- 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.