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Precalculus Help Needed
Robos A Go Go
Registered User regular
Registered User regular
I need some help preparing for a Precalc test that I'm taking next week. After some review, I've managed to do alright on the practice test I was given, but two questions are still giving me trouble.
1. h(t)= 64 - 46cos(π/5*t), where 0<t<10
(Sorry the pi symbol looks like an "n.")
The function h above gives the height above the ground, in feet, of a passenger on a Ferris wheel t minutes after the ride begins. During one revolution of the Ferris wheel, for how many minutes is the passenger at least 100 feet above the ground? Round your answer to the nearest hundredth of a minute.
Answer: 2.14
2. How many different values of x satisfy the equation:
sinx + 2sin(2x) = √x
Answer: Five
I've included the answers to help you double check your work (and prove that I'm not looking to have you do my homework). Please, just tell me how to solve these problems including the laws, formula, or identities you used to get your answers. Thanks.
1. h(t)= 64 - 46cos(π/5*t), where 0<t<10
(Sorry the pi symbol looks like an "n.")
The function h above gives the height above the ground, in feet, of a passenger on a Ferris wheel t minutes after the ride begins. During one revolution of the Ferris wheel, for how many minutes is the passenger at least 100 feet above the ground? Round your answer to the nearest hundredth of a minute.
Answer: 2.14
2. How many different values of x satisfy the equation:
sinx + 2sin(2x) = √x
Answer: Five
I've included the answers to help you double check your work (and prove that I'm not looking to have you do my homework). Please, just tell me how to solve these problems including the laws, formula, or identities you used to get your answers. Thanks.
Robos A Go Go on
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36 = - 46cos(π/5*t)
arccos(-.783) = π/5*t
{3.93, 6.07} = t
6.07-3.93 = 2.14
I can get 3.93 with my calculator, but how do you get 6.07?
https://twitter.com/Hooraydiation
In this case, when you take the inverse cosine of a number, what you are really doing is asking "the cosine of what angle will give me this number?" Then, you have to recognize that what the cosine represents is, for a given angle in the unit circle, is the x value of the point on that circle. Since it's only the x value, you need to recognize that there are two distinct angles that will both give the same result if you take the cosines of those angles. In this case, plugging cos^-1(.783) in your calculator gives you 2.47 radians, or 141.5 degrees. If you subtract that result from 360, which gives you 218.5, this angle will give you the same value when you take its cosine.
Then, you just finish the problem with this second value for the angle. To be really rigorous, you should realize that these are just boundary conditions (do I know if the ferris wheel is going up or down at these points?), and you should check a value in between those two values to make sure that the height of the ferris wheel is above the 36 feet.
The second problem involves identities and I hate those.
sin(x) + 2sin^2(x) - sqrt(x)
(I assume this is what you mean because the equation with 2sin2x doesn't have 5 values)
The points where that graph intersect the positive x axis are the solutions, since at those points
sin(x) + 2sin^2(x) - sqrt(x) = 0
which is what you want. By inspection, there are 5 values.
(Note: There are easier ways to do this than the below, but I'm trying to keep it to a pre-calc level and I think the function-shape analysis would be useful to you in the future)
If you don't want to solve it via graphing:
sin(x) + 2sin^2(x) - sqrt(x) = 0 => sin(x)(1 + 2sin(x)) - sqrt(x) = 0
The trivial solution is zero. sqrt(0) = 0, sin(0) = 0
After that, we can figure out how many times the graph (if we were to draw one) would intersect the x axis by considering the amplitude of the left hand side versus the amplitude of the right hand side.
The equation sin(x) + 2sin^2(x) has a maximum amplitude of 3:
max(sin(x)) = 1 => max(sin(x)(1 + 2sin(x)) = (1*(1 + 2*1)) = 3
Sqrt(x) = 3 => x = 9
Therefore, the graph of sin(x) + 2sin^2(x) - sqrt(x) will be entirely below the x axis once x is greater than 9.
How many times does sin(x) + 2sin^2(x) intersect the x axis between x = 0 and x = 9?
Well, by itself sin(x) would be zero at 0, pi, and 2pi (also at all higher values of n*pi, but by 3pi the sqrt will have dragged the graph down below the axis). These values correspond to the times when the sin function transits across the axis on its way to a peak or a trough. The peak is at pi/2 and then there's a trough at 3pi/2 and another peak at 5pi/2. So at this point we have 3 x's that satisfy the equation.
The sin^2 term is never negative, and because the function is sin(x) + 2*sin^2(x), the sin^2 term is going to dominate when sin(x) is much less than zero. So, without looking at the graph, we can guess that when sin(x) is a little bit less than zero, the sin term will dominate, making the value negative, because the square of a small negative number is an even smaller positive number. But once sin(x) gets past about -0.5, the sin^2 term dominates and pulls it back up above the axis. This means that our trough at 3pi/2 isn't actually a trough: it's a mini-peak. But if we go below zero briefly before rising up to the mini-peak, that's another value for our equation, bringing us up to 4. After the mini-peak at 3pi/2, the function dives back down toward the axis, dipping below again when the sin term dominates over the sin^2 term (as sin(x) approaches zero on its way toward its next peak). So that's a 5th value.
We know that the next mini-peak would correspond to the second trough in sin(x), which is at 7pi/2. We don't even make it to the zero at 3pi, so we don't need to worry about that one. There must not, then, be any more valid values.