Question 1124245
If you spin a radius vector, {{{ r }}}, at the origin,
It traces out a sine wave plotted on the time axis.
It has a -r to +r height, so the height is {{{ 2r }}}
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In your problem, {{{ 2r = 70 }}} and {{{ r = 35 }}}
If the Ferris wheel was at ground level, the 
function would be {{{ f(t) = 35*sin( ( 2pi/58 )*t ) }}},
but they want the center {{{ 40 }}} ft above ground,
So the function is: {{{ H(t) = 35*sin( ( 2pi/58 )*t ) + 40 }}}
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b)
Find {{{ t }}} when {{{ H(t) = 61 }}}
{{{ 61 = 35*sin( ( 2*pi/58 )*t ) + 40 }}}
{{{ 35*sin( ( 2*pi/58 )*t ) = 21 }}}
{{{ sin( ( 2*pi/58 )*t ) = .6 }}}
If this was:
{{{ sin( theta ) = .6 }}} then, in radians,
{{{ theta = .6435 }}} radians ( calculator )
Now I can say:
{{{ ( 2*3.14159/58 )*t = .6435 }}}
{{{ .10833*t = .6435 }}}
{{{ t = 5.9402 }}} sec
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Here’s a plot:
{{{ graph( 700,400, -50, 150, -10, 80, 35*sin( ( 2*pi/ 58 )*x ) + 40 ) }}}
Note that with this curve, you would have to get on
the Ferris wheel at the 58 sec time. If you used -cos instead of
sin, then you get on at the zero time mark
Hope all this helps