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Last updated date: 13th Jun 2024
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Hint: $0.02=\dfrac{2}{100}\Rightarrow \sqrt{0.02}=\sqrt{\dfrac{2}{100}}=\dfrac{\sqrt{2}}{10}$ .
The nearest perfect square numbers to 2 are 1 and 4.
 $\sqrt{1}<\sqrt{2}<\sqrt{4}$ ⇒ $1<\sqrt{2}<2$
We can use either the method of long division, binomial expansion or calculus to find the square root of 2.

Complete step-by-step answer:
Since, $0.01<0.02<0.04$ , we can say that $\sqrt{0.01}<\sqrt{0.02}<\sqrt{0.04}$ or $0.1<\sqrt{0.02}<0.2$ .
Let us use differentiation (calculus) to find the value of $\sqrt{0.02}$ .
Let's say $y=f(x)=\sqrt{x}$ is a function of x.
For a change of Δx in the value of x, let's say that the value of y changes by Δy.
⇒ y + Δy = f(x + Δx)
We know that $f\left( 0.01 \right)=\sqrt{0.01}=1$ .
∴ f(0.02) = f(0.01 + 0.01) which means that the change Δx = 0.01.
We also know that for small values of Δx and Δy, $\dfrac{\Delta y}{\Delta x}\approx \dfrac{dy}{dx}$ .
Now, $\dfrac{\Delta y}{\Delta x}=\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \sqrt{x} \right)$ .
Using the definition that roots are fractional powers ( $ {{x}^{\dfrac{p}{q}}}=\sqrt[q]{{{x}^{p}}}$ ):
 $\dfrac{\Delta y}{\Delta x}=\dfrac{d}{dx}\left( \sqrt{x} \right)=\dfrac{d}{dx}\left( {{x}^{\dfrac{1}{2}}} \right)$
And using the formula of derivatives $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}} $ , we get:
⇒ $\dfrac{\Delta y}{\Delta x}=\dfrac{1}{2}{{x}^{\left( \dfrac{1}{2}-1 \right)}}=\dfrac{1}{2}{{x}^{\dfrac{-1}{2}}}$
Substituting x = 0.01 and Δx = 0.01, we get:
 $\dfrac{\Delta y}{0.01}=\dfrac{1}{2}{{(0.01)}^{\dfrac{-1}{2}}}$
Using the meaning of negative powers $ {{a}^{-x}}=\dfrac{1}{{{a}^{x}}}$ , we get:
⇒ $\dfrac{\Delta y}{0.01}=\dfrac{1}{2}\times \dfrac{1}{\sqrt{0.01}}$
⇒ $\Delta y=\dfrac{1}{2}\times \dfrac{1}{0.1}\times 0.01$
⇒ Δy = 0.05
Finally, since y + Δy = f(x + Δx), we can say that:
 $\sqrt{0.02}=f(0.02)=f(0.01+0.01)=f(0.01)+\Delta y$
Substituting the values f(0.01) = 0.1 and Δy = 0.05, we get:
 $\sqrt{0.02}=0.1+0.05=0.15$ .
Hence, the value of the square root of 0.02 is approximately 0.15.

Note: The smaller the value of Δx, the better the approximation.
This process can be repeated infinitely many times to get a closer value of the function at a given point.