Question

A function f is given along with a local linear approximation l to f at a point P. Use the information given to determine the point P

(a) f (x, y) = x2+ y2; l(x, y) = 2y − 2x − 2

(b) f (x, y) = x2y; l(x, y) = 4y − 3x + 8

(c) f (x, y,z) = xy + z2;l(x, y, z) = y + 2z − 1

(d) f (x, y, z) = x yz; l(x, y, z) = x − y − z − 2

Answer 1

We know that the linear approximation has the formula

This means that the coefficients in front of x and y are the partial derivatives of f evaluated at the point P.

(a) We have

fx (x, y) = 2x

fy (x, y) = 2y

and therefore at the point P = (a, b) we must have

2a = −2

2b = −2

Hence P = (−1,1) is the only possible solution. Note that we also need to check, that the constant term in the linear approximation is correct, by computing the linear approximation of f at the computed point P

l(x, y) = 2y − 2x − 2

(b) We have

fx (x, y) = 2x y

fy (x, y) = x 2

and therefore at the point P = (a, b) we must have

2ab = −3

a 2 = 4

The second equation has two solutions x81 a = ±2 and hence we obtain two candidates P1 = (2,− 3 /4) and P2 = x81 (−2, 3 /4) . However the linear approximations at these points are

At P1 : l(x, y) = 6 − 3x + 4y

At P2 : l(x, y) = −6 − 3x + 4y

Hence the given linear function cannot be the linear approximation of f

(c) We have

(d) We have

and therefore at the point P = (a, b,c) we must have

bc = 1

ac = -1

ab = -1

The first equation gives us b = 1/ c and the second one a = − 1 /c . Hence the third equation becomes c 2 = 1, giving us the two solutions P1 = (1,−1,−1) and P2 = (−1,1,1). The linear approximations of f at these points are

At P1 : l(x, y, z) = −2 + x − y − z

At P2 : l(x, y, z) = 2 + x − y − z

Therefore P1 is the only solution.