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Differential Calculus Help desperately needed

  1. Nov 3, 2004 #1
    hey i need help with 2 differential calculus problems, i missed the lecture so i am clueless as how to how to solve this.

    i dont really want the answer, id rather someone show me the methods

    anywhere, here goes:
    2) A street light is at the top of a 16ft tall pole. A woman 6th tall walks away from the pole with a speed of 8ft/sec along a straight path. How fast is the tip of her shadow receeding relative to the base of the pole when she is 30ft from the base of the pole?
    (Note that the problem asks for the speed at which the tip of the shadow is moving along the ground, that is, the speed relative to the fixed street light)

    3) Gravel is being dumped from a conveyor belt at a rate of 10cubic ft per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 18ft high? Recall that the volume of a right circular cone with height h and radius of the base r is given by (equation of the volume of a right cirular cone)

    THANKS!!!!!!!
     
  2. jcsd
  3. Nov 3, 2004 #2
    question #3 i believe this is how it goes

    find the equation for the volume of a cone. you need to find the relationship between the radius of the cone and the height because you have no information about the radius. once you have the relationship, substitute for r in the volume equation and differentiate. you know dv/dt and h and can thus solve for dh/dt.
     
  4. Nov 3, 2004 #3

    jamesrc

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    2) You should start this problem by drawing a figure. You should have two similar triangles; use that fact to write an expression relating what is known to what is unknown.

    Once you've written that equation, take the derivative. You'll than have a relationship between the speed of the woman and the speed of the tip of her shadow.

    3) Take a look at what you're given and what you're asked to find: you know the rate of change of the volume and you know that for this cone h = r. You are asked to find the rate of change of the height for a given height.

    Write out the equation for the volume of the cone. You should notice that we are not interested in anything about the radius except for the fact that it is equal to the height. So, you should write the volume as a function of the height only.

    Now take the derivative of this expression with respect to time. You will find the rate of change of volume, [tex]\dot{V}[/tex], as a function of the height and the rate of change of height, [tex]\dot{h}[/tex].

    If you've got those steps down, all you have to do is rearrange your equation, plug in the given values of h and dV/dt, and solve for dh/dt.

    Hope this helps.
     
  5. Nov 3, 2004 #4
    sorry guys, i have tried your methods, but i dont think i have such a strong grasp on what you are saying.

    im just asking for a bit more clarification. im a total mess right now.

    and why does h=r?

    thanks for your help
     
    Last edited: Nov 3, 2004
  6. Nov 3, 2004 #5

    jamesrc

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    Woops, actually d = h
    2r = h --> r = h/2
    (from this statement):
    "It forms a pile in the shape of a right circular cone whose base diameter and height are always the same."

    So for that one, start by writing out the volume of the cone:

    [tex] V = \frac{1}{3}\pi r^2 h [/tex]

    rewrite by substituting r = h/2:

    [tex] V = \frac{1}{12}\pi h^3 [/tex]

    Take the derivative with respect to time:

    [tex] \frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt} [/tex]

    Now just solve for dh/dt and plug in the values for h and dV/dt they give in the problem.
     
    Last edited: Nov 3, 2004
  7. Nov 4, 2004 #6
    Much Thanks!!!!!!!!!!!
     
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