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Related rates of change! (2 problems)

  1. Jul 28, 2005 #1
    Hi,
    I'm having doubt about whether or not I'm approaching two related rates of change problems correctly...
    Below is my working:

    1) The diagram represents a trough. (It's drawn like a cylinder, except the ends are right, isosceles triangles) The trough is 1.5m long. Water is being poured into the trough at the rate of 4litres per minute. At what rate is the water in the trough rising after 5 minutes?

    Now, here's how I approached this one:
    I first derived the volume formula from the information given. The formula would be area of triangle times by 150cm (Converting the length) and because it's isosceles (the triangle) the height is equal to the width, so the volume V = 150 * 0.5 * xy = 150 * 0.5 * x^2 = 75x^2
    That's out of the way!
    Now, it says water is being poured into the trough at 4L per minute, which leads me to interpreting it as:
    dV/dt = 4L/min = 4000cm^3 /min (converting to centimetres cubed)
    Now, it asks for the rate the water's rising after 5 minutes, dx/dt (Again, my understanding), so we'll find the volume after 5 minutes in order to get a proper value for the height to find dx/dt:
    dV/dt * 5 = 20000cm^3 <-- The volume after 5 minutes..
    And I have a volume formula, so let's make it equal to the volume after 5 minutes:
    20000cm^3 = 75x^2
    Solving for x I get x, height & width, = 16.3299cm (To 4dp)
    Now's the question of finding dx/dt...
    I already have dV/dt, which is 4000cm^3/min, and I have dV/dx from V, which is 150x, so let's use the chain rule and some algebra:
    dV/dt = dV/dx * dx/dt
    4000 = 150x * dx/dt
    Now, I know what the height will be after 5 minutes, 16.3299cm, so I have...
    dx/dt = 4000/150x = 4000/150(16.3299) = 1.63299cm/min <-- The rate of rising of the water in the trough after 5 minutes.

    ....or so I think; did I screw up? :\

    2) A rectangular carton is to be constructed so that the length is twice the breadth. The total surface is to be 1200 square centimetres. Find the dimensions of the box that will give the greatest volume.

    I derived a surface area equation using the geometric properties of a cuboid, ending up with the formula:
    SA = 4x^2 + 6xy (Where x = breadth, y = height)
    And given the fact that the surface area will be 1200cm^2, I made it equal:
    4x^2 + 6xy = 1200
    Now, what I needed was y in terms of x so I can get myself a single-variable volume formula; using algebra:
    y = (600-2x^2)/3x
    Now, thinking about it, volume is the area of the base, 2x * x, multiplied by the height, y, so:
    V, volume, = 2x^2y
    Substituting for y, after some simplification and muliplying out of brackets, I get:
    V = 400x - 4/3 x^3
    Now, we want maximum dimensions for maximum area, so differentiating the volume formula:
    dV/dt = 400 - 4x^2 => 0
    4x^2 = 400
    x^2 = 100
    x = 10cm
    There, we have one dimension that would give the greatest volume! Now for another, 2x:
    2x = 20cm
    Now for the third, y:
    y = 400/30 = 13.333333cm (No clue how to put the recurring sign in LaTeX)
    So, I'm left with the maximum dimensions:

    10cm x 20cm x 13.33333333cm

    That's what I think the answer is.

    I don't know if what I've come up with is right or wrong, but I'd like someone to point out if I made a huge mistake...

    Can anybody please shed some light on if I've done everything correct or not? And if I have made errors, please tell me where I screwed up.

    Thanks a lot for any attempts at helping me out; I appreciate it.
     
  2. jcsd
  3. Jul 28, 2005 #2

    OlderDan

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    For #1 it appears from your work that the right angle is at the bottom with sides at 45 degrees to the horizontal. The variable x is the length of a side that is covered by water. That is NOT the depth of the water. I think you need an additional factor to find the rate at which the height of the water is rising in proportion to the increase in x.

    A quick review of #2 uncovers only one minor problem. Your equation

    dV/dt = 400 - 4x^2 => 0

    should say

    dV/dx = 400 - 4x^2 => 0
     
  4. Jul 28, 2005 #3
    Thanks for the input on number 1...
    Number 2... yup, I guess I was careless with the derivative notation; thanks for that.

