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Homework Help: Implicit applications

  1. Nov 12, 2013 #1
    So I was looking at the wording of this question and I do not know what it means that the drum is located 4 feet above the bow. Also if the rope is being pulled in at a rate of 3 feet per second wouldn't the boat be moving at 3 feet per second at all times because they are connected?
  2. jcsd
  3. Nov 12, 2013 #2


    Staff: Mentor

    It would help to draw a picture. The rope is not pulling horizontally, so 3 fps on the rope doesn't translate to 3 fps for the boat. The rope feeds onto the drum at a point 4 feet higher than where the rope is tied to the boat.

    For the purposes of this problem, you can ignore any sag in the rope, and pretend it's straight between the drum and the boat.
  4. Nov 12, 2013 #3
    What kind of equation would I need for this type of question? So far in calc we have only done rates of change of area, and volume, and things where the equations are obvious.
  5. Nov 12, 2013 #4
    Let 'x' be the horizontal distance between the dock and the boat .
    Let 'y' be the vertical distance between the drum and the boat .
    Let 'r' be the length of the rope between the drum and the boat .

    What is the relation between the above parameters ?
  6. Nov 12, 2013 #5
    Wait can I do this tan([itex]\theta[/itex]) = y/4 and then sec([itex]\theta[/itex])^2*d[itex]\theta[/itex]/dt = .25*dy/dt, I can find [itex]\theta[/itex], but I do not know how to get the d[itex]\theta[/itex]/dt, the only way I would know how to is if I was given the radius, I think it has something to do with the 3 feet/s, but I don't know what(for me y = distance between ship and dock).
  7. Nov 12, 2013 #6
    What is θ ?

    Anyways,whatever is θ ,angle of rope with the vertical or horizontal ,the relationship tan([itex]\theta[/itex]) = y/4 is not correct .

    Please do the way I have suggested .
  8. Nov 12, 2013 #7


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    Homework Helper

    The picture is actually more important here than any formulas. You should be imagining a right triangle. The distances Tanya Sharma described are sides of that triangle.
  9. Nov 12, 2013 #8
    I am imagining a triangle, hence the tan[itex]\theta[/itex], but [itex]\theta[/itex] is just [itex]\theta[/itex] = arctan(25/4), ok I will try to do it your way sharma.
  10. Nov 12, 2013 #9
    x^2 + y^2 = r^2
    x^2 +16 = r^2
    2x*dx/dt + 2y * dy/dt = 2r*dr/dt
    50*dx/dt = 6*r

    x^2 + y^2 = r^2
    sqrt(25^2 + 4^2) = r

    dx/dt = 0.32
    This doesn't seem right to me, did you get this? Or am I totally wrong?
  11. Nov 12, 2013 #10
    No..That is incorrect.

    Suppose θ is the angle which rope makes with horizontal ,then tanθ = y/x . The general angle θ will be expressed in terms of generalized parameters ,i.e x,y,r ,not in terms of values of x,y,r at a particular instant .Only when the value of parameter is a fixed quantity like y=4 ,then you can use value '4' in place of 'y' .
  12. Nov 12, 2013 #11
    Please recheck your calculations. I think you should get something like 3.04 .
  13. Nov 12, 2013 #12
    Yea I rechecked and got 3.04, thanks for your help!
  14. Nov 12, 2013 #13
    Well done!!!
  15. Nov 12, 2013 #14
    Panphobia...Now,since you have successfully got the answer ,you may try your approach .

    Let θ be the angle which rope makes with horizontal ,then tanθ = y/x . Proceed in similar manner and you will end up with the same correct answer .
  16. Nov 12, 2013 #15


    Staff: Mentor

    In this problem, the vertical distance is constant, so if θ the angle that the rope makes at the boat, relative to the horizontal, then tan(θ) = 4/x, where x is the horizontal distance between the boat and the drum.

    This is not the best place to start, though, as we should start with a relation that ties together the hypotenuse and the horizontal leg of the triangle.
    x2 + 42 = r2

    Now differentiate with respect to t to get the relationship between the rates.

    There is no need for involving θ (hence dθ/dt) here.
  17. Nov 12, 2013 #16
    Mark...This is exactly the same approach i have explained to OP.Are you referring to Panphobia or me ?
  18. Nov 13, 2013 #17


    Staff: Mentor

    Tanya, I was referring to you both. I thought you were saying that the equation needed to be x2 + y2 = r2, and that you could only put in the values at a particular instant. My point was that there is no need for a y variable, as the height value is constant. I might have misunderstood your intent.
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