Q. 73 Suppose a drinking fountain expe... [FREE SOLUTION]

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Chapter 6: Q. 73 (page 540)

Suppose a drinking fountain expels water so that it falls in the shape of the parabola $f (x) = 3 - \frac{1}{2} x^{2}$ , 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 $f (x) = 0$ ).

Short Answer

Expert verified

The distance that water travels is: $4.0545 i n c h e s$

Step by step solution

Step 1. Given information

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

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

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

Put, $f (x) = 0 and find x$

role="math" localid="1649303807741" $f (x) = 3 - \frac{1}{2} x^{2} = 0 鈬� 3 = \frac{1}{2} x^{2} 鈬� x^{2} = 6 鈬� x = \sqrt{6}$

Step 3. Finding the distance that water travels

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