# Math Help - A couple of pre cal questions need help with

1. ## A couple of pre cal questions need help with

1. A buoy bobs up an down with the height H of its transmitter in feet above sea level modeled by h=a sin b t +5. The height varies from 1 ft to 9 ft and it takes 3.5 sec from 9 ft to the next peak. What are the constants a and b?

So far I got to -5=9*sin(b*3.5)

2. A ferris wheel 50 ft in diameter makes one revolution every 40 sec. IF the center is 30 ft above the ground how long after reaching the low point (5 ft above ground) does it take to reach 50 ft above the ground.

I really don't know what to do on this one. I know the circumference and ft a second. But I can't translate that to 50 ft above the ground.

3. A tsunami hit and caused waves that travel at 540 ft sec and reach a height of 60 ft. IF the period is 30 minutes estimate the length between the crests.

In completely lost on this one..

2. Originally Posted by bilbobaggins
1. A buoy bobs up an down with the height H of its transmitter in feet above sea level modeled by h=a sin b t +5. The height varies from 1 ft to 9 ft and it takes 3.5 sec from 9 ft to the next peak. What are the constants a and b?

So far I got to -5=9*sin(b*3.5)
and what exactly are you doing?

you should recognize this from the general form of the sine graph:

a is the amplitude, is is half the length between 1 and 9.

what they described about time being 3.5 sec was describing the period. for a general sine graph of this form, $\mbox{Period } = \frac {2 \pi}b$

3. Originally Posted by bilbobaggins
2. A ferris wheel 50 ft in diameter makes one revolution every 40 sec. IF the center is 30 ft above the ground how long after reaching the low point (5 ft above ground) does it take to reach 50 ft above the ground.

I really don't know what to do on this one. I know the circumference and ft a second. But I can't translate that to 50 ft above the ground.
there are several ways to do this, but to keep in-line with the theme we have going here, let's do it this way:

try to model the height with a sine graph.

we are using: h = a sin(bt) + 30

the height varies from 5 to 55

it takes 40 sec from 55 ft to the next peak

so i made it similar to your last question. can you do it now?

if you don't understand how i set up the equation or know how to interpret the way i did, say so.

4. Originally Posted by bilbobaggins
3. A tsunami hit and caused waves that travel at 540 ft sec and reach a height of 60 ft. IF the period is 30 minutes estimate the length between the crests.
try this one yourself. based on what i did

5. Originally Posted by Jhevon
and what exactly are you doing?

you should recognize this from the general form of the sine graph:

a is the amplitude, is is half the length between 1 and 9.

what they described about time being 3.5 sec was describing the period. for a general sine graph of this form, $\mbox{Period } = \frac {2 \pi}b$
So is it 9=5.5sin(2pi/b*3.5)+5?
Originally Posted by Jhevon
there are several ways to do this, but to keep in-line with the theme we have going here, let's do it this way:

try to model the height with a sine graph.

we are using: h = a sin(bt) + 30

the height varies from 5 to 55

it takes 40 sec from 55 ft to the next peak

so i made it similar to your last question. can you do it now?

if you don't understand how i set up the equation or know how to interpret the way i did, say so.
32.5sin(2pi/b*t)+30. I still don't think I'm quite understanding the bt part. But other than that do they look okay?

6. Originally Posted by bilbobaggins
1. A buoy bobs up an down with the height H of its transmitter in feet above sea level modeled by h=a sin b t +5. The height varies from 1 ft to 9 ft and it takes 3.5 sec from 9 ft to the next peak. What are the constants a and b?
...
Hello,

1. sealevel at 5 ft. (When t = 0 you get h(0) = 5)
2. distance "traveled" by the buoy : 8 ft. That means rising or falling by 4 ft measuring from sealevel.
3. time for one periode: 3.5 s

Now plug in these values into the given equation:

$h(t) = 4 \cdot \sin\left(\frac{2\pi}{3.5} \cdot t \right)$

Thus the constants are ....

7. Originally Posted by earboth
Hello,

1. sealevel at 5 ft. (When t = 0 you get h(0) = 5)
2. distance "traveled" by the buoy : 8 ft. That means rising or falling by 4 ft measuring from sealevel.
3. time for one periode: 3.5 s

Now plug in these values into the given equation:

$h(t) = 4 \cdot \sin\left(\frac{2\pi}{3.5} \cdot t \right)$

Thus the constants are ....
So for #2 it would be 50=32.5sin(2pi/40*t)+5
where 32.5 is halfway between 5 and 50.

