
Find the square root of 0.02.
Answer
558.9k+ views
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.
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

