Question

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 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 4 m from the dock?
i don't get how the dy/dt is -1 and dx/dt -y/x??

Answer 1

You haven't defined your x and y, so I can't answer the last part of your question.

On my diagram, I have a right-angled triangle with a height of 1 m, a horizontal of x m and a hypotenuse of y m

given: dy/dt = -1 m/s
( I made it negative, because the value of y will be <b>decreasing</b> as the rope is being pulled in )

Since we are involving variables x and y, I will need some kind of equation that contains x and y.
Heh, how about Pythagoras

x^2 + 1^2 = y^2
2x dx/dt = 2y dy/dt
x dx/dt = y dy/dt
(notice I have not worried about the x=4, now I will)
4^2 + 1^2 = y^2
so y = √17

sub in our stuff into

x dx/dt = y dy/dt
4 dx/dt = √17(-1)
dx/dt = -√17/4 m/s or appr -1.03 m/s

again, my answer is consistent with our concept that if x is decreasing, then dx/dt will be negative

So the boat it approaching the dock at appr 1.03 m/s