Expert: Alon Mandes - 12/31/2008

Question
QUESTION: A canoe is being pulled toward a dock by a rope attached to its bow and passing through a docking ring on the dock. The surface of the dock is 3 ft higher than the bow of the canoe. If the rope is being pulled at 1ft/sec, how fast is the canoe approaching the dock when 14ft of rope remain between the dock and the boat?

ANSWER: Let x be the distance between the dock & the canoe.
Let z be the length of the rope.
Then, x=√(z^2-9). Hence , dx/dz=[z/√(z^2-9)]*(dz/dt). We know that:
1. dz/dt=1 ft/sec
2. z=14 ft
Therefore, dx/dt=[14/√(196-9)]*1 -> dx/dt=0.073 df/sec.

Alon.

---------- FOLLOW-UP ----------

QUESTION: Why is that you divide z with the distance formula and then multiply it by the rate of change of z? I am confused. Thank you so much

Answer
Let's 1st remember the rules of square root deriving :
1. (√x)'=1/2√x
2. [√(ax^2+bx)]'=(2ax+b)/[2√(ax^2+bx)]
Now, our function is x=√(z^2-9). The derivative of this form is
[2z/2√(z^2-9)]=[z/√(z^2-9)].
Now lets remember a fact about differentiation :
if y=f(x) then dy=f'(x)dx. & dy/dt=f'(x)*(dx/dt).
In our case :
x=√(z^2-9) so dx=[√(z^2-9)]'*dz & also we can claim that
dx/dt = [√(z^2-9)]'* (dz/dt)

Alon.