→ nickivan:積分實在很難打字所以就算了>< 06/19 11:29
: 試題 :
: 1. use the chain rule to find δz/δs and δ^2z/δtδs
: (1) z=x^2y+3xy^4, x=s^2t, y=s lnt
: (2) z=f(s+at)+g(s-at), where a is a constant and f,g are differentiable.
: (3) z=f(x,y) is differentiable, where x=cost e^t ,y=sint e^t
(1)dz=2y*x^(2y-1)*(dx/ds)+[(3y^4)*(dx/ds)+4*3xy^3*(dy/ds)
and (dx/ds)=(2t)*s^(2t-1) (dy/ds)=ln(t)
=>dz/ds =2y*x^(2y-1)*(2t)*s^(2t-1)+(3y^4)*(2t)*s^(2t-1)+12xy^3*ln(t)
(2)dz=f'(s+at)*(1+a*dt/ds)+g'(s-at)*(1-a*dt/ds)
and from y=slnt => t=[exp(y)]^(1/s)=>dt/ds=(1/s)[exp(y)]^(1/s-1)
=>dz/ds=f'(s+at)*{1+a*(1/s)[exp(y)]^(1/s-1)}+g'(s-at)
*{1-a*(1/s)[exp(y)]^(1/s-1)}
(3) dz=f'(x,y)[dx/dt+dy/dt], and dx/dt=[-sin(t)]e^t+[cos(t)e^t]
dy/dt=[cos(t)]e^t+[sin(t)e^t]
=>dz/dt=f'(x,y){[-sin(t)]e^t+[cos(t)e^t]+[cos(t)]e^t+[sin(t)e^t]}
: 2. if f is homogeneous of degree n (that is f(tx,ty)=t^n(x,y) for all t),
: show that
: f (tx,ty)= t^(n-1)f (x,y)
: x x
題目有錯: 應該是 f(tx,ty)=t^n*f(x,y) for all t
=>df(tx,ty)/dx=f x (tx,ty)*t =t^n*f x (x,y)
=>f x (tx,ty) =(1/t)*(t^n)*f x (x,y)=t^(n-1)f x (x,y) QDE
: 3. find the points on the ellipsoid x^2+2y^2+3z^2=1 where the tangent
: plane is parallel to the plane 3x-y+3z=1
Suppose tangent plane is 3x-y+3z=a 將此平面代入x^2+2y^2+3z^2=1
解出a
: 4. find the local maximum and minimum values of f=x^3-3x-y^3+12y
f(x,y)=x^3-3x-y^3+12y
f.o.c=> df/dx=3*x^2-3=0
df/dy=-3*y^2+12=0
解出 x^2=1 and y^2=4
if x=1 and y=2 代入f(x,y)=1^3-3*1-2^3+12*2=1-3-8+24=12
x=-1 and y=2 代入f(x,y)=-1+3-8+24=18
x=1 and y=-2 代入f(x,y)=1-3+8-24=-18
x=-1 and y=-2 代入f(x,y)=-1+3+8-24=-12
local maximum=18
local minimum=18
: 5. explain the method of lagrange multipliers works in finding the
: extreme values of f(x,y,z) subject to the constraint g(x,y,z)=k
Max or Min f(x,y,z) subject to g(x,y,z)=k
let L(x,y,z,a)=f(x,y,z)+a[k-g(x,y,z)],where a>0 is lagrange multipliers
: 6. use a double integral to find the volume of the tetrahedron bounded
: by the planes x+y+z=2, x=3y, x=0, and z=o
: 7. sketch the regions and evaluate the integrals:
: (1) change the order to : 1 1 ________
: ∫ ∫ √x^3 + 1 dxdy
: 0 √y
交換積分順序 先積y在積x
1 1 ________
: ∫ ∫ √x^3 + 1 dydx
: √y 0
1 ________ 1
= ∫ {[√x^3 + 1] y| }dx
: √y 0
1 ________
= ∫ {[√x^3 + 1] }dx
: √y
: 第五題是要描述這個方法 而不是解釋它的原理
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