Q. 12 - Calculus

Question

Do you think that the function $f\left(x\right)=2x^2$ has twice the arc length of $g\left(x\right)=x^2$ on the interval [0, 3]? Why or why not?

Step-by-Step Solution

Answer

No, the function $f\left(x\right)=2x^2$ has not twice the arc length of $g\left(x\right)=x^2$ on the interval $\left[0,3\right]$ because $\sqrt{1+16x^2}\ne 2\sqrt{1+4x^2}.$

Step 1. Given Information.

The given functions are $f\left(x\right)=2x^2 \mathrm{and} g\left(x\right)=x^2.$

Step 2. Explanation.

To find that the function $f\left(x\right)=2x^2$ has twice the arc length of $g\left(x\right)=x^2$, we will use the formula of arc length of the definite integral which is ${\int }_a^b\sqrt{1+{(f\text{'}(x\left)\right)}^2}dx.$

Now, let $L_1$ for the function  $f\left(x\right)=2x^2, so f\text{'}\left(x\right)=4x:$

$$\begin{gathered} L_1={\int }_0^3\sqrt{1+(4x{)}^2}dx \\ {L}_1={\int }_0^3\sqrt{1+16x^2}dx \end{gathered}$$

Let $L_2$ for the function $g\left(x\right)=x^2, so g\text{'}\left(x\right)=2x:$

$$\begin{gathered} L_2={\int }_0^3\sqrt{1+(2x{)}^2}dx \\ {L}_2={\int }_0^3\sqrt{1+4x^2}dx \end{gathered}$$

Thus, we can depict that $L_1\ne 2L_2.$