Linearization and Tangent Planes

To this point, we have considered only partial derivatives of a function f (x; y) : In this section, we introduce the concept of a total derivative of a transforma- tion. Recall that a function f(x) of one variable is said to be di¤erentiable at p if the following limit exists:

f (p + h) f (p) f 0 (p) = lim (1) h 0 h ! The problem with extending (1) to higher dimensions is that it requires h to become a vector, and yet we cannot "divide" by a vector. However, we can rewrite (1) in the form

f (p + h) f (p) lim f 0 (p) = 0 h 0 h ! and thus, we obtain also that

f (p + h) f (p) f 0 (p) h lim = 0 h 0 h h !

Consequently, the de…nition of the derivative (1) can be re-interpreted to mean that f (x) is di¤erentiable at p if there is a number f 0 (p) such that

f (p + h) f (p) f 0 (p) h lim j j = 0 h 0 h ! j j In this formulation, division is by h ; which naturally generalizes to the norm of a vector h = h; k : Indeed, if wej j denote f (x; y) by f (x), where x = x; y ; and let p = (p; qh) bei a point, then we have the following: h i

De…nition 5.1: A function f (x; y) is di¤erentiable at a point (p; q) if there exists a vector f (p) for which r f (p + h) f (p) f (p) h lim j r j = 0 (2) h 0 h ! jj jj When it exists, f (p) is the total derivative of f (x) at p. r

The vector f (p) is also called the gradient of f (x) at p: r

1 If we write f (p) = a; b , then we can determine the values of a and b by evaluating ther limit inh (2)i in two di¤erent directions. Along the x-axis, h = h; 0 and thus we have h i f (p + h; q) f (p; q) a; b h; 0 lim j h i h ij = 0 h 0 h; 0 ! jjh ijj f (p + h; q) f (p; q) ah lim = 0 h 0 h !

f (p + h; q) f (p; q) lim a = 0 h 0 h !

This implies that f (p + h; q) f (p; q) a = lim = fx (p; q) h 0 h !

Similarly, (see the exercises), it can be shown that b = fy (p; q) ; so that the gradient is given by f (p; q) = fx (p; q) ; fy (p; q) r h i This yields the following:

Theorem 5.2: If f (x; y) is di¤erentiable at a point (p; q) ; then its total derivative is given by the gradient of f at (p; q) ; which is

f (p; q) = fx (p; q) ; fy (p; q) r h i

The total derivative is also known as the Jacobian of the mapping f (x; y) for reasons that will become apparent in the next chapter.

EXAMPLE 1 Find the total derivative (i.e., gradient) of

f (x; y) = x2 + 3xy

Solution: Since fx = 2x + 3y and fy = 3x; the total derivative is

f = 2x + 3y; 3x r h i

De…nition 5.1 can be applied to a function f of any number of variables, in which case Theorem 5.2 says that f is the vector of …rst partial derivatives. For example, the gradient of a functionr of 3 variables U (x; y; z) is given by

U = Ux;Uy;Uz r h i and U is the total derivative of U (x; y; z) : r

2 Check your Reading: How is the tangent plane to the surface related to the surface itself?

Linearization and Tangent Planes

If we let x = p + h; then h = x p; so that De…nition 5.1 can be restated in the following form:

De…nition 5.1a: A function f (x; y) is di¤erentiable at a point (p; q) if there exists a vector f (p) for which r f (x) f (p) f (p) (x p) lim j r j = 0 (3) x p x p ! jj jj When it exists, f (p) is the total derivative of f (x) at p. r

The de…nition of the limit then implies that for any " > 0; there is a neighborhood of p for which f (x) [f (p) + f (p) (x p)] < " x p (4) j r j jj jj for all x in that neighborhood. Consequently, we de…ne the linearization of f (x) at p to be

