Ferris Wheel Trigonometry Problem


– A FERRIS WHEEL HAS A DIAMETER
OF 30 METERS WITH A CENTER 17 METERS
ABOVE THE GROUND. IT MAKES ONE COMPLETE ROTATION
EVERY 60 SECONDS. WE WANT TO GRAPH ONE COMPLETE
PERIOD OF THE GRAPH THAT MODELS THE HEIGHT IN RELATION TO TIME. ASSUME A RIDER STARTS
AT THE LOWEST POINT, THEN FIND THE EQUATION OF THE
GRAPH USING THE COSINE FUNCTION. WE ALSO WANT TO KNOW
WHAT THE HEIGHT IS OF THE RIDER AT 45 SECONDS AND AT WHAT TIME OR TIMES
THE RIDER IS AT A HEIGHT OF 15 METERS. LET’S START BY DIAGRAMMING
OUR FERRIS WHEEL. THE FERRIS WHEEL
HAS A DIAMETER OF 30 METERS AND THE CENTER IS 17 METERS
ABOVE THE GROUND. WELL, IF THE DIAMETER
IS 30 METERS THAT MEANS THE RADIUS
OR A DISTANCE FROM A POINT ON THE CIRCLE TO THE CENTER, THIS DISTANCE HERE,
WOULD BE 15 METERS. SO IF THE CENTER IS 17 METERS
ABOVE THE GROUND AND THE RADIUS IS 15 METERS, THAT MEANS THE LOWEST POINT
ON THE FERRIS WHEEL IS 2 METERS ABOVE THE GROUND. WHICH MEANS THE HIGHEST POINT
ON THE FERRIS WHEEL WOULD BE 30 + 2 OR 32 METERS
ABOVE THE GROUND. AND, AGAIN, IT TAKES 60 SECONDS
FOR ONE ROTATION. SO NOW WE’RE GOING
TO PLOT POINTS ON THE GRAPH USING THE LOWEST POINT AT TIMES
ZERO AFTER 1/4 OF A ROTATION OR 15 SECONDS
OR THIS POINT HERE. AFTER 30 SECONDS
OR HALF A ROTATION, WHICH WOULD BE THIS POINT HERE. AFTER 3/4 OF A ROTATION
OR 45 SECONDS, WHICH IS THIS POINT HERE,
AND THEN AGAIN AT 60 SECONDS. SO NOTICE AT TIME T=0 THE RIDER STARTS AT THE LOWEST
POINT, SO THE HEIGHT IS 2 WHICH MEANS
THE FUNCTION MUST CONTAIN THE .02, WHICH WOULD BE HERE
WHERE THE Y VALUE WOULD BE 2. AT 15 SECONDS,
OR THIS POINT HERE, THE HEIGHT WOULD BE THE SAME AS THE DISTANCE FROM THE CENTER
TO THE GROUND, WHICH IS 17 METERS, SO THE
FUNCTION MUST CONTAIN THE .1517, WHICH WOULD BE HERE. SO THE Y VALUE HERE IS 17, WHICH WILL ACTUALLY BE THE
MIDLINE OF OUR FUNCTION. AFTER 30 SECONDS,
OR HALF A ROTATION, THE HEIGHT WOULD BE 30 + 2
OR 32 METERS, SO THE FUNCTION CONTAINS THE
.3032, WHICH IS THE HIGH POINT. AT 45 SECONDS THE RIDER IS HERE,
AGAIN, AT A HEIGHT OF 17 METERS. SO THE .4517 IS ON THE FUNCTION
AND THEN, FINALLY, AFTER ONE ROTATION
OR 60 SECONDS, THE RIDER IS BACK AT THE BOTTOM,
2 METERS FROM THE GROUND. SO THE FUNCTION CONTAINS
THE .62, WHICH WOULD BE HERE. SO BECAUSE OF THE FERRIS WHEEL
WE HAVE A CIRCULAR FUNCTION AND THE GRAPH WOULD LOOK
