# Length of diagonal rope

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

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quasar987
Homework Helper
Gold Member
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.

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 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.

quasar987
Homework Helper
Gold Member
Say L is the distance Haselhof-swimmer, and D is the distance swimmer-tower. You got

$$L = \sqrt{9+D^2}$$

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

$$\frac{dD}{dt}|_{D=4}$$

quasar987 said:
Say L is the distance Haselhof-swimmer, and D is the distance swimmer-tower. You got

$$L = \sqrt{9+D^2}$$

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

$$\frac{dD}{dt}|_{D=4}$$
Sorry but I do not understand
$$\frac{dD}{dt}|_{D=4}$$

the derivative of D when D is 4 should not give us 0??

quasar987
Homework Helper
Gold Member
This is a little strange to me too... But here's what I think.

It's like when we have $f(x) = e^{ax}$. Then $df/dx = ae^{ax} = af(x)$. 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 $df/dx|_{f=c}$ to have is mean $df/dx|_{x^* \ \mbox{such that f(x^*)=c}}$. And notice that $df/dx|_{f=c} = ac \neq 0$.

Here we have that

$$\frac{dL}{dt} = \frac{dL}{dD}\frac{dD}{dt} \Leftrightarrow \frac{dD}{dt} = \frac{-1}{\frac{dL}{dD}}$$ (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

$$\frac{dD}{dt}|_{D=4}$$

SO, to sum up: calculus is a Bitc... That'll be all for today... :zzz:

Last edited:
Doc Al
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$$L^2 = 9 + D^2$$