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- Thread starter musicgold
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- #2

HallsofIvy

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No, it isn't.

- #3

SteamKing

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What happens to the exponent when we differentiate?

- #4

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No, it isn't.

Are you sure?

- #5

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What happens to the exponent when we differentiate?

Oh! How about this?

dI/dS = (- N U/ D

- #6

SteamKing

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That looks better.

- #7

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That looks better.

Thanks SteamKing. I am still struggling with one question:

The reason behind this calculation was to understand the sensitivity of I to S. I wanted to know how much I changes for a 1% change in in S. It seems the differentiation doesn't answer that.

What should I do to understand the sensitivity?

Thanks.

- #8

Mark44

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It's much simpler to write the equation in this form:Oh! How about this?

dI/dS = (- N U/ D^{2}) / ( S U/ D + 1)^{2}

I = N(US + D)

Differentiating with respect to S yields

dI/dS = -N * (US + D)

Also, to find the sensitivity of I to small changes in S, write the equation above using differentials.

dI = ##\frac{-NU}{(US + D)^2} dS ##

For small changes in S, you can approximate dS by ΔS and dI by ΔI.

- #9

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Also, to find the sensitivity of I to small changes in S, write the equation above using differentials.

dI = ##\frac{-NU}{(US + D)^2} dS ##

For small changes in S, you can approximate dS by ΔS and dI by ΔI.

How do I use this equation to find the change in I for a 1% change in S?

- #10

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- #11

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I am still a bit confused. Please see the attached file.

N=100, U= 110, S = 9 and D=120

With these values I = 9.01%

Also these values give me dI/dS = -0.009

Now, for dS = 1%, I get dI= -0.01%. However, this result doesn't make any sense.

For example, if I actually increase S by 1% from 9.00 to 9.09, I changes from 9.01% to 8.93%, a decline of -0.08%, much higher than the 0.01% indicated above.

What am I missing?

- #12

Mark44

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You didn't show what you did to get your value of dI, so it's hard to say what you're missing. Using the formula I wrote in post #8, I get dI ≈ -8.93 X 10^{-5}.

I used the values you show in the PDF:

N = 100

U =110

S = 9

D = 120

ΔS = .01

So dI ≈ ## \frac{-100(110)}{(9 * 110 + 120)^2} * .01##

Edit: Corrections to the above

When S increases by 1%, that means that ΔS = .09, not .01.

This results in a value for dI ≈ ## \frac{-100(110)}{(9 * 110 + 120)^2} * .09 ≈ -8.035 X 10^{-4}##

I used the values you show in the PDF:

N = 100

U =110

S = 9

D = 120

ΔS = .01

So dI ≈ ## \frac{-100(110)}{(9 * 110 + 120)^2} * .01##

Edit: Corrections to the above

When S increases by 1%, that means that ΔS = .09, not .01.

This results in a value for dI ≈ ## \frac{-100(110)}{(9 * 110 + 120)^2} * .09 ≈ -8.035 X 10^{-4}##

Last edited:

- #13

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You didn't show what you did to get your value of dI, so it's hard to say what you're missing. Using the formula I wrote in post #8, I get dI ≈ -8.93 X 10^{-5}.

Sorry about that. You are right, that is how I also calculated it dI= -0.0000893

See the attached Excel file.

Last edited:

- #14

Mark44

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If S increases by 1%, ΔS = .09, so the corrected value for ΔI is about -.0008035.

Calculating I directly with S = 9 results in .09009 (approx.)

Calculating I directly with S = 9.09 results in .08929

So I has decreased (making ΔI negative), and by direct calculation we see that ΔI = -.00080. This agrees reasonably well with the value obtained using differentials.

Using the values I showed in post 12, I get dI ≈ -0.0000893

Calculating I directly with S = 9, I get I ≈ .09009

Increasing S by 1% (to 9.09), I get I ≈ .089293

- #15

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Got it. Thank you Mark44. With that correction, it is clear to me now.

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