Can someone please help me with Calc?

[Syringer]

Here is the problem:

The position function of a particle is given by r(t) = < t^2, 5t, t^2 - 16t >. When is the speed a minimum?

The answer is t = 4.

I don't understand how the book arrived at that answer.

My thinking was that I would take the derivitive of the vector and find the magnitude of it, which would give me the speed and then set it to 0 to find when t would be a minimum. But that didnt' work. So I am now in need of help with this problem.

Any help is greatly appreciated.

 

[DannyLove]

this is off-topic:

i've been noticing many threads like this and I've also noticed that some say to do your own work, etc. etc. Don't you think it is best if you ask a teacher, or tutor, or classmates? i mean, i'm sure you'll get input here, but don't you feel that face to face conversation input is much better than this forum?

just a thought

danny~!

 

[MustPost]

shh, shutup danny the real answer is 42

 

[ThisIsMatt]

nevermind, I think I'm screwed up

 

[RaynorWolfcastle]

ok take the derivative of that function to get the velocity-time graph. v(t)
Take another derivative to find the acceleration-time graph. a(t)
find all t values when a(t) = 0
determine whether they are mins or maxes
plug the t value of t min into v(t) And Voil&agrave;! Minimum velocity

Btw, looks like you got yourself a piecewise function on your hands you need to get ranges for where each part of the function applies


-Ice

 

[JeremyJoe]

im not sure what you mean by that position is it a 3-dementional graph?

since that equations is the position you need to take the derivative of it to get the speed or velocity, but im not very sure how you proposed it is very clear

 

[ThisIsMatt]

I was right the first time I think...

Syringer - position is the derivative of velocity. If you were given the velocity function, to find the minimum speed you would differentiate and set equal to zero, right? Well, this is already differentiated, so set the eq's equal to zero.

 

[Syringer]

It dones't work if you take the 2nd derivative because then you are left with <2, 0, 2> which wouldn't give you anything for finding a value for t. Plus, I am trying to find t.

So I still need to know how the book got the answer t = 4. Then I will have solved the problem.

 

[RaynorWolfcastle]

Ermm... ThisIsMAtt......

position is the INTEGRAL of speed or velocity btw you need a constant of integration if you do that
velocity is the DERIVATIVE of postition

-Ice

 

[RaynorWolfcastle]

lookie here
v(t) = 2t, 5, 2t-16
a(t) = 2, 0, 2

because this is a piecewise function i assume that in this case the function joins at

t^2 = 5t
t = 5

and at

5t = t^2 - 16t
t=21

you should be able to find the solution from here, what's the fun if I give you the answer?

-Ice

 

[Syringer]

iccool83 - Now this is a position funtion. r(t) gives you the position. I'll write it in i, j, k format for you.

Then it becomes r(t) = (t^2) i + (5t) j + (t^2 - 16t) k.

I already know the answer. The answer is t = 4.

I need to find out WHY it is t = 4

I would also appreciate anyone else who can help me with this problem.

 

[quoc]

<< I was right the first time I think...

Syringer - position is the derivative of velocity. If you were given the velocity function, to find the minimum speed you would differentiate and set equal to zero, right? Well, this is already differentiated, so set the eq's equal to zero. >>



No, you've got it backwards--velocity is the derivative of position.

icecool83: I believe it's not a piecewise function, but a 3-dimensional vector.

Anyways, I was able to get the answer you gave. Here is how I arrived at that solution.

1. Ok, so you have the function for position:
r(t) = < t^2, 5t, t^2 - 16t >

2. The velocity is the derivative of this:
v(t) = < 2t, 5, 2t - 16 >

3. The speed is the magnitude of the velocity, which is the square root of the sum of the squares of the velocity vectors:
s(t) = Sqrt( 4t^2 + 25 + (4t^2 - 64t + 256) ) = Sqrt( 8t^2 - 64t + 256 )

4. So that is the function for speed. The minimum speed will occur when the derivative of the speed is equal to 0. So next, find the derivative of the speed.
s'(t) = 1/2 * (16t - 64)^(-1/2)

5. Set that equal to 0, and solve for t.
s'(t) = 1/2 * (16t - 64)^(-1/2) = 0
16t - 64 = 0
16t = 64
t = 4

There is your solution. Please think about how I arrived at the solution, and what each of the equations represents. Like, equation 2 is the derivative of equation 1, so it represents the velocity (because velocity is the derivative of position, it is the rate of change of the position). Equation 3 is the magnitude of velocity, therefore it is the speed. And to find the minimum speed, you find when the derivative is equal to 0. I hope this helps.

