Question

using differentials, find the approximate value of 1/(2.002)^2

Answer 1

Given value 1 / 2.002 ^2 is in the form y = 1/x^2

X=2

Δx = 0.002

dy/dx = d(1/x^2) /dx = d(x^-2) /dx = -2x^-3

Δy= (dy/dx) * Δx

= (-2x^-3) * 0.002

Here x =2,

Δy = -2* 1/8 * 2/1000

Δy = -0.0005

We know that Δy = f(x+ Δx) – f(x)

Δy = 1/ (x+ Δx)^2 – 1/x^2

-0.0005 = 1 / (2.0002^2) – 1/2^2

1 / (2.0002^2) = 0.25 – 0.0005 = 0.2495

Approximate value of 1 / (2.0002^2) is 0.2495