數學問題(d,in,matrix)2

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.

g(x) = x cos^4 x
g(π - x) = (π - x) cos^4 x
g(x) + g(π - x) = π = K