Find the point on the curve closest to the point (1,3)
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Find the point on the curve closest to the point (1,3)

[From: ] [author: ] [Date: 11-05-05] [Hit: ]
This enables an effective iterative procedure, similar to the Newton-Raphson method, to be employed. This converges to the required values in around 6-8 iterations, depending upon the initial estimate adopted. It requires that f(x) and f (x) be calculated for each iteration,......

m1 = -(x - 1)/[f(x) - 3]

while that obtained by differentiation is

f '(x) = 2.08*x*sin(1.05*x) + 1.092*x^2*cos(1.05*x)

The functional dependence of these two curves is fortunately quite different, so that they intersect at the required value at a large angle. This enables an effective iterative procedure, similar to the Newton-Raphson method, to be employed. This converges to the required values in around 6-8 iterations, depending upon the initial estimate adopted. It requires that f(x) and f '(x) be calculated for each iteration, but otherwise the computational burden is relatively light.

The coordinates of A were obtained as (1.644756, 2.779180) to six-figure accuracy. As a check, the slope of AP and the gradient f '(x) were multiplied to check whether they were perpendicular, yielding a value for m1.f '(x) = -1.00002.

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The equation you give corresponds to a parametrised curve (x,1.04x^2sin(1.05x))

The distance from (1,3) is ||(x,1.04x^2sin(1.05x))-(1,3)|| =sqrt((x-1)^2 +(1.04x^2sin(1.05x)-3)^2)

We may ignore the square root as the minimum of a positive function is attained at same point as the minimum of that function squared.

Hence we want to minimise (x-1)^2 +(1.04x^2sin(1.05x)-3)^2

Differentiate wrt x:

2(x-1) + 2(1.04x^2sin(1.05x)-3)[2.08 x sin(1.05x)+1.05*1.04x^2cos(1.05x)]

Solve for zero though I have to say this looks a tricky task :o
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