Using Differentials

**Bashyboy** · May 14th 2012, 01:01 PM

The problem is to estimate $y + \Delta{y}$ given that $y = x^2 - 5x$ , x = 4, and $\Delta{x} = 0.01$

I have tried this and found that the approximate change in y is dy = 0.03 and the actual change in y is 0.0301, but I am not sure what to do with this information.

**skeeter** · May 14th 2012, 02:19 PM

$y = x^2 - 5x$

$dy = (2x-5)dx$

at $x = 4$ , $y = -4$ and $\Delta y \approx dy = 0.03$

$y + \Delta y \approx y + dy = -4 + 0.03 = -3.97$

(note that y(4.01) = -3.9699 for reference)

**Bashyboy** · May 14th 2012, 03:49 PM

Oh, I see. I wonder why they didn't write $y(4) + \Delta{x}$ or something to that effect; it would have been a little easier for a simpleton like me.

**HallsofIvy** · May 14th 2012, 05:05 PM

Telling you that "x= 4" does tell you that "y" refers to "y(4)".

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