Thank You Consider the system of equations {x 1 y 2 2 - 2 x 2 y 3 = 1 x 1 y 1 +

Question

Thank You

Consider the system of equations {x 1 y 2 2 - 2 x 2 y 3 = 1 x 1 y 1 + x 2 y 2 - 4 y 2 y 3 = -9 x 2 y 1 + 3 x 1 y 2 3 = 12. Show that there is a continuously differentiable function expressing y 1, y 2, y 3 in terms of x 1, x 2 in a neighbourhood of (x 1, x 2, y 1, y 2, y 3) = (1,0,-1,1,2). That is. show that there exists an open set V (1,0,-1,1,2), an open set U (1,0) and : U rightarrow R such that (1,0) = (-1,1,2) and (x 1,x 2,y 1,y 2, y 3) is in V and satisfies all three equations above iff (x 1,x 2) epsilon U and (y 1, y 2, y 3) = ( x1, x 2). Evaluate at (1,0).

Explanation / Answer

(a) Let f : R^2 x R^3 -> R^3 defined as f( (x1,x2),(y1,y2,y3)) = (x1y2^2-2x2y3-1,x1y1^5+x2y2-4y2y3+9,x2y1+3x1y3^2-12)

First, we see that f(1,0,-1,1,2) =(0,0,0). So with x=(1,0) and y=(-1,1,2) that is f(x,y)=0.

First let's compute the matrix of partial derivate with respect to the second variable (y1,y2,y3):


Y =

0 2x1y2 -2x2

5y1^4x1 (x2-4y3) -4y2

x2 0 6x1y3

Now let's put (x1,x2)=(1,0) and (y1,y2,y3)=(-1,1,2)

So Y=

0 2 0

5 -8 -4

0 0 12

We know that Y has to be invertible, the determinant is 12 * (-5*2) = -120 so the matrix is invertible.

So the implicit function theorem can be applied (this was important to check Y was invertible !) and :

There exist an open set U of R^2 containing (1,0), an open set V of R^3 containing (-1,1,2) and a function phi : U -> V such that { (x,phi(x)) | x in U } = { (x,y) in U x V | f(x,y)=0}.

So in particular (x,y)=(x1,x2,y1,y3,y3) in V satisfy all equation <=>

f(x,y)=0 <=> y=phi(x) <=> (y1,y2,y3) = phi(x1,x2) where x=(x1,x2) and y=(y1,y2,y3)


(b) We use the formula for partial derivate see course) :

With x=(1,0) we have phi(1,0)=(-1,1,2)

dphi/dx1(1,0) = - Y ^(-1)df/dx1(x,phi(x))

df/dx1 = (y2^2,y1^5,3y3^2) => df/dx1(x,phi(x)) = (1,-1,12)

We can compute :

Y^(-1) = 1/60

48 12 4

30 0 0

0 0 5


Hence dphi/dx1(1,0) = (dphi1/dx1,dphi2/dx1,dphi3/dx1) = -1/10 ( 14, 5, 10)

I advise you to check the values, as i didn't verify ... but you have at least the methodoly.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.