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Math Help - 3 Calculus Questions

  1. #1
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    3 Calculus Questions

    If anybody could please help me with the following problems I'd greatly appreciate it.

    1. A point moves along the curve y = x^2 + 1 so that the x-coordinate is increasing at the constant rate of 3/2 units per second. The rate, in units per second, at which the distance from the origin is changing when the point has coordinates (1,2) is equal to...
    2.(multiple choice) The area bounded by the curve y = 1/x+1, the axes, and the line x = e - 1 equal to: a) 1 - (1/e^2) b) ln(e - 1) C) 1 D) 2 or E) 2 (square root e + 1) - 2
    3. A solid is cut out of a sphere of radius 2 by two parallel planes each 1 unit from the center. The volume of this solid is.....

    Thanks for any help!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by turtle View Post
    1. A point moves along the curve y = x^2 + 1 so that the x-coordinate is increasing at the constant rate of 3/2 units per second. The rate, in units per second, at which the distance from the origin is changing when the point has coordinates (1,2) is equal to...
    dx/dt = 3/2 units/s

    dy/dt = 2x*dx/dt = 3x units/s

    The overall rate of change will be:
    ds/dt = sqrt{(dx/dt)^2 + (dy/dt)^2} units/s
    (Think of this as a speed with x and y components.)

    Thus
    ds/dt = sqrt{9/4 + 9x^2} units/s

    Thus
    ds/dt(1, 2) = sqrt{9/4 + 9*1^2} units/s = sqrt{45/4} units/s = (3/2)*sqrt{5} units/s

    -Dan
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  3. #3
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    Hello, turtle!

    Here's #1 . . .


    1. A point moves along the curve y = x + 1 so that the x-coordinate
    is increasing at the constant rate of 3/2 units per second.

    The rate, in units per second, at which the distance from the origin is changing
    when the point has coordinates (1,2) is equal to __

    The distance from the origin is given by: .z .= .x + y . [1]

    Since y = x + 1, we have: .z .= .x + (x + 1) .= .x^4 + 3x + 1


    Differentiate with respect to time: .2z(dz/dt) .= .(4x + 6x)(dx/dt)

    . . . . . . . . . . . . . dz . . . x(2x + 3) . .dx
    . . and we have: . --- .= . ------------- --- . [2]
    . . . . . . . . . . . . . dt . . . . . . z . . . . . dt


    With the point (1,2), substitute into [1]: .z .= .1 + 2 . . z = √5
    And we are told that: .dx/dt = 3/2

    . . . . . . . . . . . . . . .dz . . . . 1(21 + 3) . 3 . . . . 3√5
    Substitute into [2]: .--- . = . -------------- -- . = . -----
    . . . . . . . . . . . . . . .dt . . . . . . . .√5 . . . .2 . . . . . 2

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