Approximating Squareroot

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April 27th 2010, 07:34 AM
mzkmikey

Approximating Squareroot

Can someone please show me how to approximate the square root of 172, by using calculus. thank you
April 27th 2010, 07:56 AM
maddas

You can use Newton's method. Alternatively, write $\sqrt{\frac{172}{14^2}} = \frac1{14} \sqrt{172}$ and approximate $\sqrt{\frac{172}{14^2}}$ by using the Taylor series $\sqrt{x} = 1 + \frac12 (x-1) - \frac18(x-1)^2+\cdots$ which is valid for $|x-1|<1$ . You would get $\approx13.1166181$ by this method.
April 27th 2010, 08:21 AM
Soroban

Hello, mzkmikey!

Quote:

Approximate $\sqrt{172}$ using calculus.

I believe they want us to use differentials.

We have the function: . $f(x) \;=\;\sqrt{x} \;=\; x^{\frac{1}{2}}$

The differential is: . $df \;=\;\tfrac{1}{2}x^{-\frac{1}{2}}dx \;=\;\frac{dx}{2\sqrt{x}}$

Let $x = 169,\;dx = 3$

We know that: . $f(172) \;=\;f(169 + 3) \;=\;f(169) + \Delta x$

. . where $\Delta x$ is actual difference between $\sqrt{172}$ and $\sqrt{169}$

And we know:. . $f(172) \;=\;f(169+3) \;\approx\; f(169) + dx$ .[1]

. . where $dx$ is an approximation of $\Delta x.$

We have: . $\begin{Bmatrix}f(169) &=& \sqrt{169} &=& 13 \\ \\[-3mm]<br /> \dfrac{dx}{2\sqrt{x}} &=& \dfrac{3}{2\sqrt{169}} &=& \dfrac{3}{26} \end{Bmatrix}$

Therefore: . $f(172) \;\approx\;13 + \frac{3}{26} \;=\;\boxed{13.11538462}$
April 27th 2010, 11:15 AM
mzkmikey

thank you