Question

I need help please, i'll give a medal and follow

Answer 1

i need help on the first one

Answer 2

@Preetha

Answer 3

@sleepyjess

Answer 4

@johnweldon1993

Answer 5

@yaslin.esquivel how high off the ground is point A?

Answer 6

@johnweldon1993 50 feet?

Answer 7

point A is 50 ft off the ground, yes

Answer 8

so is that the answer for 1? @jim_thompson5910

Answer 9

let me know if you agree with this drawing and all it's dimensions

Answer 10

yes, thank you

Answer 11

ok so for #1, they want the equation

Answer 12

not just the height of point A

Answer 13

I'm assuming you've learned of trig equations before?

Answer 14

yes

Answer 15

sin, cos, and tan @jim_thompson5910

Answer 16

yes, which will you use here?

Answer 17

tan

Answer 18

hint: look at problem 3

Answer 19

so it can be either sin or cos

Answer 20

yes, tan is the only one we don't use

Answer 21

we can use either sine or cosine

sine is probably the easier of the two

Answer 22

what is sin(0) equal to?

Answer 23

40/50?

Answer 24

use a calculator to compute `sin(0)`

http://web2.0calc.com/

Answer 25

make sure you click the "rad" button to convert to radian mode

Answer 26

0.745113160479

Answer 27

type in `sin(0)` to get `0`

http://prntscr.com/b83fc5

Answer 28

ok

Answer 29

for practice, what is `sin(3)` equal to? make sure you're in radian mode

Answer 30

0.14112000806

Answer 31

good

Answer 32

sin(0) = 0

we want sin(0) to be equal to 50 since the height of point A is 50 and the rider is at position A at time t = 0.


How do we fix this? We can add 50 to the result of sin(0)


So sin(0)+50 = 50


The `+50` shifts the sine curve up fifty units

Answer 33

Agreed? or no?

Answer 34

hmmhh, agreed?

Answer 35

where are you not sure?

Answer 36

I don't remember my teacher teaching us to add a number to the result of sin

Answer 37

I'm doing this so we force the output to be 50

at time t = 0, point A is 50 ft high

for y = sin(x), if x = 0 then y = 0

but I want y to be equal to 50, so that's why I'm adding on 50

Answer 38

ohhh ok

Answer 39

the midline of y = sin(x) is y = 0
the midline of y = sin(x)+50 is y = 50

so D = 50 to represent our midline

Answer 40

the radius of the circle is 40 ft

this is equal to the amplitude of the sine curve. Why? Because the furthest you can go is 40 ft in either direction (up or down). You cannot move more than 40 ft away from the center point

so the amplitude is A = 40

Answer 41

agreed so far?

Answer 42

yes i agree

Answer 43

the period is 18 seconds since it makes a full rotation every 18 seconds

so T = 18

T is used for period and NOT P. This is just common convention. T for time most likely

Answer 44

Use T = 18 to find the value of B

T = (2pi)/B

B*T = 2pi

B = (2pi)/T

B = (2pi)/18 ... plug in T = 18

B = pi/9

Answer 45

Finally, the value of C (the phase shift) is 0. So C = 0

Answer 46

we have these 4 values

A = 40
B = pi/9
C = 0
D = 50

giving us this

\[\Large y = A\sin(Bx - C)+D\]

\[\Large y = 40\sin\left(\frac{\pi}{9}x - 0\right)+50\]

\[\Large y = 40\sin\left(\frac{\pi}{9}x\right)+50\]

Answer 47

so if i plugged in y=40sin(π9x)+50 it would give me the height of the times?

Answer 48

So the answer to problem #1 is
\[\Large h = 40\sin\left(\frac{\pi}{9}t\right)+50\]
I replaced x with t
I replaced y with h

t = time in seconds
h = height the rider is off the ground

Answer 49

so if you were to plug in t = 0, you should get h = 50

you can plug in any other value of t to find the corresponding height (at time t seconds)

Answer 50

omg thank you so much

Answer 51

let me know if you're able to complete #2 or not

Answer 52

ok and thank you again

Answer 53

no problem

Answer 54

for the second one is it h=40sin(π/9*1.5)+50 = 70?

Answer 55

@jim_thompson5910

Answer 56

let me check

Answer 57

did you mean t = 1.5 is one of the solutions @yaslin.esquivel ?

Answer 58

if so you are correct. What's another solution?

Answer 59

Btw, here is an interactive applet to see how the height of the rider is connected to a sine curve

See the link. It leads you to a GeoGebra web applet. You'll need Java to be installed (I think?) for it to work. It also may take a few seconds to load, so please be patient.

http://www.geogebra.org/m/kpnbPJwY

Move the slider around (panel on the right side) to see how point R moves and how point Q moves, and to see how they are connected.