Q. 84 - Calculus

Question

Rajini bends a metal rod into a curve that lines up with the graph of $y=4x^{3/2}$ from $x=0 \mathrm{to} x=2$, as shown here. Given that the length of a curve $y=f\left(x\right)$ from $x=a \mathrm{to} x=b$ can be calculated with the formula ${\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}$, find the length of the thin metal rod.

Step-by-Step Solution

Answer

The length of the thin metal rod is $11.53$.

Step 1. Given Information

Rajini bends a metal rod into a curve that lines up with the graph of $y=4x^{3/2}$ from $x=0 \mathrm{to} x=2$, as shown here. Given that the length of a curve $y=f\left(x\right)$ from$x=a \mathrm{to} x=b$ can be calculated with the formula ${\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}$, find the length of the thin metal rod.

Step 2. We have to calculate the formula ∫ a b 1 + (   f ' ( x ) ) 2 d x .

To calculate the formula we firstly find the value of $f\text{'}\left(x\right)$.

We know that

$$\begin{gathered} y=f\left(x\right) \\ f\left(x\right)=4x^{3/2} \\ \frac{df\left(x\right)}{dx}=4\frac{\mathrm{d}}{\mathrm{d}x}{x}^{3/2} \\ \frac{df\left(x\right)}{dx}=4·\frac{3}{2}·x^{1/2} \\ \frac{df\left(x\right)}{dx}=6x^{1/2} \\ f\text{'}\left(x\right)=6\sqrt{x} \end{gathered}$$

Step 3. Now putting the value of f ' ( x ) on the formula

$$\begin{gathered} {\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}={\int }_0^2\sqrt{1+{\left(6x^{1/2}\right)}^2}dx \\ {\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}={\int }_0^2\sqrt{1+36x}dx \end{gathered}$$

Let

$$\begin{gathered} u=1+36x \\ \frac{du}{dx}=36 \\ du=36dx \\ \frac{1}{36}du=dx \end{gathered}$$

Step 4. This substitution changes the integral into

$$\begin{gathered} {\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}=\frac{1}{36}{\int }_0^2\sqrt{u}du \\ {\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}=\frac{1}{36}{\int }_0^2{u}^{1/2}du \\ {\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}=\frac{1}{36}{\left[\frac{u^{1/2+1}}{1/2+1}\right]}_0^2 \\ {\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}=\frac{1}{36}{\left[\frac{u^{3/2}}{3/2}\right]}_0^2 \\ {\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}=\frac{1}{36}·\frac{2}{3}{\left[u^{3/2}\right]}_0^2 \\ {\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}=\frac{1}{53}{\left[{(1+36x)}^{3/2}\right]}_0^2 \\ {\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}=\frac{1}{53}\left[{(1+36\times 2)}^{3/2}-{(1+36\times 0)}^{3/2}\right] \\ {\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}=\frac{1}{53}\left[{\left(73\right)}^{3/2}-{\left(1\right)}^{3/2}\right] \\ {\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}=\frac{1}{53}\left[623.71-1\right] \\ {\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}=\frac{1}{53}\times 622.71 \\ {\int }_a^b\sqrt{1+( f\text{'}{\left(x\right))}^{2}dx}=11.53 \end{gathered}$$