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#1 Mahmoud

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Posted 09 October 2008 - 08:33 PM

Okay, well I've been working on these differentiation world problems all day and I have gotten every single one of them write. So, I know how to do these problems but this problem just blew me away.

A boat is pulled into a dock by means of a winch 12 feet above the deck of the boat (see figure).
The winch pulls in rope at a rate of 4 feet per second. Determine the speed of the boat when there is 13 feet of rope out. What happens to the speed of the boat as it gets closer to the dock?

Note: I will be looking for a picture of the figure online so that I can post it here.

Can somebody please explain how to work this problem or at least provide me with the formula I should use?

Edited by Mahmoud, 14 October 2008 - 03:44 PM.


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#2 No1

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Posted 09 October 2008 - 08:41 PM

oh, smeg... I need to learn how to do these sorts of problems for a calculus test tomorrow.

I'll see if I can do anything with it. :wub: Can't guarantee I'll be of any help even if I do get it figured out, though.
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#3 Invictus

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Posted 09 October 2008 - 08:50 PM

View PostMahmoud, on Oct 9 2008, 10:33 PM, said:

Okay, well I've been working on these differentiation world problems all day and I have gotten every single one of them write. So, I know how to do these problems but this problem just blew me away.

A boat is pulled into a dock by means of a winch 12 feet above the deck of the boat (see figure).
The winch pulls in rope at a rate of 4 feet per second. Determine the speed of the boat when there is 13 feet of rope out. What happens to the speed of the boat as it gets closer to the dock?

Note: I will be looking for a picture of the figure online so that I can post it here.

Can somebody please explain how to work this problem or at least provide me with the formula I should use?

Not quite sure on how to do this using differentition, seeing as what you are given is the velocity. I'm quite certain this is an integral question.

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#4 No1

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Posted 09 October 2008 - 08:52 PM

View PostLilich Jr, on Oct 9 2008, 10:50 PM, said:

View PostMahmoud, on Oct 9 2008, 10:33 PM, said:

Okay, well I've been working on these differentiation world problems all day and I have gotten every single one of them write. So, I know how to do these problems but this problem just blew me away.

A boat is pulled into a dock by means of a winch 12 feet above the deck of the boat (see figure).
The winch pulls in rope at a rate of 4 feet per second. Determine the speed of the boat when there is 13 feet of rope out. What happens to the speed of the boat as it gets closer to the dock?

Note: I will be looking for a picture of the figure online so that I can post it here.

Can somebody please explain how to work this problem or at least provide me with the formula I should use?

Not quite sure on how to do this using differentition, seeing as what you are given is the velocity. I'm quite certain this is an integral question.
Rates of change.

It isn't.


e: I got 8. :wub: [8 what, you ask? I dunno.]
e2: wait, yeah, I did a couple things wrong.
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#5 Rashy

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Posted 09 October 2008 - 10:25 PM

I think I'll need a picture to confirm, but I'll try to answer it anyways.

Think of it like a right angled triangle. The hypotenuse is the length of the cable. The height is 12ft, which remains constant throughout. Then comes the horizontal distance (from the dock to the boat), which also changes.

Now, you know that z^2 = x^ + y^2 (where x is the horizontal distance, y is the vertical distance, z is the length of the cable. Not that y^2 is constant).

Differentiate the above in terms of TIME.
2z*dz/dt = 2x*dx/dt + 0 (as per chain rule). The 0 comes from differentiating a constant.

You know dz/dt = 4ft/s. You know z = 13ft. You know x (you can calculate it when z = 13, y = 12, using pythagoras). The only thing you don't know is dx/dt, which is what you want to find.

Classic related rates problem.

e: I assumed that it looks like a triangle, but I would like a diagram to confirm. You should just draw it on ms paint, roughly, I don't need an exact one.
e2: oops, mixed up horizontal and vertical.

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#6 Mahmoud

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Posted 10 October 2008 - 03:43 PM

View PostRashdan, on Oct 10 2008, 12:25 AM, said:

I think I'll need a picture to confirm, but I'll try to answer it anyways.

Think of it like a right angled triangle. The hypotenuse is the length of the cable. The height is 12ft, which remains constant throughout. Then comes the horizontal distance (from the dock to the boat), which also changes.

Now, you know that z^2 = x^ + y^2 (where x is the horizontal distance, y is the vertical distance, z is the length of the cable. Not that y^2 is constant).

Differentiate the above in terms of TIME.
2z*dz/dt = 2x*dx/dt + 0 (as per chain rule). The 0 comes from differentiating a constant.

You know dz/dt = 4ft/s. You know z = 13ft. You know x (you can calculate it when z = 13, y = 12, using pythagoras). The only thing you don't know is dx/dt, which is what you want to find.

Classic related rates problem.

e: I assumed that it looks like a triangle, but I would like a diagram to confirm. You should just draw it on ms paint, roughly, I don't need an exact one.
e2: oops, mixed up horizontal and vertical.

Ah okay, got it. Thanks a lot. And, by the way, the diagram was a right triangle :wub: and exactly how you explained what you thought it would look like.
One question comes to mind though. Why did the y become 0?

Edit: Wait, I think I got it. Is it because when you plug in 12 for y, you get 144. Then the derivative of 144 is 0. Is that right?

Edited by Mahmoud, 10 October 2008 - 03:44 PM.


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#7 Rashy

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Posted 10 October 2008 - 04:13 PM

Basically, yeah. Differentiate any constant and you get 0. You don't even have to do y^2 first in order to see that. if y is constant, so is any power of y, or any root (sqrt, cuberoot, etc) of y.

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#8 Mahmoud

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Posted 10 October 2008 - 11:01 PM

Okay. Thanks for the help. :wub:

Edited by Mahmoud, 10 October 2008 - 11:01 PM.


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#9 sozo91

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Posted 13 October 2008 - 02:16 PM

Alright. I don't have a calculator, but I have the concept.

Let's call the length of the rope s, the height y, and the distance from the pier or w/e x. We have s^2 = y^2 + x^2.

Then 2s ds/dt = 2y dy/dt + 2x dx/dt. What we want to find is the speed of the boat, which is dx /dt.

We know ds/dt = 4 ft/s, y = 12 feet, and s = 13 feet. Then x must be equal to (13^2 - 12^2) ^ 1/2 = 5 feet.

Plugging back in, we have 2 (13) * 4 = 2 * 12 * 0 + 2 * 5 *dx/dt.

Consequently, dx/dt = 2* 13 *4/ (2*5) = 13*4 /5 = 10.4 ft/s

I'll try to make a diagram.

No, nvm it did not work.

Edited by sozo91, 13 October 2008 - 02:19 PM.

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