By ignoring the thickness of the water pipe, show that the length of the water pipe is given by (5 csc0+3 sec0) m?

The diagram shows two corridors of width 5 m and 3 m which meet at a Fight angle. A water pipe AB is to be carried horizontally around the corner

(a)By ignoring the thickness of the water pipe, show that the length of the water pipe is given by (5 csc0+3 sec0) m(b) Calculate the length of the longest water pipe that can be carried and state the value of, in degree, at which it occurs.

Update:

(a)By ignoring the thickness of the water pipe, show that the length of the water pipe is given by (5 csc0+3 sec0) m

(b) Calculate the length of the longest water pipe that can be carried and state the value of, in degree, at which it occurs.

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1 Answer

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  • Let the point where the 2 dimensions meet be C

    AC = a

    BC = b

    a * sin(t) = 5

    b * sin(90 - t) = 3

    b * sin(90 - t) = 3

    b * (sin(90)cos(t) - sin(t)cos(90)) = 3

    b * (1 * cos(t) - sin(t) * 0) = 3

    b * (cos(t) - 0) = 3

    b * cos(t) = 3

    b = 3 / cos(t)

    b = 3 * sec(t)

    a * sin(t) = 5

    a = 5 / sin(t)

    a = 5 * csc(t)

    a + b = 5 * csc(t) + 3 * sec(t)

    L = 5 * csc(t) + 3 * sec(t)

    L' = -5 * csc(t) * cot(t) + 3 * sec(t) * tan(t)

    L' = 0

    -5 * csc(t) * cot(t) + 3 * sec(t) * tan(t) = 0

    3 * sec(t) * tan(t) = 5 * csc(t) * cot(t)

    3 * (1/cos(t)) * (sin(t)/cos(t)) = 5 * (1/sin(t)) * (cos(t)/sin(t))

    3 * sin(t) / cos(t)^2 = 5 * cos(t) / sin(t)^2

    3 * sin(t)^3 = 5 * cos(t)^3

    sin(t)^3 / cos(t)^3 = 5/3

    tan(t)^3 = 5/3

    tan(t)^3 = 45/27

    tan(t) = 45^(1/3) / 3

    t = arctan(45^(1/3) / 3)

    L = 5 * csc(t) + 3 * sec(t)

    csc(arctan(m)) =>

    sqrt(1 + cot(arctan(m))^2) =>

    sqrt(1 + 1/tan(arctan(m))^2) =>

    sqrt(1 + 1/m^2) =>

    sqrt((m^2 + 1) / m^2) =>

    sqrt(m^2 + 1) / m

    sec(arctan(m)) =>

    sqrt(1 + tan(arctan(m))^2) =>

    sqrt(1 + m^2)

    L = 5 * (1/(45^(1/3) / 3)) * sqrt(1 + (1/3)^2 * 45^(2/3)) + 3 * sqrt(1 + (1/3)^2 * 45^(2/3))

    L = ((15 / 45^(1/3)) + 3) * sqrt(1 + (1/9) * 5^(2/3) * 9^(2/3))

    L = ((15 * 45^(2/3) / 45  +  3) * sqrt(1 + (1/9) * 5^(2/3) * 3^(4/3))

    L = ((1/3) * 45^(2/3) + 3) * (1/3) * sqrt(9 + 25^(1/3) * 3 * 3^(1/3))

    L = (1/3) * (1/3) * (45^(2/3) + 9) * sqrt(9 + 3 * 75^(1/3))

    L = (1/9) * (5^(2/3) * 9^(2/3) + 9) * sqrt(9 + 3 * 75^(1/3))

    L = (1/9) * (25^(1/3) * 3^(1/3) * 3 + 9) * sqrt(9 + 3 * 75^(1/3))

    L = (1/9) * 3 * (3 + 75^(1/3)) * sqrt(9 + 3 * 75^(1/3))

    L = (3 + 75^(1/3)) * sqrt(9 + 3 * 75^(1/3)) / 3

    (3 + 75^(1/3))^2 =>

    9 + 6 * 75^(1/3) + 75^(2/3)

    3 + 75^(1/3) = sqrt(9 + 6 * 75^(1/3) + 75^(2/3))

    (9 + 6 * 75^(1/3) + 75^(2/3)) * (9 + 3 * 75^(1/3)) =>

    81 + 27 * 75^(1/3) + 54 * 75^(1/3) + 18 * 75^(2/3) + 9 * 75^(2/3) + 3 * 75^(3/3) =>

    81 + (27 + 54) * 75^(1/3) + (18 + 9) * 75^(2/3) + 3 * 75 =>

    81 + 225 + 81 * 75^(1/3) + 27 * 75^(2/3) =>

    306 + 81 * 75^(1/3) + 27 * 75^(2/3) =>

    9 * (34 + 9 * 75^(1/3) + 3 * 75^(2/3))

    sqrt(9 * (34 + 9 * 75^(1/3) + 3 * 75^(2/3))) / 3 =>

    3 * sqrt(34 + 9 * 75^(1/3) + 3 * 75^(2/3)) / 3 =>

    sqrt(34 + 9 * 75^(1/3) + 3 * 75^(2/3))

    3 * 75^(2/3) + 9 * 75^(1/3) + 34 =>

    3 * (75^(2/3) + 3 * 75^(1/3)) + 34 =>

    3 * (75^(2/3) + 3 * 75^(1/3) + 1.5^2 - 1.5^2) + 34 =>

    3 * (75^(2/3) + 3 * 75^(1/3) + 1.5^2) - 3 * 1.5^2 + 34 =>

    3 * (75^(1/3) + 3/2)^2 - 27/4 + 34 =>

    3 * (1/4) * (2 * 75^(1/3) + 3)^2 + (1/4) * (-27 + 136) =>

    (1/4) * (3 * (2 * 75^(1/3) + 3)^2 + 109)

    sqrt((1/4) * (3 * (2 * 75^(1/3) + 3)^2 + 109)) =>

    (1/2) * sqrt(3 * (2 * 75^(1/3) + 3)^2 + 109)

    11.19409976308634021002434599773...

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