# Math exam question(s).

#### Mozart

My final math exam is tomorrow, and I am having trouble with some questions as I go through this review package.

I have the answers so its very frustrating to arrive at the wrong answer over, and over.

Anyways the question is:

The position of a bucket on a circular tar sands excavator is given by the equation: p(t)= 10sin(6pit + 3pi) + 20

Where t is the time in minutes. There are 20 buckets on the excavator and each holds 500 kg

How many kilograms of tar sands are excavated in one hour?

This is what I did: p(t)=10sin(6pi(60)+3pi) + 20
P(t)= 28.69

from here I am lost so I just did 500 x 20= 10000
10000 x 28.69= 286900 kg

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#### wisredz

How many times have did one excavator come to its starting place? How many times have did 20 excavators come to its starting place? can you answer these questions?

#### FredGarvin

Are you to a calculus level yet? Can you take the derivative of the function you stated? The derivative of position is velocity...angular velocity in this case.

#### Mozart

Well I think the period is .0055555556. I assume its that since im using 6pi(60) as b. So 2pi/b gives me .0055555... When I plug in 60 minutes into the equation I get like 28.69. So I divided 28.69 by the period and got 5164.199959. Then I multiplied 5164.2 by 20 since there is 20 other buckets, and got 103284. I then multiplied this by 500 kg and got 51651999.6 kg.

Sorry Fred I havn't been exposed to any sort of calculus yet.

#### wisredz

Think of the equation this way. The constant c does not change anything actually so we need not pay any heed to it, right? That's because we are interested in the period of this function. now take $sin(6\pi*x+3\pi)$ like this $sin(a+b)$

Can you conclude anything about the function from here? The given function is actually equal to $-10sin(6\pix)$. What can you do now?

Fred, I don't understand, what does this quesation have to do with the derivatives? We do not need to find the velocity. Or am I missing something?

#### wisredz

It seems I cannot edit my posts for some unknown reason. Anyway the last latexed equation should be

$$-10sin(6\pi*x)$$

#### Mozart

I am really confused now. I don't know how you went from sin(6pix+3pi) to
-10sin(6pix). It seems I can't even answer your simple questions. What chance do I have in finding the answer.

#### wisredz

Do you know the sum formula for sine function that says

$$sin(a+b)=sina*cosb+cosa*sinb$$?

All I did was replacing

$$a=6\pi*x, b=3\pi$$

and while doing the calculations heed that

$$cos(3\pi)=-1, sin(3\pi)=0$$ .

Do you understand it now?

#### FredGarvin

wisredz said:
Fred, I don't understand, what does this quesation have to do with the derivatives? We do not need to find the velocity. Or am I missing something?
I took the derivative of the position function with respect to t and used that and the fact that there are 20 buckets and they hold 500 kg. My answer rounded to the one given. Seemed to work ok for me.

#### wisredz

Well from using the period of the function I get the exact answer. As I said, the function is actually

$$-10sin(6\pi*x)$$

The period of this function is 1/3. so, until x=60 this function repeats itself 180 times.

Multiply this by 50*20, and you get the answer.
Could you help me understanding your solution? I still have no idea how to use the deriative.

#### FredGarvin

Like I said, there's a rounding issue, but I did this:

$$P(t) = 10 \sin (6 \pi t + 3 \pi) + 20$$

$$P'(t) = (10)(6 \pi) \cos (6 \pi t + 3 \pi)$$

Evaluate the derivative at t=60 to calculate the number of times the wheel has gone around:

$$P'(60) = 188.45$$
I used absolute value here since the negative value really doesn't mean anything.

To get the total # of kg:
$$Total = (188.45 revs)(20 \frac{buckets}{rev})(500 \frac{kg}{bucket})$$

$$Total = 1.88 x 10^6 kg$$

#### wisredz

I don't see the point still. I mean P'(t) is the velocity and pluggin 60 into this we get the speed of one bucket at time t=60. As far as i am concerned this is not the number of times the wheel has gone around.

Am I wrong?

#### FredGarvin

Wisredz,
You got me to see I forgot a part of my solution!

$$P(t) = 10 \sin (6 \pi t + 3 \pi) + 20$$

$$P'(t) = (10)(6 \pi) \cos (6 \pi t + 3 \pi)$$

Evaluate the derivative at t=60 to calculate the number of ANGULAR VELOCITY of the wheel:

$$\omega = P'(60) = 188.45 \frac{rad}{min}$$
I used absolute value here since the negative value really doesn't mean anything.

Now calculate the number of revolutions the wheel has gone thru:
$$\theta = (188.5 \frac{rad}{min})(60 min)(\frac{1 rev}{2 \pi rad})$$
$$\theta = 1800 rev$$

To get the total # of kg:
$$Total = (1800 revs)(20 \frac{buckets}{rev})(500 \frac{kg}{bucket})$$

$$Total = 18.00 x 10^6 kg$$

Thanks for keeping me honest!

#### wisredz

Anytime :)

Mozart, now you have two solutions to choose from. Mine is just a simple trigonometric solution, Fred's is a much more explanatory of the motion because he uses the derivative to find the angular velocity, which gives a lot more info about the info. Both of the solutions give the correct answer so here ya go!

Cheers

#### Mozart

Hey thanks guys, and sorry for the late response, but some genius decided to put my exam on the day of the prom, and well I was kind of out of it for a few days. Anyways the exam turned out to be quite simple. Nothing was as hard as this question. I know it's not really a hard question for you guys but for me its pretty complicated. Well thanks again.

#### Gokul43201

Staff Emeritus
Gold Member
Actually, Fred's answer is high by a factor of 10.

And you don't need to take the first deriviative to find $\omega$. The equation you have is of the form :

$$x = R sin(\omega t + \phi )$$

You just have to read off the coefficient of 't' to find the angular velocity $\omega$. In this case, $\omega = 6 \pi$. The rest follows as Fred described.

Fred, you found $\dot{x} = v = R \omega$. You must divide by R = 10, to get the right answer.

#### FredGarvin

ahhhhhhhhhhhhhhhhh......crud. You're right.

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