Math 114 Midterm project
Part one: 100 points
You are the facilities engineer for Colorado Technical University
Tan 60 = 90ft / x
X = 90ft / Tan 60
X = 856.29 ft or approx 856 ft
tan 60 = 90ft / x = 90 ft/ 856.29 ft or approx. 0.105104
sin 60 = 90 / Ö(902 + 856.29282) or approx . 0.104528
cos 60 = 856.29282 / Ö(902 + 856.29282) or approx . 0.994522
cot 60 = 1 / tan 60 = 1 / 0.105104 or approx. 9.514364
sec 60 = 1 / cos 60 = 1 / 0.994522 or approx. 1.005508
csc 60 = 1 / sin 60 = 1 / 0.104528 or approx. 9.566772
Part Two A: 50 pts
Part Two B: 50 pts
Since it is stated that the BP reading is 200 over 150, then the systolic is 200 and the diastolic is 150, and since it is also stated that k is the average of the systolic and diastolic numbers, then
k = (200 + 150)/2 = 175
Also, it is stated that the amplitude is equal to the systolic minus the value of k, then
A = 200 – 175 = 25
Then, in the given equation:
f(t) = A cos(bt) + k
since A = 25, and k = 175, then
f(t) = 25cos(bt) + 175
b = 1000p/3
f(t) = 25cos(1000pt/3) + 175
Part Three: 50 pts
The graph of R(q) = (Vo)2Ö2(cosq)(sinq – cosq)/16 where Vo = 32 ft/sec and
450 £ q £ 900
Using excel, is:
q / R(q)45 / 0
45.5 / 0.782913226
46 / 1.551802908
46.5 / 2.306434835
47 / 3.046579138
47.5 / 3.772010364
48 / 4.482507537
48.5 / 5.177854235
49 / 5.857838647
49.5 / 6.522253644
50 / 7.170896839
50.5 / 7.803570648
51 / 8.420082353
51.5 / 9.020244158
52 / 9.603873249
52.5 / 10.17079185
53 / 10.72082726
53.5 / 11.25381195
54 / 11.76958355
54.5 / 12.26798497
55 / 12.74886437
55.5 / 13.21207529
56 / 13.65747663
56.5 / 14.0849327
57 / 14.4943133
57.5 / 14.88549373
58 / 15.25835484
58.5 / 15.61278305
59 / 15.94867039
59.5 / 16.26591454
60 / 16.56441889
60.5 / 16.84409249
61 / 17.10485015
61.5 / 17.34661245
62 / 17.56930574
62.5 / 17.7728622
63 / 17.9572198
63.5 / 18.1223224
64 / 18.26811971
64.5 / 18.39456731
65 / 18.50162668
65.5 / 18.58926522
66 / 18.65745623
66.5 / 18.70617893
67 / 18.73541849
67.5 / 18.745166
68 / 18.73541849
68.5 / 18.70617893
69 / 18.65745623
69.5 / 18.58926522
70 / 18.50162668
70.5 / 18.39456731
71 / 18.26811971
71.5 / 18.1223224
72 / 17.9572198
72.5 / 17.7728622
73 / 17.56930574
73.5 / 17.34661245
74 / 17.10485015
74.5 / 16.84409249
75 / 16.56441889
75.5 / 16.26591454
76 / 15.94867039
76.5 / 15.61278305
77 / 15.25835484
77.5 / 14.88549373
78 / 14.4943133
78.5 / 14.0849327
79 / 13.65747663
79.5 / 13.21207529
80 / 12.74886437
80.5 / 12.26798497
81 / 11.76958355
81.5 / 11.25381195
82 / 10.72082726
82.5 / 10.17079185
83 / 9.603873249
83.5 / 9.020244158
84 / 8.420082353
84.5 / 7.803570648
85 / 7.170896839
85.5 / 6.522253644
86 / 5.857838647
86.5 / 5.177854235
87 / 4.482507537
87.5 / 3.772010364
88 / 3.046579138
88.5 / 2.306434835
89 / 1.551802908
89.5 / 0.782913226
90 / 0
As we can see from the graph;, since sin 450 = cos 450, then at q = 450, R(450)= 0. Also since cos 900 =0, then, at q = 900, R(900)= 0. Moreover, since the graph is symmetric at the middle, and since R(450)= R(900)= 0, and has a downward opening, then the function has a maximum value at the middle, that is, at the average of 450 and 900. Therefore, the maximum value of the projectile is achieved at q = (450 + 900 )/2 = 67.50 and maximum R is at R(67.50) » 18.73541849 ft. (Note: Applying the Rolle’s theorem)