數學問題(d,in,matrix)2－hzb53jl55v的部落格

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數學問題(d,in,matrix)2

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圖片參考：http://imgcld.yimg.com/8/n/HA00613953/o/701111010011113873395240.jpg 可唔可以答埋17同47,PLZ 33(c),點解X可以係real number又可以係matrix?

最佳解答:

17) Let center of the wheel be the origin. So co-ordinates of Q is (9, - 3). Co - ordinates of P is (3 cos w, 3 sin w). The horizontal distance between P and Q = 9 - 3 cos w. The vertical distance between P and Q = 3 + 3 sin w. Let length of PQ = L, so by Pythagoras thm., L^2 = (9 - 3 cos w)^2 + (3 + 3 sin w)^2 = 9(3 - cos w)^2 + 9(1 + sin w)^2 2L dL/dt = 18(3 - cos w) sin w (dw/dt) + 18(1 + sin w) cos w (dw/dt) 2L dL/dt = 18[(3 - cos w) sin w + (1 + sin w) cos w](dw/dt) 2L dL/dt = 18(3 sin w + cos w) (dw/dt) so dL/dt = 9( 3 sin w + cos w)/L (dw/dt) When P is at the highest point, w = 90 degree, L = sqrt (9^2 + 6^2) = sqrt 117. dw/dt = 1.5 rad/s (given) so dL/dt = 9(3)(1.5)/sqrt 117 = 3.74 cm./s. ( 3 sig. fig.) 47) g(x) + g(a - x) = K f(x)g(x) + f(x)g(a - x) = Kf(x) ∫ f(x)g(x)dx + ∫ f(x)g(a - x) dx = ∫ Kf(x) dx.........(1) But ∫ f(x)g(a - x) dx = ∫ f(a - x)g(a - x) dx. Put a - x = y, so - dx = dy When x = 0, y = a When x = a, y = 0 So ∫ f(a - x)g(a - x) dx from 0 to a becomes ∫ - f(y)g(y) dy from a to 0, which is equal to ∫ f(y)g(y) dy from 0 to a. Since y is a dummy variable, it is the same as ∫ f(x)g(x) dx from 0 to a. Put the result into (1) above, we get 2 ∫ f(x)g(x)dx = K ∫ f(x) dx so ∫ f(x)g(x) dx = K/2 ∫ f(x) dx from 0 to a. (b) Put f(x) = sin^2 x = (1 - cos 2x)/2, also you can see that f(x) = f(π - x) Put g(x) = x cos^4 x = x(cos 4x + 4 cos 2x + 3)/8 ( verify yourself) so g(π - x) = (π - x) [ cos4(π - x) + 4 cos 2(π - x) + 3]/8 You can verify that g(x) + g(π - x) = π/8 = K Now using the above result ∫ x sin^2 x cos^4 x dx = π/16 ∫ sin^2 x dx from 0 to π You can verify that ∫ sin^2 x dx from 0 to π = π/2 so answer = (π/16)(π/2) = (π)^2/32. For Q33. If A^2012 really = 1006A^2 - 1005I, that mean x can be replaced by A in the polynomial as far as this matrix is concerned.

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