Lp (x) = f (p) + f (p) (x p) r which in non-vector form is

Lp (x; y) = f (p; q) + fx (p; q)(x p) + fy (p; q)(y q) It follows that f (x) Lp (x) < " x p j j jj jj which means that the graph of Lp (x; y) is a plane that is practically the same as the surface z = f (x; y) near the point (p; q; f (p; q)) :

3 We say that z = Lp (x; y) is the tangent plane to z = f (x; y) at (p; q) :

EXAMPLE 2 Find the tangent plane to f (x; y) = 9 x2 y2 when p = (1; 1) :

Solution: To begin with, fx (x; y) = 2x and fy (x; y) = 2y; so that fx (1; 1) = fy (1; 1) = 2 and

Lp (x; y) = f (1; 1) + fx (1; 1) (x 1) + fy (1; 1) (y 1) = 7 2 (x 1) 2 (y 1)

which simpli…es to Lp (x; y) = 11 2x 2y: Thus, the equation of the tangent plane to f (x; y) at the point (1; 1; 7) is

z = 11 2x 2y which is shown below.

The tangent plane is the plane that “best approximates”a surface in the neighborhood of a point. However, this does not exclude the possibility that a tangent plane may intersect a surface in an in…nite number of points.

EXAMPLE 3 Find the linearization of f (x; y) = x3 xy2 at the point (1; 2) :

4 2 2 Solution: Since fx = 3x y and fy = 2xy; we have

fx (1; 2) = 3 4 = 1; fy (1; 2) = 2 1 2 = 4 Since f (1; 2) = 3; the linearization of f (x; y) = sin x2y at (2; ) is Lp (x; y) = 3 1 (x 1) 4 (y 2) which simpli…es to Lp (x; y) = 6 x 4y: The graph of Lp (x; y) is shown versus the graph of f (x; y) below:

Check your Reading: Which of the two tangent planes below is tangent to

5 the surface?

Quadratic Approximations

The "second derivative" of a function f (x; y) is de…ned using matrices, where a matrix A is a 2 dimensional rectangular array of numbers, such as for example

1 2 5 A = 1 0 3 If the length of the rows of A is the same as the length of the columns of B; then the product AB is de…ned to be the matrix of inner products of rows of A with columns of B: For example, if

2 1 4 3 A = and B = 3 5 0 2 then their product is the matrix of inner products of rows of A with columns of B: 2 1 4 3 2 ( 4) + 1 0 2 3 + 1 2 8 8 AB = = = 3 5 0 2 3 (4) + 5 0 3 3 + 5 2 12 19 The transposition of a matrix A is a matrix At whose rows are the columns

6 of A. For example, a1 a t 2 [a1; a2; : : : ; an] = 2 . 3 . 6 7 6 a 7 6 n 7 Vectors v = a; b will be denoted in matrix form4 by5 a column vector, h i a v = b The "second derivative" of f (x; y) at a point (p; q) is given by the Hessian matrix of f (x; y) at a point (p; q) ; which is de…ned to be

fxx (p; q) fxy (p; q) Hf (p) = fxy (p; q) fyy (p; q) Also, the quadratic approximation Qp (x; y) of f at (p; q) is de…ned to be

1 t Qp (x) = Lp (x) + (x p) Hf (p)(x p) 2 where (x p)t = [x p; y q] :

EXAMPLE 4 Find the quadratic approximation of f (x; y) = x3 3x + y2 at (1; 2) :

2 Solution: First, fx = 3x 3 implies that fx (1; 2) = 0, and fy = 2y implies that fy (1; 2) = 4: Since f (1; 2) = 2; the linearization of f (x; y) at (1; 2) is

