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。