Boat pulled toward pier by rope: find speed at distance 3.2m

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A man standing on a pier pulls a small boat towards him using a rope attached to the prow. The rope is pulled from a height of 2.4m above its point of attachment to the boat. The rope is pulled at a rate of 30 cm a minute. At what rate is the boat moving towards the pier when it is 3.2m away.

Im stuck on this problem. I know it is a right angled triange, and that you have to differentiate using phythagoras, but i am stuck. help appreciated please
 
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yeah i think your going the right direction i'll just look it up for you
 
it dosn't happen to tell you the force being exerted on the rope? (in newtowns?)
 
Kingofthedamned said:
it dosn't happen to tell you the force being exerted on the rope? (in newtowns?)
no, is a maths question :smile:
 
no, is a maths question :smile:

lol. Make a right triangle with legs of length 2.4 m for the vertical leg(x), and 3.2 m for the horizontal leg(y), with z as the hypotenuse or the length of the rope. Set up your equation as:

[tex]x^2 + y^2= z^2[/tex]

now make all of your variables (x,y,z) a function of time, so when you implicitly differentiate the equation with respect to time, it will look like this:

[tex]2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt}[/tex]

now you know that dx/dt is constant, since the dude holding the rope isn't moving up and down, but is holding it steady while he pulls it in, so dx/dt cancels to zero. Just solve for dy/dt since that is the rate of change of the horizontal leg of the triangle (the rate is the boat moving towards the pier). y, z and dz/dt are all given in the problem. Plug'n solve.
 
Last edited:
Gza said:
lol. Make a right triangle with legs of length 2.4 m for the vertical leg(x), and 3.2 m for the horizontal leg(y), with z as the hypotenuse or the length of the rope. Set up your equation as:

[tex]x^2 + y^2= z^2[/tex]

now make all of your variables (x,y,z) a function of time, so when you implicitly differentiate the equation with respect to time, it will look like this:

[tex]2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt}[/tex]

now you know that dx/dt is constant, since the dude holding the rope isn't moving up and down, but is holding it steady while he pulls it in, so dx/dt cancels to zero. Just solve for dy/dt since that is the rate of change of the horizontal leg of the triangle (the rate is the boat moving towards the pier). y, z and dz/dt are all given in the problem. Plug'n solve.
greatly appreciated, but can't get dy/dt. any help please? :frown:
 
greatly appreciated, but can't get dy/dt. any help please?

It's really just a matter of algebra now. Just get what you want on the left, and everything else on the right, and plug in what you know.

[tex]\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}[/tex]

now for z, the problem told you that you wanted the value of dy/dt, when the boat was 3.2 m away. So z is just the square root of (3.2)^2 + (2.4)^2
(Pythag). dz/dt is the rate of change of z in time, which the problem gave you as "30 cm a minute."

:smile: