Another moving ship problem I'm struggling with.

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Kristy

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Jun 26, 2007
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A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/x, how fast is the boat approaching the dock when it is 8m from the dock?

(I'm trying to get my work to show up, I tried this way to draw it and it looked terrible so just a minute and you'll be seeing another post with more work.



Edit:

Okay, the vertical distance is 1 meter and is fixed.

The horizontal disstance between the dock and the boat is 8 m and is decreasing

Then I have the velocity for the hypotenuse as decreasing by 1 meter per second.

(I'm still trying to work on it.....)
 
In your drawing, did you find a right triangle?
 
A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/x, how fast is the boat approaching the dock when it is 8m from the dock?

tkhunny said:
In your drawing, did you find a right triangle?
Click to expand...

Yes.

The base is 8m, and the height is 1 m.


Edit:

...................|
...................| 1 meter fixed
...................|
...................|
____________________
8m

This program is so frustrating. I had spaces because the line is supposed to be to the right, and it completely removed my spaces.
8 m and changing. Okay, ignore the ... those are just to fill in space.

So yes I do have a right triangle and I'm just trying to figure out how to use the stuff I used on the last problem to solve this.


Let x be the distance between the boat and the dock.
8-x at any particular time.
Okay here is where I'm getting confused applying the previous problem's knowledge. I think I should take 8-x and take the derivative, but I'm obviously missing something. I know the rate of change of the rope, and I know the vertical distance isn't changing. Somehow, hte horizontal distance must be related to the hypotenuse rope pulling in. I'm lost though. :(
 
Still struggling

\(\displaystyle \L\\\frac{dz}{dt}=-1\)

\(\displaystyle \L\\\frac{dy}{dt}=0\)

\(\displaystyle \L\\z^{2}=(x)^{2}+y^{2}\)

\(\displaystyle \L\\z^{2}=(x)^{2}+1^{2}\)

\(\displaystyle \L\\z^{2}=(x)^{2}+1\)


Differentiate:

\(\displaystyle \L\\2z\frac{dz}{dt}=2x\frac{dx}{dt}+0\) this is eqn [1]




Okay, well I managed to make somethig look okay by copying and pasting from someonoe else using that "tex" notatoion. But I haven't gotten much further understanding this. I have an idea that I should solve for

\(\displaystyle \L\\\frac{dx}{dt}=???\) at x=8
 
This is a lot like the other problem.

You have another pythagoras deal.

\(\displaystyle \L\\z^{2}=x^{2}+y^{2}\)

Differentiate:

\(\displaystyle \L\\z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\)

\(\displaystyle \L\\z=\sqrt{8^{2}+1^{2}}=\sqrt{65}\)

\(\displaystyle \L\\\sqrt{65}(1)=(8)\frac{dx}{dt}+(y)(0)\)

Note dy/dt remains constant, so it is 0.

Solve the above equation for dx/dt.
 
galactus said:
This is a lot like the other problem.

You have another pythagoras deal.

\(\displaystyle \L\\z^{2}=x^{2}+y^{2}\)

Differentiate:

\(\displaystyle \L\\z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\)

So this equation z= is more to help me know what to put on the next line (like the line above and below are more related?

\(\displaystyle \L\\z=\sqrt{8^{2}+1^{2}}=\sqrt{65}\)



\(\displaystyle \L\\\sqrt{65}(1)=(8)\frac{dx}{dt}+(y)(0)\)

Note dy/dt remains constant, so it is 0.

Solve the above equation for dx/dt.
Click to expand...
 
galactus said:
This is a lot like the other problem.

You have another pythagoras deal.

\(\displaystyle \L\\z^{2}=x^{2}+y^{2}\)

Differentiate:

\(\displaystyle \L\\z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\)

\(\displaystyle \L\\z=\sqrt{8^{2}+1^{2}}=\sqrt{65}\)

\(\displaystyle \L\\\sqrt{65}(1)=(8)\frac{dx}{dt}+(y)(0)\)

Note dy/dt remains constant, so it is 0.

Solve the above equation for dx/dt.
Click to expand...

\(\displaystyle \L\\\sqrt{65}(1)=(8)\frac{dx}{dt}+(y)(0)\)


\(\displaystyle \L\\\sqrt{65}=(8)\frac{dx}{dt}\)


\(\displaystyle \L\\\frac{sqrt{65}}{8}=\frac{dx}{dt}\)



=8.062257748/8

=1.007782219


Does this seem like I'm doing the right thing at all????
It just seems like an approximately right answer (the height isn't very much), and so it would make sense that the boat is moving toward the dock at a pretty similar rate as the rope is pulling it in, but I just don't know if I've done everythin right.
 
Is this right? I feel fairly certain that I did do the right thing (after being confused on how to set it up intially.)
 
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