Chapter 6: Q. 74 (page 540)

Dr. Geek built a skate ramp in his backyard modeled after the function $f (x) = 11 \sin \frac{1}{42} x^{2}$ , measured in feet, as shown in the next figure. Set up a definite integral that measures the distance required to skate from one side of the ramp to the other. The starting and ending x-values should both satisfy $f (x) = 11 f e e t$ . Use a graphing calculator to approximate the value of this definite integral.

Short Answer

Expert verified

Distance required to skate from one side of the ramp to the other is: $28.3804 f e e t$ .

Step by step solution

Step 1. Given information

Skate ramp modeled after the function, $f (x) = 11 \sin \frac{1}{42} x^{2}$

Step 2. Findingx when f(x)=11 feet

Skate ramp curve is: $f (x) = 11 \sin \frac{1}{42} x^{2}$

Put, $f (x) = 11 and find x$

$f (x) = 11 \sin \frac{1}{42} x^{2} = 11 \Rightarrow \sin \frac{1}{42} x^{2} = 1 \Rightarrow \frac{x^{2}}{42} = \sin^{- 1} 1 \Rightarrow x^{2} = 42 \times \frac{π}{2} \Rightarrow x^{2} = 21 \times 3.1415 \Rightarrow x = \sqrt{65.9734} \Rightarrow x = \pm 8.1224$

Step 3. Finding distance required to skate from one side of the ramp to the other.

$The arc length of f (x) from x = a to x = b can be represented by the definite integral : \int_{a}^{b} \sqrt{1 + (f' {(x))}^{2}} d x Equation of skate ramp curve is : f (x) = 11 \sin \frac{1}{42} x^{2} and f' (x) = \frac{11 \times 2 x}{42} \cos \frac{1}{42} x^{2} = \frac{11 x}{21} \cos \frac{1}{42} x^{2} Distance required to skate from one side of the ramp to the other is given by, D = \int_{- 8.1224}^{8.1224} \sqrt{1 + {(\frac{11 x}{21} \cos \frac{1}{42} x^{2})}^{2}} d x Using graphing calculator, the approximate value of this definite integral is : D = 28.3804$