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With y=f(x) (2x-10)^1/2
having a virtual line joined at point Q from the origin (0,0), determine the coordinates of point Q that maximises the angle of vision of an observer, situated at the origin, following the movement of point Q along the curve.

Note that tan (theta symbol) = y/x

Will I just have to find both derivatives and from there find the zero for the maximum?

  • Calculus -

    Did you make a sketch of y = (2x-10)^(1/2) ?
    Let Q be any point on the curve
    then Q can be labelled (x, (2x-10^1/2 )
    Join OQ
    slope of oQ = (2x-10)^1/2 / x

    let the angle made by OQ with the x-axis be Ø
    tanØ = y/x = (2x-10)^1/2 / x
    sec^2 Ø dØ/dx = (x(1/2)(2x-10)^(-1/2)(2) - (2x-10)^(1/2) (1) )/x^2 using the quotient rule
    for a max of Ø , dØ/dx = 0
    so (x(1/2)(2x-10)^(-1/2)(2) - (2x-10)^(1/2) (1) )/x^2 = 0
    (x(1/2)(2x-10)^(-1/2)(2) - (2x-10)^(1/2) (1) ) = 0 , we can ignore the denominator.
    x/(2x-10)^(-1/2) - (2x-10)^1/2 = 0

    x/(2x-10)^1/2 = (2x-10)^1/2
    x = 2x-10 , after cross-multiplying
    x = 10
    so Q is (10, 10^1/2) or (10, √10)

    at x=10, tanØ = √10/10 = .31622 , Ø = 17.548°
    at x = 10.1 , tanØ = √10.2/10.1 = .3162 Ø = 17.547 ° -- a bit smaller
    at x = 9.9 , tanØ = √9.8/9.9 = .3162116, Ø = 17.547 -- again a bit smaller than at x=10
    my answer looks good!

  • Calculus -

    thank you very much, I understand my mistake!

  • Calculus -

    So, we want to maximize

    t = arctan(sqrt(2x-10)/x)
    t' = -(x-10)/[(sqrt(2x-10)/x)^2 + 1)*x^2 * sqrt(2x-10)]

    All that junk in the bottom is never zero when x>5, so we just have to find where x-10 = 0

    well, duh: x=10

    So, Q = (10,√10)

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