calculus

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They're actaully 2 question :D

1.find the rate of change of the distance between the orgin and a moving point on the graph of y=x^2+1 if dx/dt=2 centimeters per second when x=1.

2.A triangular trough is 12 feet long and 3 feet across the top. it ends are isosceles triangles with altitude of 3 feet.
a)if water is being pumped into the trough at 2 cubic feet per minute how fast is the level rising when h is 1 ft deep?
b)if the water is rising at a rate of 3/8 inch per minute when h=2, determine the rate of which the water is being pumped into the trough.

  • calculus -

    y = x^2 + 1
    dy/dt = 2x dx/dt
    when x=1 and dx/dt = 2
    dy/dt = 2(1)(2) = 4
    Also, when x = 1, y = 2

    Distance = (x^2 + y^2)^(1/2)
    d(distance)/dt = (1/2)(x^2 + y^2)^(-1/2) (2x dx/dt + 2y dy/dt)

    = (1/2)(1/√2) (2(2) + 2(2)(4))
    = 10/√2 = 5√2

  • #2 calculus -

    Let the height of the water be h ft
    let the width of the water level be 2x ft
    2x/h = 3/3
    h = 2x or x = h/2

    V = area of triangle x 12
    = (1/2)(2x)(h)(12)
    = 12xh
    = 12(h/2)(h) = 6 h^2

    dV/dt = 12h dh/dt

    a) when h = 1 and dV/dt = 2
    2 = 12(1)(dh/dt)
    dh/dt = 1/6 ft/min

    b) You do it.

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