anonymous one year ago Could anyone help with this activity I have to do? It's trigonometry

1. anonymous

2. anonymous

I'd really appreciate it if someone could explain so I can figure it out . I'm terrible at trig . I'm taking an algebra 2 class and apparently trig is apart of algebra 2

3. anonymous

@jim_thompson5910 could you help me with this ? :/

4. jim_thompson5910

Your worksheet is a bit misleading. It says that B = 2pi/(period) but that's only true for sine, cosine, cosecant, secant. It doesn't apply to tangent and cotangent. That rule would be B = pi/(period). Luckily we won't be dealing with tangent or cotangent

5. anonymous

did I write that ?

6. jim_thompson5910

no it's printed on page 1

7. anonymous

apex is confusing and doesn't do the right information

8. anonymous

doesn't give *

9. jim_thompson5910

you wrote the correct max elevation what is the min elevation?

10. anonymous

it didn't give that . Idk

11. jim_thompson5910

|dw:1440386808612:dw|

12. jim_thompson5910

|dw:1440386828053:dw|

13. jim_thompson5910

since the ride starts 10 ft off the ground

14. anonymous

oh so it'd be 10 ? lol

15. jim_thompson5910

yeah

16. jim_thompson5910

and that would make the diameter 210 - 10 = 200 and not 210

17. anonymous

I have the diameter as 200

18. jim_thompson5910

oh nvm I misread that part

19. jim_thompson5910

don't forget the units. When it asks for the time to complete 1 revolution. Don't just write "90". Write "90 seconds". Same applies for the other values too

20. jim_thompson5910

other than that, page 1 looks great

21. anonymous

do the max and min have units ? oh and the diameter ?

22. jim_thompson5910

yeah, in this case, "feet"

23. jim_thompson5910

same for the diameter

24. anonymous

okay I just wanted to be sure

25. jim_thompson5910

26. anonymous

yes

27. jim_thompson5910

I'm guessing you haven't started at all? Or do you have an idea where to start?

28. anonymous

The graph is where I got confused . I looked over it multiple times and I tried graphing but I just didn't understand

29. jim_thompson5910

At time t = 0 seconds, the height is h = 10 ft So one ordered pair is (0,10) and this is a min -------------------------------------- 45 seconds later at t = 45, the height moves to h = 210 one max point is (45,210) -------------------------------------- another 45 seconds pass by and we're now at t = 90 seconds we're back at the min height again and now at the point (90, 10)

30. jim_thompson5910

they want you to plot these mins and maxes for 2 full cycles

31. anonymous

I always forget how to graph lol . Just one second

32. jim_thompson5910

For #2, they don't want the full graph of the sinusoidal curve (ie they don't want sine or cosine plotted). They just want those discrete points.

33. anonymous

so don't plot all three of those points ?

34. jim_thompson5910

You do plot those points and any others that I didn't list until you've done 2 full cycles

35. jim_thompson5910

If you start at t = 0, 1 full cycle ends at t = 90 2 full cycles end at t = 180

36. anonymous

|dw:1440387875856:dw| so it'd look somewhat like that ?

37. anonymous

that's a horrible drawing I know lol . My mousepad sucks

38. jim_thompson5910

that's ok. You should get something like this (see attached). It looks like you did, so you are correct for #2

39. anonymous

do I connect the dots or leave them as is ?

40. jim_thompson5910

For #2, they just want the min/max points. That's all. So you leave them as is.

41. anonymous

Okay so that one is done

42. jim_thompson5910

what did you get for #3

43. anonymous

44. anonymous

wouldn't the top of the ferris wheel be the max ?

45. jim_thompson5910

yes

46. anonymous

so it'd be 210 ?

47. jim_thompson5910

they want halfway to the top though

48. jim_thompson5910

|dw:1440388468748:dw|

49. anonymous

would you divide 210 by 2 ?

50. jim_thompson5910

you would find the midpoint of h = 10 and h = 210 add the values up and divide by 2: (10+210)/2 = ??

51. jim_thompson5910

would you divide 210 by 2 ? only if the starting height was 0

52. anonymous

so 110 ?

53. jim_thompson5910

54. anonymous

so their elevation would be 110 feet ?

55. jim_thompson5910

at the midway point, yes

56. anonymous

what do they mean by "use the diameter of the Ferris wheel to determine the correct vertical shift for your modeling function?"

57. jim_thompson5910

well I guess another way to do is to divide the diameter in half 200/2 = 100 the radius is 100. Add that radius to the starting height: 100+10 = 110

58. jim_thompson5910

I like the midpoint method better

59. anonymous

Midpoint takes out a step

60. anonymous

so I'd graph it at 110 ?

61. jim_thompson5910

are you still on #3 ? or #4 now?

