A weight is oscillating on the end of a spring. The position of the weight relative to the point of equilibrium is given by y=1/12(cos8t-3sin8t), where y is the displacement (in meters) and t is the time (in seconds). Find the times when the weight is at the point of equilibrium (y=0) for 0<=t<=1.

I have it worked out as:
0=1/12 (cos8t-3sin8t)
cos8t=3sin8t
tan8t=1/3
8t=arctan 1/3
t= 1/2 arctan 1/3

And now I'm stuck and I need three times. Can anyone please help?

you are ok up to

8t = arctan 1/3
make sure your calculator is set to radians, and then
8t = .32145 or 8t = π + .32145
t = .04022 or t = .43292
we know the period of tan 8t is 2π/8 or .7854
so we could add this to any previous answer to get more correct answers, as long as we don't go outside our domain of 0≤t≤1
so t = .04022, .43292, and .82562

testing the last time of .82562
I get y = appr 10^-9 , not bad

I redone it and I did:

0=1/12(cos8t-3sin8t)
cos8t=3sin8t
8=3tan8t
8/3=tan8t
arctan8/3=8t
1.212=8t
t=.1515

I think this one is correct. However, I'm still stuck on how I can get three times. Can someone please help?

your step from

cos8t=3sin8t
to
8 = 3tan8t is totally invalid, that would be pure mathemagics

test your value, it does not work

Alright, thank you Reiny. I understand it now!

Sure, I'd be happy to help! But before we continue, I have to warn you that my answers might come with a touch of humor. So, brace yourself for some laughter! Now, let's get back to your question.

To find the times when the weight is at the point of equilibrium (y=0), we need to solve the equation:

0 = 1/12 (cos(8t) - 3sin(8t))

You've already done a great job by simplifying it to:

cos(8t) = 3sin(8t)

Now, let's move forward and solve this equation. But first, let's do a little trigonometric magic. What if we divide both sides by cos(8t)?

1 = 3tan(8t)

Now, remember that tan(8t) is equal to sin(8t) divided by cos(8t). So, we can rewrite our equation as:

1 = 3(sin(8t)/cos(8t))

Now, let's play a little trick that might help us out. Imagine you're at a circus, and I, the Clown Bot, am giving you a funny magic show. Ding-ding! I will multiply both sides of the equation by cos(8t). Here we go:

1 * cos(8t) = 3(sin(8t)/cos(8t)) * cos(8t)

cos(8t) = 3sin(8t)

Well, well, well, lookie here! We're back to where we started! But don't stress, my friend. This means that both sides of the equation are equal, and that's how we solve it. So, let's go through the steps again to refresh your memory:

cos(8t) = 3sin(8t)

Now, dividing both sides by cos(8t)...

1 = 3tan(8t)

And now, let's solve for t by taking the arctan of both sides:

t = (1/8)&bull;arctan(1/3)

Remember, we're only interested in the times when t is between 0 and 1. So, let's substitute our expression for t:

t = (1/8)&bull;arctan(1/3) = Approximately 0.133

Since we want three times, let's try some values for the arctan function:

1st time: t = (1/8)&bull;arctan(1/3) = Approximately 0.133
2nd time: t = (1/8)&bull;(arctan(1/3) + π) = Approximately 0.904
3rd time: t = (1/8)&bull;(arctan(1/3) + 2π) = Approximately 1.674

And there you have it! Three times when the weight is at the point of equilibrium (y=0) for 0 <= t <= 1. Remember, laughter is the best equilibrium, so keep smiling!

To find the times when the weight is at the point of equilibrium (y=0), we need to solve the equation y=1/12(cos8t-3sin8t) for t.

You have correctly simplified the equation to cos8t=3sin8t. Since we know that tanθ=sinθ/cosθ, we can rewrite the equation as tan8t=1/3.

To solve for t, we can take the inverse tangent (arctan) of both sides of the equation: arctan(tan8t) = arctan(1/3). This simplifies to 8t = arctan(1/3).

Now, we need to find the value of t. Divide both sides of the equation by 8 to isolate t: t = (1/8) * arctan(1/3).

To find three values of t within the specified range 0<=t<=1, we can substitute different values of t into this equation.

For example:
1. Let's substitute t = 0 into the equation: t = (1/8) * arctan(1/3) = (1/8) * 0 = 0.
2. Let's substitute t = 0.5 into the equation: t = (1/8) * arctan(1/3) ≈ (1/8) * 0.321750554 = 0.040218819.
3. Let's substitute t = 1 into the equation: t = (1/8) * arctan(1/3) ≈ (1/8) * 0.321750554 = 0.080437638.

Therefore, the three times when the weight is at the point of equilibrium (y=0) for 0<=t<=1 are approximately t = 0, t ≈ 0.040218819, and t ≈ 0.080437638 seconds.