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The questions are in the photo below: I would like them all solved correctly. 10. In submarine location problems it is often necessary to find

The questions are in the photo below: I would like them all solved correctly.

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10. In submarine location problems it is often necessary to find a submarine's closest point of approach (CPA) to a sonobuoy (sound detector) in the water. Suppose that the submarine travels on a parabolic path y = 1:2 and that the buoy is located at the point (2, %). y Submarine track in two dimensions (2, %) (Sonobuoy) (a) Show that the distance between (2, %) and a point (2:, y) on the parabola is given by D = (222)2 + (m2+%)2 (b) Show that the value of :1: that minimises the distance, D, between the submarine and the buoy is a solution of the equation 3:3 + x 1 = 0 (Note that minimising D2. rather than D, is easier and gives the same result.) (c) Use Newton's method to solve the equation in (b) (to 4 decimal places) and hence find the point on the submarine's path that is closest to the sonobuoy. 11. (3) Sketch the graph f(:c) = v'16 2:2 for 0 g :1: g 4 . Draw four rectangles (with bases of length 1) that can be used to provide an overestimate of 4 / f(:1:)dm. Compute this overestimate. 0 (b) Sketch the graph f(m) : V16 2:2 for 0 S a: S 4. Draw four rectangles (with bases of length 1) that can be used to construct an underestimate of 4 / f(2:)d$. Compute this underestimate. 0 4 (c) What can you deduce about the value of f V 16 3:2 d3 from your answers in (a) and (b)? How 0 could you improve these Stimates? 4 (d) Use geometry to find the exact value of f \\f 16 3:2 dx. 0

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