Question

One end of a rope is fastened to a boat and the other end is wound around a windlass located on a dock at a point 4m above the level of the boat. If the boat is drifting away from the dock at the rate of 2m/min, how fast is the rope unwinding at the instant when the length of the rope is 5m?

Answer 1

looks like pythagoras to me

Answer 2

yup. i think its pythagoras.. but some how.. i was confused with the question.. i cant even understand how what it want and how to start.. ><

Answer 3

if we call the distance from the boat to the dock x, and the distance from the boat to the windlass (whatever that is) y, then we have
\[4^2+x^2=y^2\]

Answer 4

draw a picture and you will have a triangle with hypotenuse the length of the rope (i called that y) one leg 4 and the other the distance from the dock to the boat, (i called that x)

Answer 5

in my use of variables you are told
\[x'=2\] because x is the distance from the boat to the dock. and you are asked for
\[y'\] the speed at which the ropes is increasing

Answer 6

starting with the equation
\[4^2+x^2=y^2\] and differentiating wrt time we get
\[2xx'=2yy'\] solving for y' gives
\[y'=\frac{xx'}{y}\]

Answer 7

now plug in the numbers. we are given

."If the boat is drifting away from the dock at the rate of 2m/min"

meaning
\[x'=2\] and also

"how fast is the rope unwinding at the instant when the length of the rope is 5m "

tells us
\[y=5\]

Answer 8

pythagoras tells us that if y = 5 then x = 3 (we have a nice 3-4-5 right triangle) and so
\[y'=\frac{3\times 2}{5}=\frac{6}{5}\]

Answer 9

meters per minute or whatever the units are

Answer 10

steps more or less clear?

Answer 11

Clear!! thx!

Answer 12

yw