I can't figure out this one question...
A ship is docked in port and rises and falls with the waves. The function d(t) = 2sin(30t)°+5 models the depth of the propeller, d(t), in metres at t seconds.
e) within the first 10 seconds, at what times is the propeller at a depth of 3m?
How do you do this problem?? The answer is 9 seconds.
Is there another way to solve this problem? Without using π
so set 2sin (30t) + 5 = 3
2sin (30t) = -2
sin (30t) = -1
We know sin (3π/2) = -1
so 30t = 3π/2
t = π/20 = .157 seconds
since the period of the function is 2π/30 = appr .209 seconds
this boat is bouncing up and down about 5 times a second !!
This question is flawed in a major way.
you define t as seconds, yet in the equation you include the ° sign to show degrees. You can't mix units like that.
If you solve it with 30t in degrees, then the wording makes no sense
I hope they did not do this:
sin(30t) = -1
30t = 270 degrees
t = 9 degrees !!!!!!
Looks like they did!!! WOW
I haven't learned any of this with π incorporated into it. I think my teacher mentioned that we do it in grade 12? Radians and such... I'm in grade 11 though.
This question is straight out of the textbook so I guess they did do that.. And they actually mix units like that a lot! Many of the questions in this section of the textbook have equations that look like the one I posted
π radians is 180°
Well, it seems like the propeller is having a little dance with the waves! Let's break it down, shall we?
We're given the equation d(t) = 2sin(30t)° + 5, where d(t) represents the propeller depth at time t. We want to find out when the propeller is at a depth of 3 meters within the first 10 seconds.
First, let's set up the equation:
2sin(30t) + 5 = 3
Now, we need to solve for t. Let's subtract 5 from both sides:
2sin(30t) = -2
Divide both sides by 2:
sin(30t) = -1
Now, if we take the inverse sine (or arcsine) of both sides, we find:
30t = arcsin(-1)
The arcsine of -1 is equal to -π/2 (or -90 degrees), but let's stick to radians here. So we have:
30t = -π/2
Now, divide both sides by 30:
t = -π/60
Hmm, we're dealing with time here, and negative time doesn't make much sense, unless our propeller is a time traveler! That can't be right.
So, let's think about what sine does. It oscillates between -1 and 1 within the interval from 0 to 2π (360°). Since we're interested in the first 10 seconds, let's focus on the interval from 0 to 2π.
Now, when does sine equal -1 within this interval? Well, there's one time when it crosses that point, and that's when it's at 225 degrees or 5π/4 radians.
So, t = 5π/4. And if we calculate that, we find that t is approximately 3.92 seconds.
Oh wait, my clown nose is tingling! It seems I made an error before. That's not the answer we're looking for.
Upon closer inspection, we realize that 5π/4 is actually outside the first 10 seconds interval. Silly me! I got carried away by the waves.
In reality, there's no solution to the equation within the first 10 seconds. So sadly, the propeller doesn't reach a depth of 3 meters within that time frame.
But hey, at least we learned something along the way! Just be careful when dancing with those trigonometric functions, they can be quite sneaky.
To solve this problem, we need to find the values of t that make the propeller depth equal to 3 meters within the first 10 seconds.
The given function is d(t) = 2sin(30t) + 5, where d(t) represents the depth of the propeller in meters at time t seconds.
To find the times at which the propeller is at a depth of 3 meters, we need to set up an equation as follows:
2sin(30t) + 5 = 3
First, let's isolate the sine term by subtracting 5 from both sides of the equation:
2sin(30t) = 3 - 5
2sin(30t) = -2
Next, divide both sides of the equation by 2 to isolate the sine term:
sin(30t) = -1
Now, we need to find the values of t that make the sine function equal to -1 within the first 10 seconds.
Recall that the sine function takes on the value of -1 at two specific angles: -π/2 and -3π/2 radians.
Converting these angles to degrees, we have -90° and -270°.
Since we are dealing with the time in seconds, we need to convert these angles to seconds by dividing them by 30 (the coefficient of t in the function):
-90° / 30 = -3 seconds
-270° / 30 = -9 seconds
Now we have two possible values for t: -3 seconds and -9 seconds.
However, we only need to consider values within the first 10 seconds, so the only valid solution is t = -9 seconds.
Therefore, within the first 10 seconds, the propeller is at a depth of 3 meters at 9 seconds.
Note: It's important to double-check the given solution to ensure it is within the defined range. In this case, 9 seconds is indeed in the range of the first 10 seconds.