62. anonymous

on number 4

63. jim_thompson5910

yeah you'll have a dashed/dotted line that is horizontal and going through y = 110

64. anonymous

|dw:1440389236868:dw| so like this ?

65. jim_thompson5910

yeah

66. jim_thompson5910

which letter represents the vertical shift? A, B, C, or D?

67. anonymous

uh , how do you figure that out ?

68. jim_thompson5910

look back on page 1

69. anonymous

D

70. jim_thompson5910

yes, D = 110

71. jim_thompson5910

for # 5, you can find the distance using two methods a) subtract the max and min height, then divide by 2 (max - min)/2 = (210-10)/2 = 200/2 = 100 OR b) Find the distance from the center (110) to either the peak or valley Center to peak: 210 - 110 = 100 Center to valley: 110 - 10 = 100 either way, the radius of this wheel is 100 ft. This is the amplitude of the sinusoidal curve the amplitude is |A|, so A = 100

72. anonymous

Yeah I just subtracted it and got 100

73. anonymous

amplitude is distance ?

74. jim_thompson5910

the distance from midline to the max or the distance from midline to the min

75. jim_thompson5910

|dw:1440389822564:dw|

76. jim_thompson5910

or half the distance from the very top to the very bottom |dw:1440389869160:dw|

77. anonymous

so it just depends on exactly what they're asking ?

78. jim_thompson5910

what do you mean?

79. anonymous

or is all just the same ?

80. jim_thompson5910

those are 2 different ways to get the same thing

81. jim_thompson5910

so it doesn't matter which path you take

82. anonymous

ooh okay , I just wanted to make sure

83. anonymous

for 6, do I keep it in terms of pi or simplify it completely ?

84. jim_thompson5910

I'd leave it in terms of pi

85. jim_thompson5910

reduce the fraction as much as possible though

86. anonymous

so it'd be $B=\frac{ 2\pi }{ 90 } = \frac{ \pi }{ 45 }$

87. jim_thompson5910

yeah

88. jim_thompson5910

which trig function will you use for #7 ? sine or cosine?

89. anonymous

sin

90. jim_thompson5910

either works, but cosine is actually easier to work with if you graph the parent function y = cos(x), you'll notice it has the peak max point at (0,1). We want the peak max point to happen when t = 45 so that's why C = 45 is the phase shift. We shift everything over 45 units to the right

91. jim_thompson5910

A = 100 B = pi/45 C = 45 D = 110 $\Large H(t) = A\cos\left(B(t-C)\right) + D$ $\Large H(t) = 100\cos\left(\frac{\pi}{45}(t-45)\right) + 110$ attached is the graph

92. jim_thompson5910

you could use sine but the phase shift would be different. Everything else would stay the same

93. anonymous

Okay , it makes sense

94. anonymous

in number 8 , what do they mean by horizontal shift ?

95. jim_thompson5910

the parent function y = cos(x) has a max when x = 0 but we want to force the max to be when x = 45 so that's why we do a horizontal shift and why C = 45

96. jim_thompson5910

|dw:1440390964651:dw|

97. jim_thompson5910

|dw:1440390975447:dw|

98. jim_thompson5910

|dw:1440390992114:dw|

99. anonymous

So it is 45 . I'm used to shifts when it comes to 2 graphs

100. jim_thompson5910

yeah it's 45 if we use cosine

101. anonymous

Since with cosine x = 0

102. jim_thompson5910

what do you mean?

103. anonymous

idk . I'm probably confusing myself

104. anonymous

so for number 9 , where are they getting at with the parent function being flipped ? I don't even know what the parent function is :/

105. jim_thompson5910

in this case, we used the parent function y = cos(x)

106. jim_thompson5910

it's just the base function that you apply the transformations to

107. jim_thompson5910

A = 100 which is positive, we aren't flipping the function over

108. anonymous

if it were to be flipped , how would I know that was the case ?

109. jim_thompson5910

well I started with y = cos(x) which has A = 1 as a positive number shifting everything over to the right 45 units doesn't flip things

110. jim_thompson5910

I played around with the graph a bit, and I am just realizing that A could be negative, but that changes the phase shift.

111. jim_thompson5910

So there are a few ways to get the h(t) function

112. jim_thompson5910

if A = +100, then the phase shift is C = 45 if A = -100, then the phase shift is C = 0

113. anonymous

But since it's dealing with a ferris wheel , it couldn't have a negative value since you can't go in circles negative times right ?

114. jim_thompson5910

t and h can't be negative, but A could be

115. jim_thompson5910

the amplitude is |A| the absolute value ensures we have a positive amplitude

116. jim_thompson5910

on your worksheet, draw vertical bars around the "A" where it says A is the amplitude on page 1

117. anonymous

oh okay , so it doesn't really matter if it was negative cause it regardless it's going to be positive since it's absolute value ?

