Q. 10 - Calculus

Question

Use definite integrals to show that the graphs of $f\left(x\right)=\frac{1}{x} \mathrm{and} g\left(x\right)=-\frac{1}{x}$ have the same arc length on [1, 3]. (Hint: You do not have to solve the integrals!) Why does this equality make sense in terms of graphical transformations?

Step-by-Step Solution

Answer

The arc length of both the function is the same because the definite integral of both the function is the same. The equality makes sense in terms of transformation because the g(x) is the reflection of the graph of f(x) on the x-axis. The transformation of reflection doesn't change the arc length.

Step 1. Given Information.

The given functions are $f\left(x\right)=\frac{1}{x} \mathrm{and} g\left(x\right)=-\frac{1}{x}.$

Step 2. Use definite integrals.

The arc length of the definite integral is given by $I={\int }_a^b\sqrt{1+{\left(f\text{'}\left(x\right)\right)}^2}dx.$

So, the arc length of the function $f\left(x\right)=\frac{1}{x}, f\text{'}\left(x\right)=-\frac{1}{x^2}$ by using the definite integral is

 $$\begin{gathered} I={\int }_1^3\sqrt{1+{\left(-\frac{1}{x^2}\right)}^2}dx \\ I={\int }_1^3\sqrt{1+\frac{1}{x^4}}dx \end{gathered}$$

The arc length of the function $g\left(x\right)=-\frac{1}{x}, g\text{'}\left(x\right)=\frac{1}{x^2}$ by using the definite integral is

$$\begin{gathered} I={\int }_1^3\sqrt{1+{\left(\frac{1}{x^2}\right)}^2}dx \\ I={\int }_1^3\sqrt{1+\frac{1}{x^4}}dx \end{gathered}$$

Thus, we can depict that the arc length of both the function has the same. The equality makes sense in terms of transformation because the g(x) is the reflection of the graph of f(x) on the x-axis. The transformation of reflection doesn't change the arc length.