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Problem with related rates already found the answer, but need help with one part

  1. Mar 2, 2008 #1
    Hey guys can anyone please help find the answer for A, I just can't figure out what they want. I mean I was still able to figure out the answer to B, but I just kept getting the answer to A wrong and I only have one try left. Please help!


    A water trough is 15 feet long, and its cross section is an equilateral triangle with sides 3 feet long. Water is pumped into the trough at a rate of 10 cubic feet per second. How fast is the water level rising when the depth of the water is 1/2 foot?

    ( Hint: First, what is the height h of an equilateral triangle of side length s? Next, what is the area of an equilateral triangle in terms of the side length s? Then write the area in terms of h. The volume of the water in the trough at time t is the product of the cross-sectional area with water and the length of the trough. )

    a) What is the height h of an equilateral triangle of side length s?

    h = ____ ft.

    Note: Answers I said were (3*sqrt(3))/2 and 1/sqrt(3) I thought all I had to do was the pythagorean theorem...but apparently its not the answer

    b) The water level is rising at a rate of __(2*sqrt(3))/3 ft./sec.__
  2. jcsd
  3. Mar 2, 2008 #2


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    A perpendicular dropped from one vertex of an equilateral triangle bisects the opposite side. The altitude is one leg of a right triangle with hypotenuse of length s and other leg of length s/2. If we call that height x, then we have (Pythagorean theorem) s2= x2+ s2/4. Then x2= s2- s2/4= 3s2/4 and x= [itex]\sqrt{3}s/2[/itex] feet. Did you forget the "s"? And what is the other answer? (a) only asks for one answer.

    When the distance from the base of the trough to water level, measured along the slant side, is s, then the height is [itex]sqrt{3}s/2[/itex] and the "base" of the triangle is s so the area is (1/2)base*height= [itex]\sqrt{3}s^2/4[/itex] square feet so the volume of water in the trough is [itex]V= 3\sqrt{3}s^2/4[/itex]cubic feet. [itex]dV/dt= 3\sqrt{3}s/2 ds/dt[/itex]. You know know that dV/dt= 10 cubic feet per second so you can find ds/dt immediately. But that is not the rate at which the water level is rising! You know that [itex]h= \sqrt{3}s/2[/itex] so [itex]dh/dt= \sqrt{3}/2 ds/dt[/itex].
    Last edited by a moderator: Mar 2, 2008
  4. Mar 2, 2008 #3
    Yeah, theres only one answer, the 2 answers I said were answers I tried to see if they were right.

    I have a question, so S is supposed to be a variable included in the answer or I cannot plug in a number for it to get an exact numerical answer? And what is "sup" supposed to mean?
  5. Mar 2, 2008 #4


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    Yes, s is the distance from the base of the trough to the water level measured along the slant side rather than straight up (the height). Since the water is rising, that is a variable. The problem asks you to find how fast the height is increasing when the height is 1/2 foot. You will have to use [itex]h= \sqrt{3}s/2[/itex] to find s when h= 1/2 and put that into the formula.

    The "sup" was supposed to be the html tag for "superscript" but I forgot the "/" to end it: s[ sup ]2[ /sup ] without the spaces is s2.
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