# How do you use linear approximation to the square root function to estimate square roots sqrt 4.400?

Feb 1, 2017

66.33, nearly.

#### Explanation:

Using binomial expansion,

$\sqrt{{a}^{2} + b} = a {\left(1 + \frac{b}{a} ^ 2\right)}^{\frac{1}{2}} = a \left(1 + \frac{1}{2} \frac{b}{a} ^ 2\right)$, nearly.

Here, choose a = 66 and b = 44.

Now,

$\sqrt{4400} = \sqrt{{66}^{2} + 44} = \sqrt{{66}^{2} + 44} = 66 {\left(1 + \frac{44}{66} ^ 2\right)}^{\frac{1}{2}}$

$66 \left(1 + \frac{1}{2} \left(\frac{44}{66} ^ 2\right)\right) = 66 \left(1 + \frac{1}{198}\right) = 66 + \frac{1}{3} = 66.33$, nearly.

Note that the error is of order 66(1/198)^2=O(.002), nearly.

Feb 1, 2017

#### Answer:

The linear approximation is a form of (or a way of thinking about) the equation of a tangent line.

#### Explanation:

For function, $f$, the linear approximation at $a$ is given by

$L \left(x\right) = f \left(a\right) + f ' \left(a\right) \left(x - a\right)$.

Note that the equation of the line tangent to the graph at $\left(a , f \left(a\right)\right)$ has slope $m = f ' \left(a\right)$ and it has point-slope equation

$y - f \left(a\right) = f ' \left(a\right) \left(x - a\right)$ Solving for $a$ gives us the linear approximation to $f$ at $a$.

We are not told what to use for $a$, but we want to eventually use $x = 4.4$.

We want $a$ to be a number "close to" 4.4 for which it is relatively easy to calculate $f \left(a\right)$.

In this case, we want a number close to $4.4$ whose square root is relatively easy to find.

The 'obvious' (once you see it) choice is $a = 4$.

So,

$f \left(x\right) = \sqrt{x}$ and $f \left(a\right) = f \left(4\right) = 2$.

$f ' \left(x\right) = \frac{1}{2 \sqrt{t} x}$ so $f ' \left(a\right) = f ' \left(4\right) = \frac{1}{4}$

$L \left(x\right) = 2 + \frac{1}{4} \left(x - 4\right)$.

And we finish with

$L \left(4.4\right) = 2 + \frac{1}{4} \left(4.4 - 4\right) = 2.1$

If there is a typo in the question

If the question should ask us to approximate $\sqrt{4 , 400}$ we should find a different $a$.

${60}^{2} = 3600$ and ${70}^{2} = 4900$ so, for a rough approximation, use $a = 70$.

A bit more arithmetic will show that

${66}^{2} = 4356$ and ${67}^{2} = 4489$.

Since $4356$ is close to $4400$, use $a = 4356$.

OR

To get a rough estimate use $\sqrt{4400} = \sqrt{4 \times 100 \times 11} = 20 \sqrt{11}$ and approximate $\sqrt{11}$ using $a = 9$.