Lp (x; y) = 2 + 0 (x 1) + 4 (y 2) = 2 + 4 (y 2) Since fxx = 6x; fxy = 0; and fyy = 2; the Hessian at (1; 2) is 6 0 H (1; 2) = f 0 2 Since (x p) = [x 1; y 2]t ; we have 1 6 0 x 1 Qp (x; y) = Lp (x; y) + [x 1; y 2] 2 0 2 y 2 Matrix arithmetic then leads to 1 6 (x 1) Qp (x; y) = Lp (x; y) + [x 1; y 2] 2 2 (y 2) 1 1 = Lp (x; y) + (x 1) 6 (x 1) + (y 2) 2 (y 2) 2 2 2 2 = Lp (x; y) + 3 (x 1) + (y 2)

7 Substitution of Lp (x; y) = 2 + 4 (y 2) then leads to 2 2 Qp (x; y) = 2 + 4 (y 2) + 3 (x 1) + (y 2)

It can be show that for all " > 0; there is a neighborhood of p on which

2 f (x) Qp (x) < " x p j j jj jj Thus, a quadratic approximation is an even better approximation than is a linearization.

EXAMPLE 5 Find the quadratic approximation of f (x; y) = sin x2 + xy + y2 at (0; 0)

2 2 2 2 Solution: First, fx = cos x + xy + y (2x + y) and fy = cos x + xy + y (x + 2y) : Thus, f (0; 0) = fx (0; 0) = fy (0; 0) = 0 and similarly, L (x; y) = 0: However, @ f = cos x2 + xy + y2 (2x + y) xx @x = 2 cos x2+ xy + y2 (2x + y)2sin x2 + xy + y2

from which it follows that fxx (0; 0) = 2: Likewise,

2 2 2 2 fxy = cos x + xy + y sin x + xy + y (x + 2y) (2x + y) 2 2 2 2 2 fyy = 2 cos x + xy + y sin x + xy + y (x + 2y)

Thus, fxy (0; 0) = 1, fyy (0; 0) = 2 and the Hessian is

2 1 H (0; 0) = f 1 2 Consequently, the quadratic approximation is

1 2 1 x 0 Qp (x; y) = L (x; y) + [x 0; y 0] 2 1 2 y 0 Matrix arithmetic and L (x; y) = 0 then leads to

1 2x + y 1 1 Q (x; y) = [x; y] = x (2x + y) + y (x + 2y) p 2 x + 2y 2 2 2 2 However, this in turn simpli…es to Qp (x; y) = x + xy + y :

8 2 2 2 2 Below we have plotted Qp (x; y) = x +xy+y versus f (x; y) = sin x + xy + y .

They are practically the same at (0; 0) ; and indeed, notice that both surfaces have a minimum at (0; 0) :

Check your Reading: Is the product of a row of length n and a column of length n the same as an inner product?

Di¤erentiability and Di¤erentials

If in De…nition 5.1 we instead let h = x and f = f (p + x) f (p) ; then f f (p) x lim j r j = 0 x 0 x ! jj jj Thus, for any " > 0; there is a neighborhood of 0 such that f f (p) x j r j < " x k k for all nonzero x in that neighborhood. Equivalently, f f (p) x < " x j r j k k and if we further allow that " (x) is a continuous function for which " (0) = 0; then we obtain the following:

9 De…nition 5.1b: A function f (x) of n-variables is di¤erentiable at a point p with total derivative f (p) if there is a continuous function " (x) with " (0) = 0 suchr that

f f (p) x < " (x) x j r j k k on some neighborhood of 0.

If we let dz = f (p) x and let x = dx; dy ; then Theorem 5.2 implies that r h i dz = fx (p; q) dx + fy (p; q) dy De…nition 5.1b then implies that

z dz whenever dx and dy are su¢ ciently close to 0.

