gave me fractions!
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    Initial and final power supplied by a person pushing a stone

    Why doesn't my calculator give me the same value?
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    Initial and final power supplied by a person pushing a stone

    Wait, so 1,333 is not valid? Because if I use 1.3 the different values of t don't match. But if I use 1.333 they match.
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    Initial and final power supplied by a person pushing a stone

    Okay, I think I got it now. The thing is when I put 24/18 in my calculator it gives me 1.3, not 1.333. That's the origin of the confusion I guess
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