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Math Help - 3 Questions from Calc 3

  1. #1
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    3 Questions from Calc 3

    So, I've been studying all day, but am really thrown off by these three practice problems. One is:

    1.) True/False: If f is a function of one variable and u=f(x-y), then uxx=uyy. (those x and y's are subscripts)

    My question is why would a function of one variable even have partial derivatives.

    2.) True/False: if w=(sin(y))e^x, x=s-t, y=st, then wxt (the x and t are subscript) at s=1, t=1, is 3.

    Is this some form of chain rule? I don't understand how the x=s-t changes this.

    3.) Use the linear approximation or differentials to find an approximation to [26.98]^1/3 times [36.04]1/2

    There are absolutely NO problems in the book like this, and we never discussed it, so I have NO idea how to approach this.

    Please, anyone answer as many as you can. I'm really frustrated and have no idea how to approach these problems. Thank you.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by GreenMachine View Post
    So, I've been studying all day, but am really thrown off by these three practice problems. One is:

    1.) True/False: If f is a function of one variable and u=f(x-y), then uxx=uyy. (those x and y's are subscripts)

    My question is why would a function of one variable even have partial derivatives.

    2.) True/False: if w=(sin(y))e^x, x=s-t, y=st, then wxt (the x and t are subscript) at s=1, t=1, is 3.

    Is this some form of chain rule? I don't understand how the x=s-t changes this.

    3.) Use the linear approximation or differentials to find an approximation to [26.98]^1/3 times [36.04]1/2

    There are absolutely NO problems in the book like this, and we never discussed it, so I have NO idea how to approach this.

    Please, anyone answer as many as you can. I'm really frustrated and have no idea how to approach these problems. Thank you.
    If we have z=f(x,y) and x=x(s,t) and y=y(s,t)

    Then

    \frac{\partial{f(x,y)}}{\partial{t}}=\frac{\partia  l{z}}{\partial{x}}\cdot\frac{\partial{x}}{\partial  {t}}+\frac{\partial{z}}{\partial{y}}\cdot\frac{\pa  rtial{y}}{\partial{t}}
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  3. #3
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    Actually, that doesn't have to due with t being the second time your deriving it. That just gives the first partial derivative with respect to t.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by GreenMachine View Post
    Actually, that doesn't have to due with t being the second time your deriving it. That just gives the first partial derivative with respect to t.
    Just do it twice my friend
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