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Thread: Calc III help

  1. #1
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    Calc III help

    Hey, need some help with problem 3.

    i know how to take the gradient if its the form of grad-f(x, y), but this is just saying that u and v are functions...

    any help would be appreciated. thanks


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  2. #2
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    Re: Calc III help

    Helping on someone's homework sounds dodgy.
    A Fedora user

  3. #3
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    Re: Calc III help

    lol you could just point him in the right direction, or give him an example of a similar question. I can't help though only 16, haven't learned that yet lol.

  4. #4
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    Re: Calc III help

    I haven't a bloody clue either.
    A Fedora user

  5. #5
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    Re: Calc III help

    Karl's Math Help may be able to steer you in the right direction. It's what I did for a couple of my Calculus 1 quizzes. Here's his email address: helpwmath@karlscalculus.org
    The moon will illuminate my room and soon I'm consumed by my doom.

  6. #6
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    Re: Calc III help

    It says that u and v are functions of x and y so if it helps then you can write it out u(x,y) and v(x,y). They just wrote u and v to save space.

    For a)
    (Pretend what follows has partial derivatives since i'm too lazy to look for the right symbol)
    grad(u(x,y)) = du/dx*i + du/dy*j
    grad(v(x,y)) = dv/dx*i + dv/dy*j

    So what is grad(u+v)? Calculate as above, then rearrange some terms (group the u and v seperately) and go back from i+j terminology to grad terminology. You should end up with what is on the right of the property.

    Do the same for the b)-d).

  7. #7
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    Re: Calc III help

    First, I think there is a typo in 3b, I think. It really should read:
    Code:
    grad(uv) = u*grad(v) + v*grad(u)
    You should have learned that differentiation is linear, meaning it is distributive. That should solve 3a.

    Once you have that, you can solve 3b. First hold u constant, then v. Or look up a proof for a one-dimensional case, it should be identical.

    3c is the same as 3b: grad (u/v) = grad (u * 1/v). Apply 3b, then simplify.

    For 3d, start with expressing u^n = u*u^(n-1) and then apply 3b again. You'll probably end up with some recursive expression.

    Dealing with n-dimensional derivatives as if it were normal just takes some getting used to. In the end, you'll realize they all behave the same, whether f() is a function of one x or infinitely many.

    It gets extra fun when each variable can be an n-dimensional vector by itself.

  8. #8
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    Re: Calc III help

    Quote Originally Posted by Erdaron View Post
    First, I think there is a typo in 3b, I think. It really should read:
    Code:
    grad(uv) = u*grad(v) + v*grad(u)
    You should have learned that differentiation is linear, meaning it is distributive. That should solve 3a.

    Once you have that, you can solve 3b. First hold u constant, then v. Or look up a proof for a one-dimensional case, it should be identical.

    3c is the same as 3b: grad (u/v) = grad (u * 1/v). Apply 3b, then simplify.

    For 3d, start with expressing u^n = u*u^(n-1) and then apply 3b again. You'll probably end up with some recursive expression.

    Dealing with n-dimensional derivatives as if it were normal just takes some getting used to. In the end, you'll realize they all behave the same, whether f() is a function of one x or infinitely many.

    It gets extra fun when each variable can be an n-dimensional vector by itself.
    I have to agree about the typo. Now I feel silly for not seeing it.

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