The Distance the Particle Travels with Given Position Function x(t) = t^4 - 8t^2 Date of Exam May 2016 Subject Mathematics, Surveying and Transportation Engineering Problem Given the position function x(t) = t^{4} - 8t^{2}, find the distance that the particle travels at t = 0 to t = 4. A. 160 C. 140 B. 150 D. 130 Answer Key Click here to show or hide the answer key [ A ] Solution Click here to expand or collapse this section $\displaystyle s = \int_{t_1}^{t_2} \sqrt{1 + \left( \dfrac{dx}{dt} \right)^2} \, dt$ $\displaystyle s = \int_0^4 \sqrt{1 + (4t^3 - 16t)^2} \, dt$ $s = 160.9$ ← Answer: [ A ] Solution by Simpson's One-third Rule: Click here to expand or collapse this section $f(t) = \sqrt{1 + (4t^3 - 16t)^2}$ Let $\Delta = \dfrac{4 - 0}{10} = 0.4$ n t f(t) C C * f(t) 1 0 1.00 1 1.00 2 0.4 6.22 4 24.90 3 0.8 10.80 2 21.60 4 1.2 12.33 4 49.31 5 1.6 9.27 2 18.54 6 2 1.00 4 4.00 7 2.4 16.92 2 33.85 8 2.8 43.02 4 172.08 9 3.2 79.88 2 159.76 10 3.6 129.03 4 516.11 11 4 192.00 1 192.00 Sum = 1193.15 Simpson's One-third Rule may be written as $s = \dfrac{\Delta}{3} \Big[ f(t)_1 + 2\Sigma f(t)_{odd} + 4 \Sigma f(t)_{even} + f(t)_n \Big]$ From the table above $f(t)_1 + 2\Sigma f(t)_{odd} + 4 \Sigma f(t)_{even} + f(t)_n$ = Sum Hence, $s = \dfrac{0.4}{3}(1193.15) = 159.09$ ← answer Category Integral Calculus Length of Arc Log in or register to post comments