Question

Find the approximate value of [0.98^2+2.01^2+1.94^2 ]^(^1⁄2).

(a) 1.96

(b) 2.96

(c) 0.04

(d) -0.04

I had been asked this question during an interview.

My question is taken from Errors and Approximations topic in section Partial Differentiation of Engineering Mathematics

Answer 1

The correct choice is (b) 2.96

For explanation I would say: Let f(x,y,z) = (x^2+y^2+z^2 )^(^1⁄2)  ……………..(1)

Hence, x = 1, y = 2, z = 2 so that, dx = -0.02, dy = 0.01, dz = -0.06

From (1),

^∂f⁄∂x = ^x⁄f

^∂f⁄∂y = ^y⁄f

^∂f⁄∂z = ^z⁄f

And df =  ^∂f⁄∂x dx + ^∂f⁄∂y dy + ^∂f⁄∂z dz = ((xdx + ydy + zdz))/f = (-0.02 + 0.02 – 0.12)/3 = -0.04

Hence,

[0.98^2+2.01^2+1.94^2 ]^(^1⁄2) = f(1, 2, 2) + df = 3-0.04 = 2.96.