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Math Help - Calculus Problem

  1. #1
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    Calculus Problem

    Hello, this is my first time posting here. I was just wondering if anyone on this board can help me with the solution to this problem:

    A cone of radius 5 cms and height 10 cms is lowered, vertex first, into water in a cylindrical tank of radius 20 cms. The vertex of the cone approaches the bottom of the tank at a speed of 1 cm per second. How fast is the water level rising in the tank when half the height of the cone is submerged?

    It would be really helpful if you could guide me through each step, as I honestly donít even know where to start in the first place.
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  2. #2
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    Quote Originally Posted by Ophelia View Post
    Hello, this is my first time posting here. I was just wondering if anyone on this board can help me with the solution to this problem:

    A cone of radius 5 cms and height 10 cms is lowered, vertex first, into water in a cylindrical tank of radius 20 cms. The vertex of the cone approaches the bottom of the tank at a speed of 1 cm per second. How fast is the water level rising in the tank when half the height of the cone is submerged?

    It would be really helpful if you could guide me through each step, as I honestly don’t even know where to start in the first place.
    Note that the water level rises when the cone enters the water because water gets displaced by the submerging cone. The volume of water displaced by the cone is equal to the the volume of the cone that is submerged.

    The first thing you should do is draw a diagram. Let the water level rise by X cm after t seconds and let x cm be the distance travelled by the cone after t seconds. Note that \frac{dx}{dt} = 1 cm/s.

    I'll outline the more general case where the cone has radius r cm and the cylinder has radius R cm (R > r) and height h cm.

    a) Show that the volume of water displaced by the cone after t seconds is given by \frac{\pi}{3} \cdot \frac{r^2}{h^2} \cdot x^3.

    b) Show that an alternative expression for the volume of water displaced by the cone after t seconds is given by \pi R^2 X - \frac{\pi}{3} \cdot \frac{r^2}{h^2} \cdot (x + X)^3 + \frac{\pi}{3} \cdot \frac{r^2}{h^2} \cdot x^3.

    c) Hence show that 3R^2 h^2 X = r^2 (x + X)^3.

    d) Hence show that \frac{dX}{dt} = \frac{3r^2(x + X)^2}{3R^2 h^2 - 3r^2 (x + X)^2}. Recall that \frac{dx}{dt} = 1 cm/s.

    e) Hence find in terms of r and R how fast the water level is rising at the instant the cone is half submerged. Note: When the cone is half submerged, x + X = \frac{h}{2}.
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  3. #3
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    Quote Originally Posted by Ophelia View Post
    Hello, this is my first time posting here. I was just wondering if anyone on this board can help me with the solution to this problem:

    A cone of radius 5 cms and height 10 cms is lowered, vertex first, into water in a cylindrical tank of radius 20 cms. The vertex of the cone approaches the bottom of the tank at a speed of 1 cm per second. How fast is the water level rising in the tank when half the height of the cone is submerged?

    It would be really helpful if you could guide me through each step, as I honestly donít even know where to start in the first place.
    The water level in the cylinder will rise because the water displaced by the cone has to go somewhere- up. You need to do two things.

    First, in t-seconds, a cone of height t cm will be submerged in water. What is the volume of that cone? To answer that you will need the height, t, and the radius at the surface of the water. Look at the triangle forming the cone as seen from the side and use "similar triangles".

    Second, the surface of the water will be a "washer" with outer radius the radius of the tank, 20 cm, and the inner radius the radius of the cone at the surface of the water. The height will be the volume of water displaced, equal to the volume of the portion of the cone under water, divided by that surface area.
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