Approximating Change in x for y = 6 to y = 6.01 in y = 2 + (8/x)

[songoku]

Homework Statement

Given that y=2+\frac{8}{x}. Find the approximate change in x which will cause y to increase from 6 to 6.01

Homework Equations

\frac{\delta y}{\delta x}=\frac{dy}{dx}

The Attempt at a Solution

I did it with two ways :

1.
when y = 6 --> x = 2
when y = 6.01 --> x=\frac{800}{401}

approximate change in x = \frac{800}{401}-2 = -\frac{2}{401}


2.
\delta y = 0.01

\frac{\delta y}{\delta x}=\frac{dy}{dx}

\frac{\delta y}{\delta x}=\frac{-8}{x^2}~ \text{where x = 2}

\delta x = -\frac{1}{200}


My question is : Can I use the first method? I don't think I can but I don't know the reason why I may not use it.

 

[rock.freak667]

I think your first method assumes 'y' changed linearly with 'x'. But it's been a long time since I've some small changes. So it is always best to use the second method.


Also it is δy/δx ≈ dy/dx, it is only equal when you take the limit δx→0

 

[Dick]

Your first method gives you the EXACT value of the change. The second method gives you the APPROXIMATE value of the change. 1/200=0.005. 2/401 is about 0.00499. They are nearly equal, which is what you expect.

 

[songoku]

Hi rock.freak667 and Dick

Thanks a lot for you both !