SOMETHING LIKE THIS.   WE SHOULD RECOGNIZE THIS AS
RESEMBLING THE COSINE FUNCTION. SO LET’S TAKE THIS GRAPH
ON TO THE NEXT SLIDE SO THAT WE CAN FIND THE EQUATION
OF THIS GRAPH. I DID REPRODUCE IT USING SOME
GRAPHING SOFTWARE TO MAKE A MORE ACCURATE GRAPH. AGAIN, THE MINIMUM VALUE HERE
WAS 2 AND THE MAXIMUM VALUE WAS 32, WITH THE MIDLINE HERE BEING
AT Y=17. WE’RE ASKED TO FIND THE EQUATION
USING THE COSINE FUNCTION, SO WE’LL BE USING THIS FORM
OF THE COSINE FUNCTION. MEANING WE’LL HAVE TO FIND
THE VALUE OF “A,” B, D, AND C BASED UPON THE TRANSFORMATIONS
OF THE BASIC COSINE FUNCTION. SO TO FIND OUR EQUATION WE’LL
USE THIS PIECE OF THE GRAPH. BECAUSE WE’RE USING THE COSINE
FUNCTION AND AT T=0 WE HAVE A MINIMUM VALUE WE WON’T HAVE A PHASE SHIFT
OR HORIZONTAL SHIFT, WHICH MEANS D IS GOING
TO BE EQUAL TO ZERO. WHICH MEANS OUR EQUATION
WILL BE IN THE FORM Y=”A” x COSINE BX + C. BECAUSE WE WANT A HEIGHT
FUNCTION IN TERMS OF T WE’RE GOING TO FIND H OF T, WHICH WILL BE EQUAL TO “A”
COSINE BT + C. SO BEFORE WE FIND OUR EQUATION
LET’S REVIEW HOW “A,” B AND C AFFECT THE GRAPH OF THE BASIC
COSINE FUNCTION. THE ABSOLUTE VALUE OF “A”
IS EQUAL TO THE AMPLITUDE AND “A” CAN BE BOTH POSITIVE
OR NEGATIVE. IF “A” IS NEGATIVE WE HAVE
A REFLECTION ACROSS THE X AXIS. TO FIND THE VALUE B, 2PI DIVIDED
BY B IS EQUAL TO THE PERIOD. C PRODUCES A VERTICAL
TRANSLATION OR VERTICAL SHIFT. IF C IS POSITIVE THE GRAPH
IS SHIFTED UP. IF C IS NEGATIVE THE GRAPH
IS SHIFTED DOWN AND WE ALREADY SAID
WE DON’T HAVE A HORIZONTAL SHIFT
OR TRANSLATION, SO D IS GOING TO BE ZERO. SO GOING BACK TO OUR GRAPH, LET’S FIRST DETERMINE THE VALUE
OF “A.” WELL, THE AMPLITUDE
IS THE DISTANCE FROM THE MIDLINE TO THE MAXIMUM OR FROM
THE MIDLINE TO THE MINIMUM. WELL, FROM 17 TO 32
IS EQUAL TO 15, THE SAME AS THE DISTANCE
FROM 17 TO 2. SO THE ABSOLUTE VALUE OF “A”
MUST BE EQUAL TO 15, WHICH IS ACTUALLY THE RADIUS
OF THE FERRIS WHEEL. BUT BECAUSE AT T=0
WE HAVE A MINIMUM AND FOR THE BASIC COSINE
FUNCTION AT X=0 WE HAVE A MAXIMUM, WE HAVE A
REFLECTION ACROSS THE X AXIS, WHICH MEANS “A”
IS GOING TO BE -15 NOT 15. NEXT, THE VALUE OF B IS AFFECTED
BY THE PERIOD WHICH WE KNOW WOULD BE 60
SECONDS, SO 60 MUST BE EQUAL TO 2PI
DIVIDED BY B. SO TO SOLVE FOR B
WE WOULD MULTIPLY BOTH SIDES OF THE EQUATION BY B,
THAT WOULD GIVE US 60B=2PI, DIVIDE BOTH SIDES BY 60. B=2PI DIVIDED BY 60,