8. Originally Posted by bilbobaggins
So for #2 it would be 50=32.5sin(2pi/40*t)+5
where 32.5 is halfway between 5 and 50.
I'm sorry, but: No

first you have to calculate genrally the height above ground with respect to time:

$h(t) = 25 \cdot \sin\left(\frac{2\pi}{40} \cdot t-\frac{\pi}{2}\right) + 30$

The $-\frac{\pi}{2}$ is necessary because you start to measure the rotation angle when the passenger is at his lowest position.

Now you can calculate the time when a passenger has reached a height of 50 ft:

$50 = 25 \cdot \sin\left(\frac{2\pi}{40}\cdot t-\frac{\pi}{2}\right) + 30~\iff~\frac{50-30}{25}= \sin\left(\frac{2\pi}{40}\cdot t-\frac{\pi}{2} \right)$

Solve this equation for t. (I've got approximately 15.9 s)

9. Originally Posted by bilbobaggins
...
3. A tsunami hit and caused waves that travel at 540 ft sec and reach a height of 60 ft. IF the period is 30 minutes estimate the length between the crests.

...
You are supposed to know that the distance d traveled at a speed v in the time t can be calculated by:

$d = v \cdot t$

Transcribe 30 min in secondes: 1800 s

Then the distance from peek to peek is:

$d = 540\ \frac{ft}{s} \cdot 1800 s = 972,000 ft \approx 184.1 \ mi$

10. Originally Posted by earboth
I'm sorry, but: No

first you have to calculate genrally the height above ground with respect to time:

$h(t) = 25 \cdot \sin\left(\frac{2\pi}{40} \cdot t-\frac{\pi}{2}\right) + 30$

The $-\frac{\pi}{2}$ is necessary because you start to measure the rotation angle when the passenger is at his lowest position.

Now you can calculate the time when a passenger has reached a height of 50 ft:

$50 = 25 \cdot \sin\left(\frac{2\pi}{40}\cdot t-\frac{\pi}{2}\right) + 30~\iff~\frac{50-30}{25}= \sin\left(\frac{2\pi}{40}\cdot t-\frac{\pi}{2} \right)$

Solve this equation for t. (I've got approximately 15.9 s)
Could you show me the steps on how to solve for t. I havent done sin equations in a while.

11. Originally Posted by bilbobaggins
Could you show me the steps on how to solve for t. I havent done sin equations in a while.
Hello,

1. You have to use a calculator in radians mode.
2. You'll get only approximate results.
3. The final result will be rounded to a tenth of a second (with a ferris wheel I think that's accurate enough)

$\frac{50-30}{25}= \sin\left(\frac{2\pi}{40}\cdot t-\frac{\pi}{2} \right)~\iff~0.8 = \sin\left(\frac{2\pi}{40}\cdot t-\frac{\pi}{2} \right)$.......Now use the $\sin^{-1}$-function of your calculator:

$0.927295... = \frac{2\pi}{40}\cdot t-\frac{\pi}{2}$....... Add $\frac{\pi}{2}$ and multiply afterwards by $\frac{40}{2\pi}$

You'll get:

$t \approx 15.9033447...~\implies~t \approx 15.9\ s$

12. Originally Posted by earboth
Hello,

1. You have to use a calculator in radians mode.
2. You'll get only approximate results.
3. The final result will be rounded to a tenth of a second (with a ferris wheel I think that's accurate enough)

$\frac{50-30}{25}= \sin\left(\frac{2\pi}{40}\cdot t-\frac{\pi}{2} \right)~\iff~0.8 = \sin\left(\frac{2\pi}{40}\cdot t-\frac{\pi}{2} \right)$.......Now use the $\sin^{-1}$-function of your calculator:

$0.927295... = \frac{2\pi}{40}\cdot t-\frac{\pi}{2}$....... Add $\frac{\pi}{2}$ and multiply afterwards by $\frac{40}{2\pi}$

You'll get:

$t \approx 15.9033447...~\implies~t \approx 15.9\ s$
Thanks