
How do you estimate the quantity using the linear approximation of ${{\left( 3.9 \right)}^{\dfrac{1}{2}}}$?
Answer
519k+ views
Hint: We will use the linear approximation formula to find the linear approximation of the given square root number. The linear approximation formula is given by $L\left( x \right)=f\left( a \right)+f'\left( a \right)\left( x-a \right)$ where a is the value we need to consider close to the given value.
Complete step-by-step solution:
We have been given a quantity ${{\left( 3.9 \right)}^{\dfrac{1}{2}}}$.
We have to estimate the given quantity using the linear approximation.
We know that the linear approximation of a function is given as $L\left( x \right)=f\left( a \right)+f'\left( a \right)\left( x-a \right)$.
Here in this question we have to estimate the function $f\left( 3.9 \right)=\sqrt{3.9}$
Now, first we need to choose a value closer to the given number which is the value of $a$.
Let us consider $a=4$ so we will get
$\begin{align}
& \Rightarrow f\left( a \right)=f\left( 4 \right) \\
& \Rightarrow f\left( 4 \right)=\sqrt{4} \\
& \Rightarrow f\left( 4 \right)=2 \\
\end{align}$
Now, we know that if $f\left( x \right)=\sqrt{x}$ then $f'\left( x \right)=\dfrac{1}{2\sqrt{x}}$
So we get $f'\left( 4 \right)=\dfrac{1}{2\sqrt{4}}$
Now, substituting the values in the formula of linear approximation we will get
$\Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{2\sqrt{4}}\left( 3.9-4 \right)$
Simplifying the above obtained equation we will get
$\begin{align}
& \Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{2\times 2}\left( -0.1 \right) \\
& \Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{4}\left( -0.1 \right) \\
& \Rightarrow L\left( 3.9 \right)=2+0.25\left( -0.1 \right) \\
& \Rightarrow L\left( 3.9 \right)=2-0.025 \\
& \Rightarrow L\left( 3.9 \right)=1.975 \\
\end{align}$
Hence we get ${{\left( 3.9 \right)}^{\dfrac{1}{2}}}\approx 1.975$
Note: The point to be noted is that always choose the value of $a$ closer to the given number such as the square root of that number can be easily calculated. The method of approximation is useful to find the value of a function at any point. We can easily find the value of a function at any particular point.
Complete step-by-step solution:
We have been given a quantity ${{\left( 3.9 \right)}^{\dfrac{1}{2}}}$.
We have to estimate the given quantity using the linear approximation.
We know that the linear approximation of a function is given as $L\left( x \right)=f\left( a \right)+f'\left( a \right)\left( x-a \right)$.
Here in this question we have to estimate the function $f\left( 3.9 \right)=\sqrt{3.9}$
Now, first we need to choose a value closer to the given number which is the value of $a$.
Let us consider $a=4$ so we will get
$\begin{align}
& \Rightarrow f\left( a \right)=f\left( 4 \right) \\
& \Rightarrow f\left( 4 \right)=\sqrt{4} \\
& \Rightarrow f\left( 4 \right)=2 \\
\end{align}$
Now, we know that if $f\left( x \right)=\sqrt{x}$ then $f'\left( x \right)=\dfrac{1}{2\sqrt{x}}$
So we get $f'\left( 4 \right)=\dfrac{1}{2\sqrt{4}}$
Now, substituting the values in the formula of linear approximation we will get
$\Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{2\sqrt{4}}\left( 3.9-4 \right)$
Simplifying the above obtained equation we will get
$\begin{align}
& \Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{2\times 2}\left( -0.1 \right) \\
& \Rightarrow L\left( 3.9 \right)=2+\dfrac{1}{4}\left( -0.1 \right) \\
& \Rightarrow L\left( 3.9 \right)=2+0.25\left( -0.1 \right) \\
& \Rightarrow L\left( 3.9 \right)=2-0.025 \\
& \Rightarrow L\left( 3.9 \right)=1.975 \\
\end{align}$
Hence we get ${{\left( 3.9 \right)}^{\dfrac{1}{2}}}\approx 1.975$
Note: The point to be noted is that always choose the value of $a$ closer to the given number such as the square root of that number can be easily calculated. The method of approximation is useful to find the value of a function at any point. We can easily find the value of a function at any particular point.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

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

The camels hump is made of which tissues a Skeletal class 11 biology CBSE

