Understanding sensitivity of a variable using derivatives

1. Oct 29, 2015

musicgold

Hi,

1. The problem statement, all variables and given/known data

I am trying to figure out K’s sensitivity to x and y. I wish to know which variable K is more sensitive to. Both x and y are positive decimals less than 1.

I want to be able to say a 1% increase in y results in a p% change in K. As the attached file shows, the derivatives are not helping me much here.

2. Relevant equations
See the attached file.

3. The attempt at a solution
See the attached file.

Attached Files:

• Sensitivity of K.pdf
File size:
109.3 KB
Views:
58
2. Oct 29, 2015

Ray Vickson

If you type up your work I will be happy to look at it. Otherwise, not (along with many other helpers in this forum).

3. Oct 29, 2015

Staff: Mentor

Calculate the differential of K, which is $dK = \frac{\partial K}{\partial x}dx + \frac{\partial K}{\partial y}dy$
To get a value for the differential, you will need values for x, y, dx, and dy.

BTW, everything you did in longhand and posted as a PDF can be done using LaTeX, and posted directly into this pane. We prefer to see the work here, rather than having to open a PDF or open a window on another site. Posting work as an image makes it difficult for us to insert comments at the precise point where an error appears. We have a LaTeX tutorial here -- https://www.physicsforums.com/help/latexhelp/

4. Oct 29, 2015

musicgold

Please see the problem in Latex form.

$K = \frac {1-y} {1-y-x}$ ---------------------- 1

differentiating K by y

$\frac {dK} {dy} = \frac {(-1). (1-y-x) - (1-y) . (-1)} { (1-y-x)^2 }$

$\frac {dK} {dy} = \frac {x} { 1 + y^2 +x^2 + 2xy -2x-2y }$ ---------------------- 2

differentiating eqn 1 by x

$\frac {dK} {dx} = \frac {0 - (1-y) . (-1)} { (1-y-x)^2 }$

$\frac {dK} {dx} = \frac {1-y} { 1 + y^2 +x^2 + 2xy -2x-2y }$ ---------------------- 3

5. Oct 29, 2015

Staff: Mentor

This is really a partial derivative, which in LaTeX looks like this: $\frac{\partial K} {\partial y}$, and renders like this $\frac {\partial K} {\partial y}$.
Also, there's really no point in expanding $(1 - y - x)^2$. Same comments for the work below.

Otherwise, both partials look fine.
Now plug things into the equation for the differential of K (dK) that I gave in post #3. Since you are holding x fixed, then dx = 0, but you'll need some values for x and y. If all you know is that x < 1 and y < 1, then you might be able to get an upper bound on the size of dK relative to a dy.

6. Oct 29, 2015

Ray Vickson

I don't think you can make any convincing general statement like "a $1\%$ change in $y$ produces a $p\%$ change in $K$". The effect of a $1\%$ change in $y$ depends very much on the base values of $x$ and $y$, and things become especially tricky when the denominator is small (close to vanishing, but not quite), because in that case the derivatives may present a very misleading picture of the actual changes for given $\Delta x$ and/or $\Delta y$.