請問有誰知道這題工程數學~

第7題是偏微分的應用

第8題是求質心的專章，應屬於重積分的範圍

7.
令F(x,y,z)=x^2-4y^2+z^2-16=0
▽F(2,1,4)=(2x,-8y,2z)│(2,1,4)=(4,-8,8)→取(1,-2,2)為曲面在(2,1,4)之法向量

令切平面上任一點(x,y,z)，則((x-2),(y-1),(z-4))和法向量垂直
((x-2),(y-1),(z-4))●(1,-2,2)=0
(x-2)-2(y-1)+2(z-4)=0

得切平面 x-2y+2z-8=0


8.
density function：ρ(x,y)=y, 以下∫(a,b)表積分下限a上限b
m=∫∫ρ(x,y) dA
=∫(0,π/4)∫(sin x, cos x) y dy dx (以下的上下限省略)
=∫(1/2)y^2│(sin x, cos x) dx
=(1/2)∫(cos^2 x - sin^2 x) dx
=(1/2)∫cos 2x dx
=(1/4)∫cos 2x d2x
=(1/4)sin 2x│(0,π/4)
=(1/4)[sin (π/2) - sin 0]
=1/4

My=∫∫ xy dy dx
=∫(1/2)xy^2│(sin x, cos x) dx
=(1/2)∫ x(cos^2 x - sin^2 x) dx
=(1/2)∫ x cos 2x dx
u=x, dv=cos 2x dx, du=dx, v=(1/2)sin 2x
=(1/2)[(1/2)x sin 2x - (1/2)∫ sin 2x dx]
=(1/4)[x sin 2x - ∫ sin 2x dx]
=(1/4)[x sin 2x + (1/2)cos 2x]
=(1/4)x sin 2x + (1/8)cos 2x│(0,π/4)
=[(1/4)(π/4)sin π/2 +(1/8)cos π/2] - [0+(1/8)cos 0]
=π/16 - 1/8

Mx=∫∫ y*y dy dx
=∫ (y^3)/3│(sin x, cos x) dx
=(1/3)∫ (cos^3 x- sin^3 x) dx
=(1/3)[∫ cos^3 x dx - ∫ sin^3 x dx
=(1/3)[∫ cos^2 x(cos x dx) - ∫ sin^2 x(sin x dx)]
=(1/3)[∫ (1- sin^2 x) d sin x - ∫ (1 - cos^2 x) d cos x]
=(1/3)[sin x - (sin^3 x)/3 + cos x - (cos^3 x)/3]│(0,π/4)
=(1/3)[sin π/4 - (sin^3 π/4)/3 + cos π/4 - (cos^3 π/4)/3]
- (1/3)[sin 0 - (sin^3 0)/3 + cos 0 - (cos^3 0)/3]
=(1/3)(√2 - √2/6] - (1/3)(2/3)
=(5√2 - 4)/18

x-bar = My/m = (π/16 - 1/8)/(1/4) = π/4 - 1/2
y-bar = Mx/m = [(5√2 - 4)/18]/(1/4) = (10√2 - 8)/9

7.Find an equation of the tangent plane to the graph of F=x^2-4y^2+Z^2-16=0 at (2,1,4)Ans:The Normal vector isN=▽F=(2x,-8y,2z)=2*(x,-4y,z)Thus tangent plane passing (2,1,4) becomes0=(x-2)-4(y-1)+(z-4)=x-4y+z-2
8.A lamina=1(meter) has the shape of the region in the first quadrant is bounded by the graphs of y1=sinx, y2=cosx, between x=0 and x=pi/4Find its center of mass if the density is p(x,y) = y. (15%)Ans:
The mass for z=1(meter) can be computed asm=∫(P2-P1)*dV=∫(y2-y1)dV=∫(y2-y1)(y2-y1)dx=∫(cosx-sinx)^2*dx=∫(cosx^2-2sinx*cosx+sinx^2)dx=∫(1-sin2x)dx=∫dx-∫sin2x*d(2x)/2=x+(cos2x)/2.....x=0~pi/4=pi/4+(cos90)/2-0-(cos0)/2=pi/4-1/2=(pi-2)/4 (kg/m)m*Xbar=∫x*dm=∫x(P2-P1)dV=∫x(P2-P1)^2*dx=∫x(1-sin2x)dx=∫xdx-∫(2x)*sin2x*d(2x)/4=x^2/2+(1/4)*∫(2x)*d(cos2x)=x^2/2+(1/4)*[2x*cos2x-∫cos2x*d(2x)]=x^2/2+(1/4)*(2x*cos2x-sin2x).....x=0~pi/4=pi^2/32+(1/4)*(0.5*pi*cos90-sin90-0+0)=(pi^2-8)/32=> Xbar=(pi^2-8)/8(pi-2)=0.204714
m*Ybar=∫(y2*dm-y1*dm)=∫(y2*dy2-y1*dy1)dx.....dy1=cosx*dx=∫∫(-cosx*sinx-sinx*cosx)(dx)^2.....dy2=-sinx*dx=-∫∫sin2x(d(2x))^2/4=∫cos(2x)*d(2x)/4=(sin2x)/4.....x=0~pi/4=sin90/4-0=1/4Ybar=1/(pi-2)=0.876
3.工程數學哪個範圍?Ans:第8題屬於微積分求法線部份: N=Grad(F)第8題屬於應用力學重心部份 or 微積分求重心部份.工數談更高的問題


2014-01-21 07:12:02 補充：
第7題修改如下:

N(2,1,4)=▽F

=2*(x,-4y,z)

=2*(2,-4,4)

Tangent plane:

0=2(x-2)-4(y-1)+4(z-4)

=2x-4y+4z-4+4-16

=2x-4y+4z-16

=x-2y+2z-8.....ans

2014-01-21 07:36:23 補充：
第8題修改如下: 使用應用力學公式 =>

m=m2-m1

=∫p2*dV2-∫p1*dV1

=∫y2*(y2*1*dx)-∫y1*(y1*1*dx)

=∫(y2^2-y1^2)dx

=∫((cosx)^2-(sinx)^2)dx

=∫cos2x*dx

=∫cos2x*d(2x)/2

=(sin2x)/2.....x=0~pi/4

=(sin90)/2

=1/2

2014-01-21 07:37:09 補充：
Xbar*m=Σx*m2-Σx*m1.....應用力學公式

=∫x*dm2-∫x*dm1

=∫x*p2*dV2-∫x*p1*dV1

=∫x*p2*(y2*1*dx)-∫x*p1*(y1*1*dx)

=∫x*[(cosx)^2-(sinx)^2]dx

=∫x*cos2x*dx

=∫(2x)*cos2x*d(2x)/4

=∫(2x)*d(sin2x)/4

=(2x*sin2x-∫sin2x*d(2x))/4.....部份積分

=0.5*x*sin2x+0.25*cos2x

=pi*sin90/8+cos90-0-0.25*cos0

=pi/8-1/4

=(pi-2)/8

2014-01-21 07:38:01 補充：
=> Xbar=(pi-2)/8m

=(pi-2)/4

=0.2854

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