Question

Use algebra to show that the distance of the kth segment can be rewritten as$\sqrt{1+{\left(\frac{\Delta y_k}{\Delta x}\right)}^2}\Delta x$.

Short answer

The $kth$segment can be written as$\sqrt{1+{\left(\frac{\Delta y_k}{\Delta x}\right)}^2}\Delta x$.

Step-by-step solution

Step 1. Given Information.

The given expression is$\sqrt{(\Delta x{)}^2+{\left(\Delta y_k\right)}^2}$.

Step 2. Conclusion.

The expression can be written as,

$$\begin{gathered} \sqrt{(\Delta x{)}^2+{\left(\Delta y_k\right)}^2} \\ =\frac{∆x}{∆x}\sqrt{(\Delta x{)}^2+{\left(\Delta y_k\right)}^2} \\ =∆x\sqrt{\frac{(\Delta x{)}^2+{\left(\Delta y_k\right)}^2}{{(∆x)}^2}} \\ =∆x\sqrt{\frac{{(∆x)}^2}{{(∆x)}^2}+\frac{{\left(\Delta y_k\right)}^2}{(\Delta x{)}^2}} \\ =∆x\sqrt{1+\frac{{\left(\Delta y_k\right)}^2}{(\Delta x{)}^2}} \end{gathered}$$