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Advanced Mathematics 2 asks for expert answers!
6。 It is known that the function z=xy has a maximum value under the condition of x+y= 1, then this maximum value is _ _ _ _ _ _ _ _ _.

Solution: z=xy=x( 1-x)=-x? +x=-(x? -x)=-[(x- 1/2)? - 1/4]=-(x- 1/2)? + 1/4≦ 1/4

That is, when x= 1/2 and y= 1/2, z gets the maximum value of 1/4.

7。 On the curve x=t, y=-t? ,z=t? In the tangent line, the tangent equation perpendicular to the plane x-2y+3z=6 is _ _ _ _ _ _ _ _

Solution: The direction number of the normal of the plane x-2y+3z=6 is (1,-2,3).

Substituting the curve parameter equation into the plane equation, we get: t+2t? +3t? =6, thus obtaining t =1; So the coordinates of the intersection m between the curve and the plane

It is (1,-1, 1).

dx/dt= 1,dy/dt=-2t,dz/dt=3t? ; The corresponding derivatives of t= 1 are XO ′ =1,yo ′ =-2 and zo ′ = 3.

Therefore, the tangent equation perpendicular to the plane and passing through m is: (x-1)1= (y+1)/(-2) = (z-1)/3.

8. Let ∑ be the surface z=√(x? +y? ) is cut by z= 1, then ()? (x? +y? )ds = _ _ _ _ _ _ _ _ _ _ _ _

Solution: the whole domain ∑: x? +y? = 1, which is a circle with a radius of 1, so we use polar coordinates to calculate the double integral:

(∑)? (x? +y? )ds=(∑)? (x? +y? )dxdy=(∑)? (r? Because? θ+r? Sin? θ)rdrdθ=(∑)? r? drdθ

=[0,2π]∫dθ[0, 1]∫r? dr=2π/4=π/2。

9. The McLaughlin series of the function f(x)=x/(2-x) is _ _ _ _ _ _, and the convergence domain is _ _ _ _ _.

Solution: f(x)=x/(2-x)=- 1+2/(2-x)

f′(x)= 2/(2-x)? ; f ”( x)= 4(2-x)/(2-x)? =4/(2-x)? ; f ""( x)= 12(2-x)? /(2-x)^6= 12/(2-x)? ;

f? (x)=48/(2-x)^5; ..........

f(0)=0,f′(0)= 1/2,f″(0)= 1/2,f ″( 0)= 12/ 16 = 3/4,f? (0)=48/32=3/2; ......

So x/(2-x) = (1/2) x+(1/4) x? +( 1/8)x? +( 1/ 16)x? +........+( 1/2? )x? + ........

Convergence interval: a? n+ 1? /a? n? =[ 1/2^(n+ 1)]/[( 1/2? )x? ]= 1/2, so the convergence radius R=2 and the convergence interval [-2,2].

10. Triple integral I =ω∫∫∫(x? +y? +z? ) d nu, where the integer domain ω is defined as z=√(4-x? -Really? ) and z=0, converted into rectangular coordinates.

Cubic integration under _ _ _ _ _ _ _

Solution: by z=√(4-x? -Really? ) get an x? +y? +z? =4, this is a ball with the center at the origin and the radius R=2, and the integral domain ω is above this ball.

Half.

I =ω∫∫∫(x? +y? +z? )dυ= I =ω∫∫∫(x? +y? +z? )dxdydz

=[0,π]∫dφ[0,2π]∫dθ[0,2]∫r? (sin? φcos? θ+sin? φsin? θ+cos? φ)dr

=[0,π]∫dφ[0,2π]∫dθ[0,2]∫r? dr=(8/3)×2π? = 16π? /3

1 1 exchange the integration order and calculate the quadratic integral [1, 5] ∫ (1/y) dy [y, 5] ∫ (1/lnx) dx.

Solution: The integer value has nothing to do with the integral order (subject), but the problem is that ∫( 1/lnx)dx cannot be expressed in finite form.

12。 Set the function u=z√(x? -Really? ) and point A (5, 4, -3), Q: In what direction is the derivative of the function U at point A maximum? Maximum direction

What is the derivative?

Solution:? u/? x︱[A]=xz/√(x? -Really? )︱[a]=- 15/3; ? u/? y︱[A]=-yz/√(x? -Really? )︱[a]= 12/3;

u/? z︱[A]=√(x? -Really? )︱[A]=3

So (? u/? l)? Answer? =-( 15/3)cosα+( 12/3)cosβ+3c OSγ

Where is cos? α+cos? β+cos? γ= 1

Obviously, when α=π/2, β=γ=π/4 (? u/? l)? Answer? Maximum, [(? u/? l)? Answer? ]max =[( 12/3)+3](√2/2)=(7/2)√2。

13. calculate I = [l] ∫ (sinx-ey) dy+[-ln (1+x? ) +ycosx]dx where l is the curve x=√( 1-y? ) The arc from point A (0,-1) to point B(0, 1).

Solution: Is curve L a circle X? +y? In the right half of = 1, the equation is changed to a parametric equation: let x=cost, y=sint, -π/2≦t≦π/2.

When dx = dcost =-sindt, dy=dsint=costdt, t=-π/2, x=0, y =-1; When t=π/2 and x=0, y= 1. Therefore:

i=[l]∫(sinx-e^y)dy+[-ln( 1-x? )+ycosx]dx

=[-π/2,π/2]∫{[sin(cost)-e^(sint)]cost-[-ln( 1-cos? T)+sintcos (cost) ]sint}dt

=[-π/2,π/2]∫{[sin(cost)]cost-sin? Tcos (cost)-cost (sint)+sintln (sin? t)}dt

For simplicity, I calculate the above four integrals separately:

The first one: ∫ sin (cost) costdt = ∫ sin (cost) d (Sint) = sin (cost) Sint-∫ Sint cos (cost) (-Sint) dt.

=sin (cost) sin+∫ sin? Tcos (cost) dt ......................................... (1)

The second one: ∫ [-sin? T cos (cost) dt=-∫sin? Tcos (cost) dt...................(2)

(1)+(2) and substitute into the upper and lower limits, sin(cost)sint︱[-π/2, π/2] = 0 .............................. (1).

Third:-∫ Cost E (Sint) DT =-∫ E (Sint) D (Sint) =-E (Sint) ............ (3)

-e (Sint) ~ [-π/2,π/2]=-e+ 1/e=( 1-e? ) /e ............................... (2)

The fourth one: ∫sint ln(sin? t)dt=-∫ ln(sin? T)d (cost) =-[cost ln(sin? t)-∫(2sintcos? t/sin? t)dt]

=-cost ln(sin? t)+2∫[(cos? t)/sint]dt=-cost ln(sin? t)+2∫[( 1-sin? t)/sint]dt

=-cost ln(sin? t)+2[∫csctdt-∫sin TDT]=-cost ln(sin? t)+2[ln︱csct-cot t︱-cost]

{-cost ln(sin? T)+2 [ln ︱ CSCT-COT ︱-cost]} ︱ [-π/2, π/2] = 0 ... (3)

Therefore, I = [l] ∫ (sinx-ey) dy+[-ln (1+x? )+ycosx]dx=A+B+C=( 1-e? )/e。