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

This question was addressed to me in an interview for internship.

My question is taken from Partial Differentiation topic in division Partial Differentiation of Engineering Mathematics

Answer 1

The correct option 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=\(\frac{∂f}{∂x} dx+\frac{∂f}{∂y} dy+\frac{∂f}{∂z} dz=\frac{(xdx+ydy+zdz)}{f}=\frac{-0.02+0.02-0.12}{3}\)= -0.04

Hence,

\([0.98^2+2.01^2+1.94^2]^{\frac{1}{2}}\)=f(1,2,2)+df=3-0.04=2.96