OR B=PI DIVIDED BY 30. SO WE HAVE THE VALUE OF “A.” WE HAVE THE VALUE OF B. NOW WE NEED TO FIND
THE VALUE OF C. WELL, NOTICE HOW OUR MIDLINE IS
NOT THE X AXIS, IT’S Y=17, WHICH MEANS OUR GRAPH HAS BEEN
SHIFTED UP 17 UNITS AND, THEREFORE, C=17. AND NOW WE HAVE
ALL THE INFORMATION WE NEED TO WRITE THE EQUATION
OF OUR FUNCTION. WE HAVE H OF T MUST EQUAL “A,”
WHICH IS -15 x COSINE OF B x T, WHICH WOULD BE PI/30 x T
OR PI T DIVIDED BY 30 + 17. THIS EQUATION WILL GIVE US
THE HEIGHT OF THE RIDER AT ANY TIME T IN SECONDS. NOW, FOR THE NEXT QUESTION
WE’RE ASKED TO FIND THE HEIGHT OF THE RIDER
AT 45 SECONDS, WHICH MEANS WE WANT TO FIND
H OF 45. AND WE’LL SUBSTITUTE 45 FOR T, SO WE’LL HAVE -15 COSINE
OF 45PI/30 + 17. SO WE’LL HAVE -15 x COSINE OF–
THIS’LL BE 3PI/2 + 17. WE CAN DETERMINE THIS COSINE
FUNCTION VALUE FROM THE UNIT CIRCLE,
IT’S ACTUALLY EQUAL TO ZERO, SO WE HAVE -15 x 0 + 17, SO AT 45 SECONDS THE HEIGHT
WOULD BE 17 METERS. OF COURSE, FOR MANY TIMES
IT WOULDN’T COME OUT SO NICE. BUT IF YOU GO BACK AND TAKE
A LOOK AT OUR FERRIS WHEEL JUST FOR A SECOND IF IT TAKES
60 SECONDS FOR ONE ROTATION, 45 SECONDS WOULD BE 3/4
OF THE ROTATION, WHICH WOULD BE THIS POINT HERE WHERE THE HEIGHT WOULD BE THE
SAME AS THE DISTANCE FROM THE CENTER TO THE GROUND, WHICH WE WERE GIVEN
AS 17 METERS. IF THIS WASN’T A NICE
TRIG FUNCTION VALUE WE’D HAVE TO APPROXIMATE
IT USING OUR CALCULATOR. AND NOW FOR THIS LAST QUESTION
WE WANT TO KNOW AT WHAT TIME OR TIMES THE RIDER
IS AT A HEIGHT OF 15 METERS. IF WE LOOK AT OUR GRAPH
JUST FOR A MOMENT, THE HEIGHT IS 15 METERS RIGHT
HERE JUST BELOW 16. NOTICE IF WE DRAW A HORIZONTAL
LINE AT Y EQUALS 15 THERE ARE ACTUALLY 2 POINTS
ON THE FUNCTION FROM ZERO TO 60 SECONDS THAT
HAVE A HEIGHT OF 15 METERS, HERE AND HERE. SO WE NEED TO MAKE SURE
THAT WE GET TO TWO SOLUTIONS WHEN WE SOLVE THIS PROBLEM. SO TO START WE’RE GOING
TO REPLACE H OF T WITH 15 METERS AND THEN SOLVE
FOR T. SO WE’LL HAVE 15=-15 COSINE PI
T/30 + 17. SO WE WANT TO ISOLATE THE COSINE
FUNCTION SO WE’LL SUBTRACT 17
ON BOTH SIDES, THAT WOULD GIVE US -2,
SINCE 15 – 17 IS -2 AND THEN WE’LL DIVIDE BY -15. A NEGATIVE DIVIDED BY A NEGATIVE
IS POSITIVE, SO WE HAVE 2/15=COSINE OF PI T
DIVIDED BY 30. AND NOW WE’RE GOING TO TAKE THE ARC COSINE ON BOTH SIDES
OF THE EQUATION TO HELP US SOLVE FOR T. SO WE’LL HAVE ARC COSINE OF 2/15=ON THE RIGHT SIDE ARC COSINE
