withiout-using-calculator-find-approximate-for-9-01-

Question Number 44291 by peter frank last updated on 25/Sep/18

w i t h i o u t u s i n g c a l c u l a t o r f i n d a p p r o x i m a t e f o r

\sqrt{9.01}

Commented by maxmathsup by imad last updated on 25/Sep/18

\sqrt{9, 01} = \sqrt{9 + 0, 01} = \sqrt{9 + 10^{- 2}} = 3 \sqrt{1 + \frac{10^{- 2}}{9}} b u t f o r x \in v (0)

\sqrt{1 + x} \sim 1 + \frac{x}{2} \Rightarrow \sqrt{1 + \frac{10^{- 2}}{9}} \sim 1 + \frac{10^{- 2}}{18} \Rightarrow \sqrt{9, 01} \sim 3 (1 + \frac{10^{- 2}}{18}) \Rightarrow

\sqrt{9, 01} \sim 3 + \frac{10^{- 2}}{6} \dots

Commented by Joel578 last updated on 26/Sep/18

Sir please explain \sqrt{1 + x} \sim 1 + \frac{x}{2}

Commented by Necxx last updated on 26/Sep/18

i t s l i n e a r a p p r o x i m a t i o n

Commented by abdo.msup.com last updated on 26/Sep/18

f (x) = f (0) + \frac{x}{1!} f^{^{'}} (0) + \frac{x^{2}}{2} ξ (x) w i t h

{l i m}_{x \to 0} ξ (x) = 0 \Rightarrow f (x) \sim f (0) + {x f}^{^{'}} (x) (x \to 0)

l e t t a k e f (x) = \sqrt{1 + x} \Rightarrow f (0) = 1 a n d

f^{^{'}} (x) = \frac{1}{2 \sqrt{x + 1}} \Rightarrow f^{^{'}} (0) = \frac{1}{2} \Rightarrow

f (x) \sim 1 + \frac{x}{2} .

Commented by maxmathsup by imad last updated on 26/Sep/18

f (x) \sim f (0) + x f^{^{'}} (0) (x \to 0)

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Sep/18

y = \sqrt{x}

\frac{d y}{d x} = \frac{1}{2 \sqrt{x}}

h e r e x = 9 x + △ x = 9.01 △ x = 0.01

y = \sqrt{9} = 3 y + △ y = t o f i n d

\frac{△ y}{△ x} \approx \frac{d y}{d x}

\frac{d y}{d x} = \frac{△ y}{△ x}

△ y = \frac{d y}{d x} △ x

△ y = \frac{1}{2 \sqrt{9}} \times 0.01 = \frac{0.01}{6} = \frac{1}{6} \times \frac{1}{100} \approx 0.1666 \times \frac{1}{100} = 0.0016666

\sqrt{9.01} = 3.0016666