 

[quoc]

<< I need to find out WHY it is t = 4 >>



Yes, knowing WHY is what is most important in math. Please post if you didn't understand any of the steps I took in finding the solution, and I'd be happy to explain it...

 

[RaynorWolfcastle]

Hmmmm... 3-d vector, makes sense now that you say it. Just force of habit of solving 2-d problems i guess. Plus my teachers always put vectors in square brackets...

Anyways, I didn't look in great detail but the method quoc used seems right.

-Ice

 

[quoc]

<< Hmmmm... 3-d vector, makes sense now that you say it. Just force of habit of solving 2-d problems i guess. Plus my teachers always put vectors in square brackets...

Anyways, I didn't look in great detail but the method quoc used seems right.

-Ice >>



Yeah, I think vectors should be in square brackets... I was following Syringer's convention without really thinking about it.

 

[Syringer]

Thanks quoc for the help.

I see how you arrived at the answer. So you are saying that I need to find the derivative of speed, which is also called the acceleration, and set the acceleration function equal to 0 to find the answer. Correct?


And icecool38 I understand how you got confused. Maybe I should have been clearer. Thanks for taking the time to help.

 

[quoc]

<< Thanks quoc for the help.

I see how you arrived at the answer. So you are saying that I need to find the derivative of speed, which is also called the acceleration, and set the acceleration function equal to 0 to find the answer. Correct? >>


Correct, although the derivative of the speed here is not exactly the same as acceleration. Just make sure you understand the difference between velocity and speed. Velocity is a vector, and each component of the vector is a speed in a different direction (i, j, k). Speed is a scalar--it is the magnitude of the velocity vector. It would be easier to explain with a picture, but it seems like you already understand it. Acceleration would actually be the derivative of the velocity, and it would also be a vector. That's why you couldn't find minimum speed from the derivative of velocity ( < 2, 0, 2 > ), because this only tells you what is happening to the velocity in each direction.

 

[Syringer]

quoc
What would the derivative of the speed function be called? If derivative of velocity is acceleration... then what would be derivative of speed be? I don't know why we need derivateive of speed to find when the speed would be a minimum.

 

[SUBxWRX]

<< Here is the problem:

The position function of a particle is given by r(t) = < t^2, 5t, t^2 - 16t >. When is the speed a minimum?

The answer is t = 4.

I don't understand how the book arrived at that answer.

My thinking was that I would take the derivitive of the vector and find the magnitude of it, which would give me the speed and then set it to 0 to find when t would be a minimum. But that didnt' work. So I am now in need of help with this problem.

Any help is greatly appreciated. >>




lol syringe wat school u go to?....I go to Whitney by the way

 

[RaynorWolfcastle]

quoc, that's interesting, they never taught me anything about speed vs velocity. They always use them interchangeable and use i,j,k notation, instead of using scalars they just assign it a new vector like r or something. Or they just say "the magnitude of the velocity is xyz m/s"

-Ice

 

[ThisIsMatt]

Sorry, it's been too long for me

 

[quoc]

<< quoc
What would the derivative of the speed function be called? If derivative of velocity is acceleration... then what would be derivative of speed be? I don't know why we need derivateive of speed to find when the speed would be a minimum. >>



I would just call it... the derivative of speed. You need it to find the minimum, because that's the way it is for any function. When the derivative of a function is equal to zero, then the function is at either a maximum or minimum. Think about what the derivative represents--it is the rate of change. So if the rate of change is zero, it has reached some kind of peak. Again, easier to explain with a picture. Do you get how that works now?

 

[Syringer]

quoc
Okay, I will accept it as true since it obviously worked the problem and will work for future ones. I will take your word for it. It makes some sense now. Thanks for the help



And SUBxWRX I go to Cerritos....I know of a school called Whitney. No one at Cerritos likes the school much for some reason.

 

[SUBxWRX]

<< quoc
Okay, I will accept it as true since it obviously worked the problem and will work for future ones. I will take your word for it. It makes some sense now. Thanks for the help



And SUBxWRX I go to Cerritos....I know of a school called Whitney. No one at Cerritos likes the school much for some reason. >>




fair enough....but at least we stole your best calc teacher....mr murray

 

[RossGr]

velocity is always a vector quantity, Speed is the magnitude of the velocity. Physics stuff.