Asked by anna
okay sorry last two questions! I promise!
The sun always illuminates half of the moon’s surface, except during a lunar eclipse. The illuminated portion of the moon visible from Earth varies as it revolves around Earth resulting in the phases of the moon. The period from a full moon to a new moon and back to a full moon is called a synodic month and is 29 days, 12 hours, and 44.05 minutes long. Write a sine function that models the fraction of the moon’s surface which is seen to be illuminated during a synodic month as a function of the number of days, d, after a full moon. [Note: full moon equals illuminated.]
I am really confused on this one I know I have to make a sine function.
so would it be like y= 29sin(pi/6*t)+44.05
sorry that was probably a really bad guess.
2)A bicycle tire has a diameter of 20 inches and is revolving at a rate of 10 rpm. At t = 0, a certain point is at height 0. What is the height of the point above the ground after 20 seconds?
y=20sin(pi/5*20) Is that right?
The sun always illuminates half of the moon’s surface, except during a lunar eclipse. The illuminated portion of the moon visible from Earth varies as it revolves around Earth resulting in the phases of the moon. The period from a full moon to a new moon and back to a full moon is called a synodic month and is 29 days, 12 hours, and 44.05 minutes long. Write a sine function that models the fraction of the moon’s surface which is seen to be illuminated during a synodic month as a function of the number of days, d, after a full moon. [Note: full moon equals illuminated.]
I am really confused on this one I know I have to make a sine function.
so would it be like y= 29sin(pi/6*t)+44.05
sorry that was probably a really bad guess.
2)A bicycle tire has a diameter of 20 inches and is revolving at a rate of 10 rpm. At t = 0, a certain point is at height 0. What is the height of the point above the ground after 20 seconds?
y=20sin(pi/5*20) Is that right?
Answers
Answered by
bobpursley
period=29 days, 12hrs, 44.05 min.
ok, change that to days: 29+12/24+44.05/(60*24), I get 29.53 days.
y=.5cos(2PI/29.53 * t)+.5 check it. Graph it.
2. it goes up and down over a 20 inch altitude. at t=0, it is at zero.
h=-10cos(2PI*10t)+10 is one solution, there are others, depending on where you call h=0. Height at ground level is not always called zero, you could call zero at the top.
ok, change that to days: 29+12/24+44.05/(60*24), I get 29.53 days.
y=.5cos(2PI/29.53 * t)+.5 check it. Graph it.
2. it goes up and down over a 20 inch altitude. at t=0, it is at zero.
h=-10cos(2PI*10t)+10 is one solution, there are others, depending on where you call h=0. Height at ground level is not always called zero, you could call zero at the top.
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