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Related rates problem (3)

  1. Oct 21, 2008 #1

    lim

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    1. The problem statement, all variables and given/known data
    A conical tank has a base radius of 6 feet and a height of 10 feet. Initially the tank is empty. Water is poured into the tank at a rate of 75 ft/min. How fast is the depth of the water in the tank changing when the water in the tank reaches a height of 5 ft? 8 ft? when the tank is half full?


    2. Relevant equations
    V= 1/3πr^2h
    (π=pi)

    3. The attempt at a solution
    r/h=6/10
    6h= 10r
    r=3/5h.
    My teacher told us to set up a radius and height relation to get ride of differentiating the radius later.

    V= 1/3πr^2h
    dv/dt= 1/3π(9/25h^2)h =1/3π(9/25)h^3
    dv/dt= 1/3π(9/25)3h^2
    dv/dt= π(9/25)100(5)= 180π ft/s @ 5 ft
    π(9/25)100(8)= 280π ft/s @ 8 ft
    π(9/25)100(5)= 180π ft/s @ 1/2 tank (since half of the tank is also 5 ft)
    Are these correct?

    2. The problem statement, all variables and given/known data
    A rectangle has a constant area of 200 sq meters and length, L, is increasing at 4 meters/s.
    a. Width, W, at instant the width is decreasing at .5 m/s?
    b. At what rate is the diagonal, D, of the rectangle changing at the instant the width is 10 m?

    2. Relevant equations
    A=lw

    3. The attempt at a solution
    A=lw
    Da/dt= l(dw/dt) + w (dl/dt)
    200= l (-.5) + 4w
    I set it up, but couldn’t figure out how to substitute for l or w. Is there another way?

    3. The problem statement, all variables and given/known data
    Water is draining from a conical tank with height 12 feet and diameter 8 feet into a cylindrical tank that has a base with area square feet. The depth, h, in feet, of the water in the conical tank is changing at the rate of (h-12) feet per minute.

    Equations:
    V= 1/3πr^2h
    A) Write an expression for the volume of the water in the conical tank as a function of h.
    r/h = 4/12
    4h= 12r
    r=1/3h
    V= 1/3π1/9 h^3
    correct?

    B) At what rate is the volume of the water in the conical tank changing when h=3?
    Dv/dt= 1/3 π 3h^2 dh/dt
    Π/3 (1/9)3 (12)^2 dh/dt
    Π(144)3/27
    432 π/27= 16 π ft/min
    correct?

    C) Let y be the depth, in feet, of the water in the cylindrical tank. At what rate is y changing when h = 3?

    2. Relevant equations

    v=bh

    attempt

    y=bh, letting y be the depth
    dy/dt= b dh/dt + h db/dt
    dy/dt= 3 db/dt + 400 π dh/dt, ? Not sure what to do here.
     
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