Homework Help Forum: Maths

Posted by Melody on Tuesday, October 13, 2009 at 8:02am.

I have a trig question regarding a ferris wheel, radius 35m, axles 37m, clockwise direction, rotates twice every 10 mins. I have calculated the following function: h = -35cos(2pi*X/5)+37 I have calculated at what height I would be after 3 minutes by substituting X=3 in the above equation. However, I need to calculate at what times I would be 30m off the ground. Can you please tell me if I am on the right track and how to calculate the time? Many thanks

• Maths - Reiny, Tuesday, October 13, 2009 at 8:28am

  I agree with your equation, so let' set it equal to 30 30 = -35cos(2pi*X/5)+37 -7 = -35cos(2pi*X/5) .2 = cos(2pi*X/5) take the inverse cosine to get 2pi*X/5 = 1.369438 x = 1.09 minutes check it by repeating the same steps you did finding the height at 3 minutes.

  
  

• Maths - Melody, Wednesday, October 14, 2009 at 6:26am

  Many thanks, I understand that now. I have another question I am stuck on also. How do I get an equation for the height of the car on the ferris wheel over a 10 minute period if we start the analysis one minute into the ride. Can I still use the original formula h = -35cos(2pi*X/5)+37 I don't understand what part changes to make it one minute into the ride. Appreciate any help.

  
  

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