EXAMPLE 6 Compute z and dz for f (x; y) = x2 + xy when (p; q) = (1; 2) ; dx = x = 0:01; and dy = y = 0:03:

Solution: To begin with, f (1; 2) = 12 +1 2 = 3 and f (1:01; 2:03) = 2 (1:01) + 1:01 2:03 = 3: 0704, so that z = f (1:01; 2:03) f (1; 2) = 3: 0704 3 = 0:0704

Moreover, fx (x; y) = 2x + y and fy (x; y) = x; so that

fx (1; 2) = 2 1 + 2 = 4 and fy (1; 2) = 1 As a result, we have

dz = fx (1; 2) dx + fy (1; 2) dy = 4 0:01 + 1 0:03 = 0:07 which is within 0:0004 of z:

Let’slook at an application of the di¤erential, one in which dx and dy are both interpreted to be tolerances due to the inaccuracy of measuring devices.

EXAMPLE 7 A large co¤ee can has a height of 6:500 to within an accuracy of 1=1600 and a base with a radius of 3:2500 to within an

10 accuracy of 1=1600: Find the volume V and the approximate error dV in the volume of the can.

Solution: If h denotes the height and r denotes the radius of the base of the can, then V = r2h When h = 6:5 and r = 3:25; then V = (3:25)2 6:5 = 215: 69 cubic inches. Moreover,

dV = Vh (6:5; 3:25) dh + Vr (6:5; 3:25) dr

2 where Vh = r and Vr = 2rh: Since dh = dr = 1=1600, we thus have 1 1 dV = (3:25)2 + 2 (3:25) (6:25) = 10:05 in3 16 16 Thus, the volume of the co¤ee can is 215:7 in3; give or take approx- imately 10 in3:

Exercises Find the linearization L (x; y) of f (x; y) at the given point in the xy-plane. Then graph L (x; y) and f (x; y) to illustrate that z = L (x; y) is tangent to z = f (x; y).

1. f (x; y) = x2 + y3 at (1; 2) 2. f (x; y) = x2y + y2 at (1; 2) 3. f (x; y) = x2 + xy + 3x at (0; 0) 4. f (x; y) = (x + 2y)2 at (1; 3) 5. f (x; y) = ln x2 + y2 at (1; 1) 6. f (x; y) = x sin (xy) at (1; ) 7. f (x; y) = x2 sin (y) at (0; ) 8. f (x; y) = ex ln y2 + 1 at (1; 0) 9. f (x; y) = 3x+ 7y + 1 at (1; 2) 10. f (x; y) = 3x + 7y + 1 at (1; 2) 11. f (x; y) = x tan (xy) at (1; ) 12. f (x; y) = csc (x + y) at 2 ; 3

13. f (x; y) = 1 at (0; 1) 14. f (x; y) = 1 at (0; 0) (x2+y2)3=2 1 xy

11 Find the Hessian matrix and the quadratic approximation of f at the given point. 15. f (x; y) = x2 + y3 at (1; 2) 16. f (x; y) = x2 + 2xy + y3 at (2; 2) 17. f (x; y) = x2 + xy + 3x at (0; 0) 18. f (x; y) = (x + 2y)2 at (1; 3) : 19. f (x; y) = 3x 2y + 1 20. f (x; y) = 5 21. f (x; y) = ln x2 + y2 at (1; 1) 22. f (x; y) = x sin (xy) at (1; ) 23. f (x; y) = x2 sin (y) at (0; ) 24. f (x; y) = ex ln y2 + 1 at (1; 0) Find z and dz at the given point in the xy-plane.

25. z = x2 + 2y at (1; 2) 26. z = x2y + y2 at (1; 1) dx = x = 0:1; dy = y = 0:01 dx = x = 0:1; dy = y = 0:01 2 27. z = x tan (xy) at (1; ) 28. z = (x + 2y) at (1; 1) dx = x = 0:01; dy = y = 0:01 dx = x = 0:01; dy = y = 0:1 1 y x2 y2 29. z = tan x at (1; 1) 30. z = e at (0; 0) dx = x = 0:01; dy = y = 0:01 dx = x = 0:1; dy = y = 0:2 31. A rectangular box has a height of 4 feet to within an accuracy of 1 in and a square base with width 2 feet to within an accuracy of 1 in. Find the volume V and the approximate error dV in the volume of the can. 32. The length of a pendulum is measured to be l = 5 feet to within 1 in of accuracy. The period of the pendulum as it swings is measured to be T = 2:5 seconds to within 0:1 seconds. The acceleration due to gravity is related to the swinging of the pendulum by 42l g = T 2 What is the acceleration due to gravity with respect to these measurements and about how accurate is the computed value of the acceleration? 33. The diameter of the top of a soup can is measured to be 3 inches to within 1/16 of an inch. The height of the can is measured to be 4 inches to within 1/16 of an inch.

