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Length of diagonal rope

  • Thread starter brad sue
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  • #1
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Hey,
I need some suggestions to approach the problem:

A lifeguard standing on a tower throws a buoy to a swinner 5m from the tower. The lifeguard, positioned 3 m above the water, pulls in the rope at a speed of 1m/s.
How fast is the swinner coming to the shore when he is at 4m from the water edge?

thanks
 

Answers and Replies

  • #2
quasar987
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Very cute calc prob.

If David Haselhoff is pulling the rope at 1m/s, this means that the rate of change of the distance btw the Hasellhoff and the swimmer (lenght of the rope btw swimmer and the Hasellhoff's hairy palms) is of -1m/s. How is the distance btw the Hasellhoff and the swimmer related to the distance btw the swimmer and the tower? hint: old crazy greek cult leader.
 
  • #3
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quasar987 said:
Very cute calc prob.

If David Haselhoff is pulling the rope at 1m/s, this means that the rate of change of the distance btw the Hasellhoff and the swimmer (lenght of the rope btw swimmer and the Hasellhoff's hairy palms) is of -1m/s. How is the distance btw the Hasellhoff and the swimmer related to the distance btw the swimmer and the tower? hint: old crazy greek cult leader.

Crazy hint :smile:

we have a right trangle-swimmer- bottom of tower - Haselhof. (pythagoras )

but I have still no idea how to solve it since we have two position 5m and 4 m.
 
  • #4
quasar987
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Say L is the distance Haselhof-swimmer, and D is the distance swimmer-tower. You got

[tex]L = \sqrt{9+D^2}[/tex]

And what are you looking for here? You're looking for

[tex]\frac{dD}{dt}|_{D=4}[/tex]
 
  • #5
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quasar987 said:
Say L is the distance Haselhof-swimmer, and D is the distance swimmer-tower. You got

[tex]L = \sqrt{9+D^2}[/tex]

And what are you looking for here? You're looking for

[tex]\frac{dD}{dt}|_{D=4}[/tex]
Sorry but I do not understand
[tex]\frac{dD}{dt}|_{D=4}[/tex]

the derivative of D when D is 4 should not give us 0??
 
  • #6
quasar987
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This is a little strange to me too... But here's what I think.

It's like when we have [itex]f(x) = e^{ax}[/itex]. Then [itex]df/dx = ae^{ax} = af(x)[/itex]. df/dx is a function of f and x. Actually it is only a function of x since f is a function of x. But we can still write [itex]df/dx|_{f=c}[/itex] to have is mean [itex]df/dx|_{x^* \ \mbox{such that f(x^*)=c}}[/itex]. And notice that [itex]df/dx|_{f=c} = ac \neq 0[/itex].

Here we have that

[tex]\frac{dL}{dt} = \frac{dL}{dD}\frac{dD}{dt} \Leftrightarrow \frac{dD}{dt} = \frac{-1}{\frac{dL}{dD}}[/tex] (since dL/dt = -1)

So dD/dt is a function of D in the same sense as df/dx above is a function of f. Hence is makes sense to write

[tex]\frac{dD}{dt}|_{D=4}[/tex]

SO, to sum up: calculus is a Bitc... That'll be all for today... :zzz:
 
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  • #7
Doc Al
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brad sue said:
A lifeguard standing on a tower throws a buoy to a swinner 5m from the tower. The lifeguard, positioned 3 m above the water, pulls in the rope at a speed of 1m/s.
How fast is the swinner coming to the shore when he is at 4m from the water edge?
The easiest way to solve this is to define the variables L and D just like quasar987 did. Their relationship is:
[tex]L^2 = 9 + D^2[/tex]

Now just take the derivative of both sides with respect to time. You'll get an easy expression in terms of L, D, dL/dt, and dD/dt. The only unknown is dD/dt, which is what you want to find. Try it.
 

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