Use differentials (linear approximation) to approximate #25/(3.1^2+3.8^2)#?

1 Answer
Ultrilliam
May 17, 2018

See below

Explanation:

You want to evaluate this [in a really roundabout way :-) ] :

#25/(3.1^2+3.8^2)#

Well:

#25/(3^2+4^2) = 1#

So you can define:

  • #f(x,y) = 25/(x^2 + y^2) qquad implies f(3,4) = 1#

Differentials next:

  • #df = f_x dx + f_y dy#

# =25( - (2x)/(x^2 + y^2)^2 dx - (2y)/(x^2 + y^2)^2 dy) #

# = -50 ( (x\ dx+ y\ dy )/(x^2 + y^2)^2 )#

Plug in the numbers [ which means we are losing the guaranteed exactness of the differentials , but the outcome will be as accurate as a linear expansion <- the whole point of this exercise, I guess ]:

#df = -50( (3( 0.1) + 4(- 0.2) )/(25)^2) = 0.04#

And so as an approximation:

#25/(3.1^2+3.8^2) = 1 + 0.04 = 1.04#

Your calculator will tell you that:

#25/(3.1^2+3.8^2) = 1.0395.......#

Worked quite well in the end

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