I'm stuck on several trig questions:
Let x= -2sin(2PIt) + cos(4PIt) +5 be in metres and t in minutes. Use calculus to find the rate of change of x with respect to time t, and find the value of this rate of change when t=4 with an answer correct to one decimal place. Find the first minimum value of x after t=0 correct to the nearest one hundredth of a minute. Find this minimum value of x correct to the nearest millimetre.
(PI= well, pi. I don't know how to type the symbol :S)
I found dx/dy= -4PIcos(2PIt)-4PIsin(4PIt)
I tried to find the minimum, but got stuck...
Solve the following equation for 0<x<2PI (the one of the right is a 'less than or equal to'): sin(x) = cos^2(x)
A ferris wheel has been proposed. Such wheels rotate at a constant (slow) rate. You get on and off as the car moves slowly past the entrance platforms: the wheel does not stop at any time. Your tirp is one revolution. Your feet must be kept on the floor at all times for safety reasons. The height, h metres of your feet above the platform at time, t hours is given by
h(t) = 51 + 50sin ((5t-1)PI)/2)
where t is measured from the time you get on.
Find the time of the first ocasion after 10.30am, that your feet are at the height 76m above the platform.
Find the number of minutes during one rotation that your feet are at least 76m above the platform.
I got stuck somewhere whilst trying to do this...
Let x= -2sin(2PIt) + cos(4PIt) +5 be in metres and t in minutes. Use calculus to find the rate of change of x with respect to time t, and find the value of this rate of change when t=4 with an answer correct to one decimal place. Find the first minimum value of x after t=0 correct to the nearest one hundredth of a minute. Find this minimum value of x correct to the nearest millimetre.
(PI= well, pi. I don't know how to type the symbol :S)
I found dx/dy= -4PIcos(2PIt)-4PIsin(4PIt)
I tried to find the minimum, but got stuck...
Solve the following equation for 0<x<2PI (the one of the right is a 'less than or equal to'): sin(x) = cos^2(x)
A ferris wheel has been proposed. Such wheels rotate at a constant (slow) rate. You get on and off as the car moves slowly past the entrance platforms: the wheel does not stop at any time. Your tirp is one revolution. Your feet must be kept on the floor at all times for safety reasons. The height, h metres of your feet above the platform at time, t hours is given by
h(t) = 51 + 50sin ((5t-1)PI)/2)
where t is measured from the time you get on.
Find the time of the first ocasion after 10.30am, that your feet are at the height 76m above the platform.
Find the number of minutes during one rotation that your feet are at least 76m above the platform.
I got stuck somewhere whilst trying to do this...