0
$\begingroup$

We have $$W = 10 + 0.895m -12.33c^{0.18} + 12.33c^{0.18}$$

we calculate the differential of m, which is $0.895$ and the differential of $c$, which is something like $$-12.33/c^{0.82} + 12.33/c^{0.82}$$ and what then?

what do we do with $c$. I am confused...

  • 2
    The way you've written it, the terms involving $c$ cancel out.2012-11-10

1 Answers 1

1

I'm guessing that you have bounds on the errors in your measurements of $m$ and $c$, and you want to approximate the error in your estimate of $W$.

Use this:

\begin{equation} W(m + \Delta m, c + \Delta c) \approx W(m,c) + \frac{\partial W(m,c)}{\partial m} \Delta m + \frac{\partial W(m,c)}{\partial c} \Delta c. \end{equation}

This equation lets you see how large the quantity $|W(m + \Delta m, c + \Delta c) - W(m,c)|$ might be.

  • 0
    can you explain what that equation is? the 2 rational terms are they derivatives?2012-11-10
  • 0
    Yes, they are the partial derivatives of $W$ with respect to $m$ and $c$. For example, $\frac{\partial W(m,c)}{\partial m} = .895$ in your problem.2012-11-10
  • 0
    why W(mc)?? someone told me you only need to compute the partial derivative of c and m2012-11-10
  • 0
    I wrote $W(m,c)$, not $W(mc)$. $W$ is a function of $m$ and $c$.2012-11-10
  • 0
    in my book there iw no w(m,c), it's just the addition of two partial derivatives2012-11-11
  • 0
    Bring $w(m, c) $ over to the left hand side, and you see the sum of two partials approximates the error in $w$.2012-11-11