118. jim_thompson5910

exactly

119. jim_thompson5910

eg: y = -2*cos(x) has an amplitude of |-2| = 2

120. anonymous

okay , I get it

121. anonymous

in the equation on number 10, what's "Af" ?

122. jim_thompson5910

I don't see #10 in any of the attachments

123. anonymous

one moment

124. anonymous

125. jim_thompson5910

oh, that's A*f

126. jim_thompson5910

f is the trig function look back at page 1

127. anonymous

so it's the amplitude times the trig function ? 100cosx ? or ...

128. jim_thompson5910

yeah like how I wrote above when I wrote out H(t)

129. anonymous

where did you write that ?

130. jim_thompson5910

it's buried a bit when I wrote $\Large H(t) = 100\cos\left(\frac{\pi}{45}(t-45)\right) + 110$

131. anonymous

oooh okay , I remember

132. anonymous

and t would b 90 , correct ?

133. anonymous

be *

134. jim_thompson5910

t is just a variable that represents the time in seconds

135. jim_thompson5910

you plug in values of t to get corresponding values of h eg: plug in t = 0 to get h = 10 plug in t = 45 to get h = 210 etc etc

136. anonymous

wait so how exactly is the equation set up ?

137. jim_thompson5910

what do you mean?

138. anonymous

I plug everything into that equation right ?

139. jim_thompson5910

I'd just leave it as $\Large H(t) = 100\cos\left(\frac{\pi}{45}(t-45)\right) + 110$

140. jim_thompson5910

They want the function or equation that models the curve

141. anonymous

So no solving ?

142. jim_thompson5910

nope, they just want you to write the equation

143. anonymous

and for number 11 , i use that same equation, I'm just solving for t ?

144. jim_thompson5910

correct. replace h(t) with 100 and solve for t

145. anonymous

To start I'd subtract 110 over to 100 right ?

146. jim_thompson5910

yes

147. anonymous

what do I do from there ?

148. jim_thompson5910

you should have this $\Large 100\cos\left(\frac{\pi}{45}(t-45)\right) = -10$

149. jim_thompson5910

now divide both sides by 100

150. anonymous

so it'd be equal to -1/10 ?

151. jim_thompson5910

yeah or -0.1

152. anonymous

I feel like decimal is better

153. jim_thompson5910

Then apply the arccosine function to both sides $\Large \Large \cos\left(\frac{\pi}{45}(t-45)\right) = -0.1$ $\Large \Large \arccos\left(\cos\left(\frac{\pi}{45}(t-45)\right)\right) = \arccos(-0.1)$ $\Large \Large \frac{\pi}{45}(t-45) = ??$

154. anonymous

what's arccos ?

155. jim_thompson5910

arccos is $$\Large \cos^{-1}$$, ie inverse cosine

156. anonymous

so csc ?

157. jim_thompson5910

no, that's a common misconception

158. jim_thompson5910

if you want to say 1/cos, you say secant

159. jim_thompson5910

you don't say $$\Large \cos^{-1}$$ to mean secant $$\Large \cos^{-1} \ne \sec$$

160. anonymous

oh okay , so I haven't learned arccos

161. jim_thompson5910

your calculator should have a cos^(-1) or arccos button

162. anonymous

I have cos^-1

163. jim_thompson5910

ok use that

164. anonymous

so I find arccos-0.1

165. jim_thompson5910

yeah what is that numeric value?

166. anonymous

95.74 ?

167. jim_thompson5910

make sure you're in radian mode

168. anonymous

1.67

169. jim_thompson5910

yes or -1.67

170. jim_thompson5910

we will use the negative form to figure out when the height reaches 100 ft for the first time (the +1.67 is used to figure out when the height reaches 100 ft for the second time)

171. jim_thompson5910

so solve for t $\Large \Large \frac{\pi}{45}(t-45) = -1.67$

172. anonymous

do I distribute or multiply pi/45 by the recipricol on both sides ?

173. jim_thompson5910

I'd multiply by the reciprocal

174. anonymous

so it'd be $t-45=\frac{ -75.15 }{ \pi }$

175. jim_thompson5910

-75.15/pi = -23.9209879 approximately

176. jim_thompson5910

now just one more step to go

177. anonymous

178. jim_thompson5910

yes

179. anonymous

so it would equal 21.08 ?

180. jim_thompson5910

I'm getting 21.07901205 with a calculator and geogebra is telling me 21.06521 so it's pretty close

181. jim_thompson5910

so roughly 21 seconds

182. anonymous

I rounded the 21.07901205

183. jim_thompson5910

that works

184. anonymous

thank you so much !

185. jim_thompson5910

and desmos says the same thing more or less https://www.desmos.com/calculator/swknd4tau1

186. jim_thompson5910

you'll have to click on the first intersection point between the curve and the line

187. anonymous

You really helped a lot lol . I can finally release some stress . Thank you

188. jim_thompson5910

you're welcome