1. ## Approximating Squareroot

Can someone please show me how to approximate the square root of 172, by using calculus. thank you

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

3. Hello, mzkmikey!

Approximate $\displaystyle \sqrt{172}$ using calculus.
I believe they want us to use differentials.

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

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

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

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

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

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

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

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

Therefore: .$\displaystyle f(172) \;\approx\;13 + \frac{3}{26} \;=\;\boxed{13.11538462}$

4. thank you