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 T O P I C    R E V I E W uberclay Posted - 03/01/2011 : 00:32:34 Greetings. I'm wrapping up a precalc refresher and need some help getting through this problem.... A truck is moving so that the axle, which is 6cm in diameter, rotates once per second. As the truck hits a bump, the shocks depresses by 20cm and then continue to depress by 80% of the previous bounce as the truck bounces every half-second. If the middle of the axle is 30cm from the bottom of the truck bed construct a graph illustrating the distance from a point at the top of the axle when the truck hits the bump, to the bottom of the truck bed.What is the equation of the graph?For the solution I have come up with two functionsf(x) = 3sin[2(x + 1/4)]models the rotation of the point on the axle (starting at the top by the horizontal translation of x by 1/4), moving about a circle with a diameter of 6cm, and having a period of 1 second. (sin used here by requirement)andg(x) = 30 - (20 * 0.8^2x)models the distance from the center of the axle to the bottom of the truck bed (30 cm) from the initial depression (-20cm), and through each successive bounce, every 1/2 second.As I need f(x) to follow the graph of g(x) I have combined them byy = g(x) - f(x) ory = [30 - (20 * 0.8^2x)] - 3sin[2(x + 1/4)]However, I am not entirely confident that this is the proper solution - although it looks good on paper.*edited question for readability 4   L A T E S T    R E P L I E S    (Newest First) uberclay Posted - 05/17/2011 : 21:21:12 I wish I would have thought to google this question in the first place. Although there are many places to find an "answer"; I preferred to construct my own.The behavior of the point on the axle can be modeled accordinglyf(x) = 3sin[2(x + 1/4)]+30models the rotation of the point on the axle (starting at the top by the horizontal translation of x by 1/4), moving about a circle with a diameter of 6cm, having a period of 1 second, rotating about the center of an axle which is 30cm from the truck bed. (sin used here by requirement)g(x) = -10cos{4x)-10models the maximum depression of the springs as a periodic event, every 1/2 second (assuming the springs depress, return to their original position, depress again, and so on). This gives negative values for the depressions over time.h(x) = (0.8)^2xIs the exponential decay (by 20%) of the bouncing, every half-secondThe solution can be found by I(x) = f(x) + g(x)*h(x) uberclay Posted - 03/15/2011 : 00:27:04 I suppose my question is about the bounce of the truck.I have produced an exponential function to predict the distance of the bounce at time, but isn't bouncing periodic? And would it better be represented using a trigonometric function?Assume that the wheel remains in contact with the road and the "bounce" is that of the truck bed sinking downward toward the axle. At time 0, the truck bed would be 30-20 cm from the axle center. As the period is given as 1/2 second per bounce, I would assume that for all (2k+1)/4 {k Z} seconds the truck bed would return to 30 cm above the center of the axle and then sink back down over the next 1/4 second.The data from the bouncing would look like:(x, y)0, -20.25, 0.5, -16.75, 01, -12.81.25, 0 1.5, -10.24When graphed, this is clearly a periodic function - but I am currently unable to construct a function that will produce the behavior:{y| 0 >= y >= -20 * 0.8^(2x)} uberclay Posted - 03/03/2011 : 18:27:25 That is what I have taken "illustrating the distance from a point at the top of the axle when the truck hits the bump" to mean. I believe the data for y = g(x) - f(x) should look like(x, y)(0, 7cm)(0.5, 17)(1, 14.2)(1.5, 22.76)(2, 18.808)...I am uploading the graph and will post it when processing completes.Thank you. royhaas Posted - 03/03/2011 : 12:39:58 What happens at x=0? As that when the truck hits a bump?

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