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April 17, 2014

April 17, 2014

Posted by **fawn** on Monday, January 28, 2013 at 7:24pm.

- math (trig.) -
**fawn**, Monday, January 28, 2013 at 7:30pmI answered 17.5*sin((pi/4)*t)+17.5 but it's wrong I'm not sure what I'm doing incorrectly

- math (trig.) -
**Reiny**, Monday, January 28, 2013 at 8:25pmYou will need a phase shift

your amplitude is correct at 17.5

your vertical shift of 17.5 is also correct

period = 2π/k = 8

8k = 2π

k = 2π/8 = π/4

so your k for the period is correct

so let's adjust:

height = 17.5 sin (π/4)(t + d) + 17.5 , where d is a phase shift

So when you are at the 3:00 position and going downwards you must be 3/4 through a rotation and

t = 6 at a height of 17.5

sub in our equation:

17.5sin(π/4)(6+d) + 17.5 = 17.5

sin (π/4)(6+d) = 0

but I know that sin0 = 0 and sin π = 0

so (π/4)(6+d) = 0 or π/4(6+d) = π

d = -6

or

(1/4)(6+d) = 1

6+d = 4

d = -2

Just realized that if at 3:00 position you are going down, the wheel must be going clockwise (in math counterclockwise is a positive rotation)

But all is not lost, lets' just change our equation to

height = 17.5 sin (π/4)(-t + d) + 17.5

so our values of d would change:

d = 6 or d = 10

let's try d = 6

height = 17.5sin(π/4)(-t+6) + 17.5

testing:

t = 0 , height = 17.5 sin(π/4)(6) + 17.5 = 0 that's good

t - 2 , height = 17.5sin(π/4)(-2+6) + 17.5 = 17.5 , ok

t = 4 , height = 17.5sin(π/4)(-4+6) +17.5 = 35 , we are at the top

t = 6 , height = 17.5sin(π/4)(-6+6) +17.5 = 17.5 YEAHH, we are at 3:00 and coming back down

My equation is

**f(t) = 17.5 sin (π/4)(-t+6) + 17.5**

remember this equation is not unique,

if you recall sin(-x) = -sinx

so we could also write our equation as

f(t) = -17.5 sin (π/4)(t - 6) + 17.5

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