a. Write the surface area S as a function of the height and diameter of the can.

b. Find S when the height is 400 and the diameter is 300:

c. Find dS when the height is 400; the diameter is 300; and the di¤erentials of height and diameter are both 1=1600: About how much variation in the area computation is possible given that height and diameter are accurate only to a sixteenth of an inch?

34. Find the volume V and the di¤erential dV for the can in exercise 33. About how much variation in the volume computation is possible given that height and diameter are accurate only to a sixteenth of an inch? 35. Show that the tangent plane to any plane of the form z = ax + by + c is the plane itself.

12 36. Show that the quadratic approximation at (0; 0) of

Q (x; y) = ax2 + bxy + cy2 is Q (x; y) itself. 37. The mirror for a telescope should be in the shape of a paraboloid with focus (0; 0; p), which is the graph of a function of the form

x2 + y2 Q (x; y) = (5) 4p For example, a telescope mirror with focus at (0; 0; 1) is shown below:

However, if the telescope mirror is small enough, it more economical to manufacture a spherical cross-section that approximates (5). The graph of a spherical cross-section of radius R is given by

f (x; y) = R R2 x2 y2 Find the Quadratic approximation of fpat (0; 0) and show that it is of the form (5). What is R in terms of p? How does this relate to the manufacture of spherical mirrors as approximations to parabolic mirrors? 38. In problem 37, what value of R leads to its quadratic approximation having a focus of (0; 0; 2)? 39. Use the de…nition of the total derivative to show that if f (p; q) = a; b ; r h i then b = fy (p; q) : 40. Show that if f (x) is di¤erentiable at p; then there exists a neighborhood of p on which

f (p)(x p) " x p < f (x) f (p) < f (p)(x p) + " x p r jj jj r jj jj for all x in that neighborhood. Use this to explain why f (x) is continuous at p . (that is, why lim f (x) = f (p)) x p ! 41. Show that the tangent plane to the graph of f (x; y) = x2 y2 is of the form z = 2px 2qy p2 q2 (6) 13 Then show that the intersection of z = x2 y2 with its tangent plane (6) results in a pair of straight lines. How many lines does the surface z = x2 y2 contain? 42. Write to Learn: Write a short essay which explains why iff (x; y) is a 2 2 polynomial such as f (x; y) = x y + y ; then fx and fy are the coe¢ cients of h and k in the expansion of f (x + h; y + k) : How does this relate to the concept of di¤erentiability? 43. Write to Learn: In a short essay, explain why the function f (x; y) = x2 + y2 is not di¤erentiable at (0; 0) : (Hint: consider the graph of f (x; y) and also consider the de…nition of di¤erentiability). p 44. Show that fx (0; 0) exists but fy (0; 0) does not exist when

f (x; y) = x4 + y2

Does the total derivative of f (x; y) existp at (0; 0)? Explain. 45. Explain how De…nition 5.1 applies to functions of 3 variables U (x; y; z) and then mimic the derivation which follows to show that

U (p) = Ux (p) ;Uy (p) ;Uz (p) r h i 46. For f (x) di¤erentiable at a point p;let us de…ne a function A (x) by

A (x) = f (p) x r Show that A (x) is linear (i.e., that A (x + y) = A (x) + A (y) and A (kx) = kA (x) for any scalar k ). Also, explain the signi…cance of vectors x for which A (x) = 0: