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Math Help - Related Rates II (?)

  1. #1
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    Cool Related Rates II (?)

    Here's another one that apparently I can't get it.
    A water trough is 4 m long and its cross-section is an isosceles trapezoid which is 80 cm wide at the bottom and 120 cm wide at the top, and the height is 40 cm. The trough is not full. Give an expression for V, the volume of water in the trough in cm3, when the depth of the water is d cm.
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  2. #2
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    A water trough is 4 m long and its cross-section is an isosceles trapezoid which is 80 cm wide at the bottom and 120 cm wide at the top, and the height is 40 cm. The trough is not full. Give an expression for V, the volume of water in the trough in cm3, when the depth of the water is d cm.
    draw two altitudes from the lower corners of the trapezoid to the upper base.

    sketch in a horizontal water level of depth d.

    note the similar right triangles on either end ... let x = horizontal water level distance from the altitude to the trapezoid's nearest side

    \frac{x}{d} = \frac{20}{40}

    x = \frac{d}{2}

    water level cross-sectional trapezoid area ...

    A = \frac{d}{2}\left[.8 + (.8 + d)\right]

    A = .8d + \frac{d^2}{2}

    Volume in m^3 ...

    V = 4\left(.8d + \frac{d^2}{2}\right)
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  3. #3
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    We need to express the volume in terms of the height, h, of the water in the trough.

    By using similar triangles, \frac{s}{d}=\frac{20}{40}=\frac{1}{2}

    s=\frac{d}{2}

    Therefore, the width of the water at depth h is 80+2s\Rightarrow 80+2(\frac{d}{2})=80+d

    Be careful, the length of the trough is given in meters, whereas the other dimensions are in cm. The trough is 400 cm long.

    V=\frac{400}{2}\left(80+80+d\right)d=200(160+d)d=2  00d^{2}+32000d

    We could just as easily express it in terms of meters instead of cm as well.
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  4. #4
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    Draw a picture! A picture of the trapezoidal ends is sufficient since the volume is just the end area times the constant length 4m = 400 cm.

    Your picture should be a trapezoid with bottom base 80 cm long, top base 120 cm long and height 40 cm. Now draw another line at height "h" representing the water level. The area formed by the "water" is also a trapezoid and its area is 1/2 times the height h times the sum of the two bases, 80 cm and the length of that "water" line. At this point there are two ways to proceed.

    1) Draw vertical lines from the bottom corners of the trapezoid to the top. That divides the trapezoid into three parts, two triangles and a rectangle. The length of the "water line" across that rectangle is the same as the base of the trapezoid, 80 cm. To find the length of the two parts that cross the triangles, use "similar triangles". The larger triangles have height 40 cm and base (1/2)(120- 80)= 20 cm. The height of the smaller triangles is "h" and, calling the height x, we have \frac{x}{h}= \frac{20}{40}= \frac{1}{2} so x= \frac{h}{2}. The upper base of the trapezoid is the sum of the three parts: \frac{h}{2}+ 80+ \frac{h}{2}= h+ 80. The water forms a trapezoid with height h and bases 80 and h+ 80. Its area is h(h+80+ 80)/2= \frac{1}{2}h^2+ 80h.

    2) Since all the sides are straight lines, the base width will be a linear function of height: w= Ah+ B. When h= 0, the width is 80 cm and when h= 40, the width is 120: 80= A(0)+ B= B and 120= 40A+ B= 40A+ 80 so 40A= 40 and A= 1: the base width is h+ 80 and the area is (1/2)(h)(80+ h+ 80)= \frac{1}{2}h^2+ 80, as before.

    As I said, the volume is that area times the length of the trough: (400)(\frac{1}{2}h^2+ 80)= 200h^2+ 32000.

    (I took too long! Everyone in the world answered before I did! Well, wdma, I hope this helps. Oh, this was titled "related rates". To find the rate at which the volume increases as h increases (or vice-versa) differentiate both sides of V(h)= (400)(\frac{1}{2}h^2+ 80)= 200h^2+ 32000 with respect to time, t.
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