OF COSINE PI T DIVIDED BY 30 IS JUST PI T DIVIDED BY 30. AND NOW TO SOLVE FOR T
WE’RE GOING TO MULTIPLY BOTH SIDES BY 30/PI. NOTICE ON THE RIGHT SIDE PI/PI
30/30 SIMPLIFY TO 1. SO THE RIGHT SIDE IS JUST T. SO WE’RE GOING TO WRITE THIS AS T=30 DIVIDED BY PI x ARC
COSINE OF 2/15. NOW FROM HERE LET’S GO AHEAD
AND CALL THIS ANGLE THETA AND WE’LL GO TO THE CALCULATOR
TO DETERMINE WHAT THIS ANGLE WOULD BE. REMEMBER, WE’RE LOOKING
FOR TWO ANGLES AND SINCE THE COSINE FUNCTION
VALUE HERE IS POSITIVE, IF WE LOOK AT THE UNIT CIRCLE THE X COORDINATE
WOULD HAVE TO BE POSITIVE, THEREFORE, THE TWO ANGLES MUST BE TERMINAL IN THE FIRST
AND FOURTH QUADRANTS. SO NOW WE’LL GO
TO THE CALCULATOR. LET’S MAKE SURE THAT WE’RE IN
RADIAN MODE, WHICH WE ARE. SO FROM THE HOME SCREEN
WE’RE GOING TO PRESS 2nd, COSINE FOR ARC COSINE OR INVERSE
COSINE TO 2/15 AND ENTER. SO ONE ANGLE IS APPROXIMATELY
1.437 RADIANS. LET’S GO AHEAD AND WRITE THAT
DOWN. WE’LL CALL THIS THETA SUB 1. BUT, AGAIN, WE STILL
HAVE TO FIND ANOTHER ANGLE THAT HAS THE SAME
COSINE FUNCTION VALUE. WELL, 1.437 RADIANS
IS IN THE FIRST QUADRANT, WHICH WOULD BE THIS QUADRANT
HERE, WHICH MEANS THE OTHER ANGLE
WHERE THE X COORDINATE IS STILL POSITIVE WOULD HAVE THE
SAME REFERENCE ANGLE IN THE FOURTH QUADRANT. SO THIS REFERENCE ANGLE HERE
WOULD STILL BE 1.437, WHICH MEANS THE ANGLE THAT WE
WANT BETWEEN ZERO AND 2PI WOULD HAVE TO BE 2PI,
ONE ROTATION – 1.437 RADIANS. SO THETA SUB 2 IS 2PI – 1.437, SO THETA SUB 2 IS APPROXIMATELY
4.846. SO NOW THE TWO TIMES WILL BE T SUB 1=30 DIVIDED
BY PI x THETA SUB 1, WHICH IS 1.437 AND x SUB 2 IS GOING TO BE=30
DIVIDED BY PI x 4.846. SO ONE TIME
IS GOING TO BE APPROXIMATELY– I’VE ALREADY DETERMINED THIS
VALUE, IT’S APPROXIMATELY 13.7 SECONDS, AND x SUB 2 IS 30 DIVIDED BY PI
x 4.846, WHICH IS APPROXIMATELY
46.3 SECONDS. NOW, IF WE GO BACK AND TAKE A
LOOK AT THE FERRIS WHEEL JUST FOR A MOMENT, NOTICE HOW 15 METERS
WOULD BE JUST BELOW THE CENTER OF THE FERRIS WHEEL THAT IS 17
METERS ABOVE THE GROUND. SO IF WE DREW
A HORIZONTAL LINE HERE NOTICE HOW THE FIRST TIME
ON THE LEFT WOULD BE JUST LESS
THAN 15 SECONDS, WHICH WE FOUND, AND THE TIME ON THE RIGHT
WOULD BE JUST PAST 45 SECONDS, WHICH WE FOUND AS WELL–
THIS EXPLANATION HELPS. THANK YOU FOR WATCHING.  

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