What (approximately) is the square root of $26$?

What (approximately) is the square root of $26$?

Question

By making use of the relation $\delta y \approx \dfrac{dy}{dx} \times \delta x$, and without using tables, prove that $5.1$ is an approximate value of $\sqrt{26}$.

Suggestion

There is a convention in maths that $\delta y$ signifies ‘a small increase in $y$’.

We need to specify $y=f(x)$. Can we think of an appropriate $f(x)$, given that we’re trying to find $\sqrt{26}$?

A further suggestion

What happens if we choose $y = \sqrt{x}$?

A final suggestion

If we take $y = \sqrt{x}$, what happens if we then use $x = 25$ and $\delta x = 1$?

Solution

Let’s consider the function $y = \sqrt{x}$. We know that $y(25) = \sqrt{25} = 5$ and that $\frac{dy}{dx} = y' = \frac{1}{2} x^{-1/2} \implies y'(25) = \frac{1}{2\sqrt{25}} = \frac{1}{10}.$

Now using $\delta y \approx \dfrac{dy}{dx} \times \delta x$ at the point $x = 25, y = 5$, with $\delta x = 1$, we see that $\delta y = 0.1 \times 1 \implies y + \delta y = 5.1.$

Thus $\sqrt{26} = \sqrt{25+1} \approx 5.1.$

We can’t immediately say how accurate our approximation is here.

If we turn to our calculators, they tell us $\sqrt{26} = 5.0990\!\ldots$, and so our approximation is surprisingly accurate ($0.02$% out).

With a problem like this, such accuracy cannot always be guaranteed!