Q. 73 - Calculus

Question

Suppose a drinking fountain expels water so that it falls in the shape of the parabola$f\left(x\right)=3-\frac{1}{2}{x}^2$, measured in inches, as shown in the given figure. Set up and solve a definite integral that measures the distance that water travels starting from when it leaves the fountain (at$x=0$) and ending when it hits the surface of the fountain (when$f\left(x\right)=0$).

Step-by-Step Solution

Answer

The distance that water travels is: $4.0545 inches$

Step 1. Given information

Path of water from a drinking fountain follows the path of parabola, $f\left(x\right)=3-\frac{1}{2}{x}^2$.

Step 2. Finding x when f ( x ) = 0

Given path of water is a parabola, $f\left(x\right)=3-\frac{1}{2}{x}^2$

Put, $f\left(x\right)=0 \mathrm{and} \mathrm{find} x$

  $$\begin{gathered} f\left(x\right)=3-\frac{1}{2}{x}^2=0 \\ \Rightarrow 3=\frac{1}{2}{x}^2 \\ \Rightarrow x^2=6 \\ \Rightarrow x=\sqrt{6} \end{gathered}$$

Step 3. Finding the distance that water travels

$$\begin{gathered} \mathrm{The} \mathrm{arc} \mathrm{length} \mathrm{of} f\left(x\right) \mathrm{from} x=a \mathrm{to} x=b \mathrm{can} \mathrm{be} \mathrm{represented} \mathrm{by} \mathrm{the} \\ \mathrm{definite} \mathrm{integral}:{\int }_{\mathrm{a}}^{\mathrm{b}}\sqrt{1 + ( \mathrm{f}\text{'}{\left(\mathrm{x}\right))}^2}d\mathrm{x} \\ \mathrm{The} \mathrm{distance} \mathrm{that} \mathrm{water} \mathrm{travels} \mathrm{starting} \mathrm{from} \mathrm{when} \mathrm{it} \mathrm{leaves} \mathrm{the} \mathrm{fountain} \\ (\mathrm{at} \mathrm{x}=0) \mathrm{and} \mathrm{ending} \mathrm{when} \mathrm{it} \mathrm{hits} \mathrm{the} \mathrm{surface} \mathrm{of} \mathrm{the} \mathrm{fountain} (\mathrm{at} \mathrm{x}=\sqrt{6}) \\ \mathrm{is} \mathrm{given} \mathrm{by}, \mathrm{D}={\int }_0^{\sqrt{6}}\sqrt{1 + {(-\mathrm{x})}^2}d\mathrm{x} \\ \mathrm{D}={\int }_0^{\sqrt{6}}\sqrt{1 + {\mathrm{x}}^2}d\mathrm{x} \\ \mathrm{D}={\overline{)\frac{\mathrm{x}}{2}\sqrt{1 + {\mathrm{x}}^2}+\frac{1}{2}\ln \left|\mathrm{x}+\sqrt{1 + {\mathrm{x}}^2}\right|}}_0^{\sqrt{6}} \\ \mathrm{D}=\left[\frac{\sqrt{6}}{2}\sqrt{1 + {\left(\sqrt{6}\right)}^2}+\frac{1}{2}\ln \left|\sqrt{6}+\sqrt{1 + {\left(\sqrt{6}\right)}^2}\right|\right] \\ -\left[\frac{0}{2}\sqrt{1 + {\left(0\right)}^2}+\frac{1}{2}\ln \left|0+\sqrt{1 + {\left(0\right)}^2}\right|\right] \\ \mathrm{D}=\left[\frac{\sqrt{42}}{2}+\frac{1}{2}\ln (\sqrt{6}+\sqrt{7})\right]-\left[0\right] \\ \mathrm{D}=3.2404+0.8141 \\ \mathrm{D}=4.0545 \end{gathered}$$