    Now, here's my understanding on what you told me about number 1..

    1) The diagram represents a trough. (It's drawn like a cylinder, except the ends are right, isosceles triangles) The trough is 1.5m long. Water is being poured into the trough at the rate of 4litres per minute. At what rate is the water in the trough rising after 5 minutes?

    The volume of the water in there would be V = 75xz (Where z is the height of water)
    Given dV/dt = 4000cm^3/min... V after 5 min. would be 4000cm^3/min * 5min = 20000cm^3
    So 20000 = 75xz; z = 20000/75x = 4000/15x = 800/3x; x = 20000/75z = 800/3z
    dV/dz = 75x
    dV/dt = dV/dz * dz/dt = 75x * dz/dt
    4000 = 75x * dz/dt

    I probably took a wrong turn here... and I'm stuck; I could substitute x for z yielding:

    4000 = 20000/z * dz/dt
    dz/dt = z/5

    Or:

    dz/dt = 160x/3

    Oh God.. :(
    Anyone?
    Thanks.
     
  5. Jul 28, 2005 #4

    OlderDan

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    This is not the way I interpret your discription of the trough. From your original post I interpreted your x and y variables as the legs of the right triangle. When you said
    V = 150 * 0.5 * xy = 150 * 0.5 * x^2 = 75x^2
    that told me that the right angle of the triangle points downward so that equal lengths of the two sides are covered by water at any time. There is nothing wrong with that equation for the volume of water if the angle points downward. What is wrong is identifying x as the height of the water. If x = y the height of the water is the altitude from the right angle to the surface of water. The volume of the water by your new equation is correct only if you change your original definition of x to be the width of the rectangle that is the water/air surface. There is a definite relationship between this width and the depth of water determined by the geometry of the triangle. You can use that relationship to express the volume in terms of the single variable (depth). So far you have not done that, and you cannot do it by soving your volume equation for x and for z. You need to find another relationship. Draw the triangle and look for it.
     
  6. Jul 29, 2005 #5
    Here's a rough ASCII drawing I made of the diagram:
    http://www.brokendream.net/xh4/probpic
    It's just like that, except the trough is being tilted a bit to the left so that the right triangle is what's touching the ground. (Sorry, best description I can give)
    Anyway, I really don't know what to do from this point, and below the diagram it says:
    Hint: Call the height 'x' cms. Consider the geometrical properties of the triangular ends.

    And... now I'm lost.
     
  7. Jul 29, 2005 #6

    OlderDan

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    Ah.. that's a completely different thing than what your equations represent. I did not understand your original reference to a cylinder, but I dismissed it based on the equations you had written and because the word cylinder does not necessarily mean a circular cylinder. I thought you were actually talking about a trough of uniform length with a triangular cross-section as suggested by your equations.

    What you apparently have is a more difficult shape to deal with than the one I thought you were talking about. The x in the verbal description you gave is not the x in your diagram. The x in your diagram is a constant; it is the radius of the cylinder (assuming it is a circular cylinder), so lets call that R and use x to represent the distance from the bottom of the tank to the top of the water. Let me also use L to represent the given 1.5 m length of the top of the trough.

    You can think of the volume of the water in the trough as layer upon layer of water stacked higher and higher. Call the thickness of each layer dx and the volume of each layer dV. Then

    dV = Adx

    Every layer is a sheet of thickness dx and area A. The area A has a rectangular shape whose dimensions increase as x increases. If the trough were completely filled the dimensions of the top layer would be L by 2R. At the bottom of the trough a layer would have dimensions of (L - 2R) by zero. When the water has depth x, the dimensions of the top surface would be (L - 2*(R-x)) by the length of a chord of a circle of radius R at a distance of (R-x) from the center of the circle.

    Draw a picture of the trough as it appears from one end (a semicircle) and see if you can figure out the distance from where the water touches one side of the trough to the other side (the length of the chord). Then draw a rectangle that represents the top surface of the water when the depth is x, with both dimensions of the rectangle expressed in terms of x. Once you have that you can write the area as a function of x, and then you will have dV in terms of x and dx. The volume of water of final depth D is the integral of dV = A(x)dx from zero to D.

    See if you can take it from here and check back again.
     
  8. Jul 29, 2005 #7
    Alright... sadly I've failed to explain my problem properly and gave you the wrong impression yet again... sorry for that.
    Let me start again. :P
    The base is actually a right isosceles triangle, not a semicircle or circle...
    So "I thought you were actually talking about a trough of uniform length with a triangular cross-section as suggested by your equations."
    Your thought was correct, if what you mean by length is what the actual triangle is prolonged by. (I really can't tell the difference between height and length)

    So here's what the thing is:
    ______________
    |\ ........................... / | <--- side 'x' cm (the ...s are there to fix the diagram)
    |_\___________/_| <-0 the other side of the triangle = 'x'cm as well
    <-constant 1.5m->

    And that's pretty much where I'm stuck at as I struggle to form a proper volume formula using the variable 'z' as the height of the water in centimetres...

    The trough is being held so that the right angle is what's touching the ground, so turn your head pi/4 radians clockwise and you should have an idea how my diagram looks.

    I hope I'm making a lot more sense now.

    Thanks for the attempt at helping, I appreciate it.
     
    Last edited: Jul 29, 2005
  9. Jul 29, 2005 #8

    OlderDan

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    Well then the difficulty level is reduced to a point somewhere between my first impression and last impression. You have no curved surfaces to deal with. Everything I said is my last post is relevant except for saying to replace the x in the diagram by R and that the one dimension of the area is the length of a chord. Let me start over. There is more than one way to do this, but I will look at what is probably the most direct way to express the volume of water in terms of the depth.

    You should not be using x as you have used it in your diagram because you were told to use x as the depth of the water. Imagine taking a saw and cutting through the trough perpendicular to its length at its center. Based on what you are telling me now the shape of the cross-section is a right triangle with the right angle at the bottom (this is what I thought originally). The hypotenuse of the triangle is horizontal, and the legs are at 45 degrees to the horizontal. Let me represent the length of each leg by S, and the height from the right angle to the hypotenuse as H. H and S are constants determined by the size of the trough and they are proportional by trigonometry or the Pythagorean theorem. The length of the hypotenuse of this crosss-section is twice the altitude, or 2H. What was missing from the first interpretation was the end caps of the trough being at an angle instead of being vertical. If you cut the trough near an end, the cross-section will be a trapezoid instead of a triangle.

    In your diagram, what you have called x should be called H. The length of the top of the trough is L. The length of the bottom of the trough is (L - 2H). If you pour water into the trough to some depth x, the top surface of the water will be a rectangle. Both dimensions of that rectangle will increase as x increases. The long dimension will be (L - 2H + 2x). The other dimension is the length of the hypotenuse of a right triangle of altitude x, which is 2x. So the area of the top-surface rectangle can be expressed in terms of the single variable x. The volume of any layer of water is

    dV = A(x)dx

    Once you have A(x) expressed in terms of x, you find the total volume by integrating from x = 0 to a total depth D (you need another letter to distinguish between the integration variable and the total depth; once the integral is done you can replace D by x and write V = V(x) instead of V = V(D))

    Now see if you can express A in terms of x and do the integral to find V.
     
  10. Jul 30, 2005 #9
    Thanks for the attempt at explaining...
    I got lost half way through your explanation (I apologize for my ignorance) and came up with this: (This time I have a diagram for the picture)
    [​IMG]
    When the water is filling up, won't the lengths of both sides of the corner always be equal, = x?
    Then the hypotenuse of the water from the left end to the right end would be root(x^2 + x^2) = x*root(2).... and so the depth of the water would be x*root 2 all over 2, right?
    Or wrong?
    If so the area of that sector is 1/2 * xroot2 / 2 * x root 2, right? Or wrong? :\
    Please tell me if I'm thinking the right way or not. :D
    Thanks a lot.
     
    Last edited: Jul 30, 2005
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