4 Partial Differentiation

Transcription

1 4 Partial Differentiation Many equations in engineering, physics and mathematics tie together more than two variables. For example Ohm s Law (V = IR) and the equation for an ideal gas, PV = nrt, which gives the relationship between pressure (P), volume (V ) and temperature (T). If we vary any two of these then the behaviour of the third can be calculated: P = nrt V, V = nrt P, T = PV nr. How P varies as we change T and V is easy to see from the above, but we want to adapt the tools of one-variable calculus to help us investigate functions of more than one variable. For the most part we shall concentrate on functions of two variables such as z = x 2 + y 2 or z = xsin(y + e x ). Graphically z = f(x, y) describes a surface in 3D space varying the x- and y-coordinates gives the z-coordinate, producing the surface: z 15 z 0 = f(x 0, y 0 ) 10 x (x 0, y 0 ) y 2 x 00 y 2 One task of interest will be to maximise or minimise such functions, where we may have to take into account limitations on the domain of definition, i.e. those points (x, y) for which we calculate z = f(x, y). Restrictions on the domain can come from both mathematical and physical reasons. For example above the function T = PV nr makes mathematical sense for negative P or V, but, physically, negative pressure or volume has no obvious meaning. Exercise 4.1. What is the domain of the function z = f(x, y) = 1 x 2 y 2? Consider the function z = ln(x + y). From its definition we see that it is defined only when x + y > 0, that is only for points (x, y) R 2 lying above the line y = x. Moreover on any line x + y = a for a > 0 we have z = lna, that is z maintains a constant value, so we have a contour line of the surface: 59

2 y x x + y = a x + y = 0 Note also that on any line with equation y = mx + c for constants m 1 and c, we have z = ln ( (m + 1)x + c ) = ln ( x + c/(m + 1) ) + ln(m + 1) for all x > c/(m + 1), so that we get a copy of the graph of the logarithm curve when not travelling parallel to the lines of constancy. For example on y = x, z = ln(2x) = lnx + ln 2. As another example, consider the function z = x 2 +y 2. If we choose a positive value for z, for example z = 4, then the points (x, y) that can give rise to this value are those satisfying x 2 + y 2 = 4 = 2 2, i.e. those on the circle centred on the origin of radius 2. On the other hand if we fix a value for x, for example x = 0, then we have z = y 2. If we fix x = 1 then z = y Both of these are parabolas, and indeed fixing any value of x produces such a curve. Symmetrically, fixing a value for y also produces a parabola, e.g. z = x 2 + ( 3) 2 = x Note that at (x, y) = (0, 0), z = 0, but if x 0 or y 0, then x 2 > 0 or y 2 > 0, and it follows that z > 0. Thus the minimum value taken by this function is z = 0, at the origin. This contrasts with our earlier example z = ln(x + y) where if we move along y = x we have z = ln(2x) = lnx + ln 2, which diverges to as x 0, and diverges to + as x +. Thus there is no overall maximum or minimum value. Unfortunately in general it is harder to picture what is happening with less simple multivariable functions, such as z = sin(x 2 + y) + e xy. One useful technique illustrated above for z = x 2 + y 2 is to hold either x or y constant. For example consider z = x 2 (1 y) xy 2 + y 3. Setting x = 0 gives and setting x = 1 gives z = y 3 dz dy = 3y2, z = y 3 y 2 y + 1 = (y 1) 2 (y + 1) dz dy = 3y2 2y 1 = (y 1)(3y + 1). On the other hand if y = 2 then z = 3x 2 4x 8 dz = 6x 4 = 2(3x 2). dx 60

3 All of these slices through the surface give us an insight into the behaviour of the function: z z z y y x z = y 3 z = y 3 y 2 y + 1 z = 3x 2 4x Definition of partial derivatives Suppose that z = f(x, y) is a function of two variables. We define partial derivatives taken with respect to x and with respect to y by: x = lim h 0 f(x + h, y) f(x, y), h y = lim k 0 f(x, y + k) f(x, y), k whenever these limits exist. These definitions mirror those for the one variable case. For x we are holding the value of y fixed, altering x by a small amount h to get the point (x+h, y), and calculating the slopes of straight line approximations to the tangent in the x-direction. Similarly calculating involves holding x fixed and finding the limit of approximations to y the tangent in the y-direction. Applying this definition to the function z = x 2 (1 y) xy 2 + y 3 above we have [ (x + h) 2 x = lim (1 y) (x + h)y 2 + y 3] [ x 2 (1 y) xy 2 + y 3] h 0 h (2xh + h 2 )(1 y) hy 2 = lim h 0 h [ = lim (2x + h)(1 y) y 2 ] = 2x(1 y) y 2, h 0 and this limit exists at all points (x, y) in the plane. A similar calculation shows that y = x2 2xy + 3y 2. 61

4 Definition of partial derivatives However, in practice it is rarely necessary to go back to the definitions as we did above. Indeed, since all we are doing is holding one variable fixed, we can treat this variable as a constant in our calculations. For example if z = x 2 + xy 5 6x 3 y + y 4 then Similarly, x = d dx (x2 ) + y 5 d d (x) 6y dx dx (x3 ) + y 4 d dx (1) = 2x + y 5 1 6y 3x 2 + y 4 0 = 2x + y 5 18x 2 y. y = d x2 dy (1) + x d dy (y5 ) 6x 3 d dy (y) + d dy (y4 ) = x x 5y 4 6x y 3 = 5xy 4 6x 3 + 4y 3. Using this technique we can make use of known results from one-variable theory such as the product and quotient rules, and the chain rule if we are considering a function of one variable. (The general chain rule for partial derivatives is a little more complicated.) For example if z = sin(xy)e x+y then x = ( ) sin(xy) e x+y + sin(xy) x x ex+y Product rule = cos(xy) x (xy) ex+y + sin(xy) e x+y (x + y) x Chain rule 2 = y cos(xy)e x+y + sin(xy)e x+y A similar calculation shows that y = xcos(xy)ex+y + sin(xy)e x+y Alternative notations If z = f(x, y), then x is often written as f x(x, y), and y as f y(x, y). A further alternative is to write these as f 1 (x, y) and f 2 (x, y) respectively, where the number indicates whether the derivative is being taken with respect to the first or second variable. Exercise 4.2. For which points (x, y) is the function z = ln(y x 2 ) + sin(x + y 2 ) defined? Sketch the domain of definition. Calculate and x y. 62

5 Exercise 4.3 (S03 6(ai)). Compute x and y when z = x2 y + 3xsin(x 2y) Functions of more variables We can extend the notion of partial derivatives to functions of any (finite) number of variables in a natural way. For example if w = sin(x + y) + z 2 e x then w = cos(x + y) (x + y) + z2 x x x ex = cos(x + y) + z 2 e x w y = cos(x + y) (x + y) + 0 = cos(x + y) y w = 0 + (z2 )e x = 2ze x. The geometrical significance of such functions is not so immediate as for functions of only two variables. 4.2 Tangent planes By holding one variable fixed in the definition of partial derivatives, for example setting y = b, we are taking a surface z = f(x, y), intersecting it with the plane y = b, and then taking derivatives of the resulting curve in this plane: z 1 f x (a, b) a x For one-variable calculus we are interested in approximating a function y = f(x) by finding 63

6 Tangent planes a tangent line to the curve at some point ( a, f(a) ). For two variables the appropriate object is the tangent plane. n t y t x By taking partial derivatives in the orthogonal directions corresponding to the x- and y-axes we can produce vectors that are in the direction of the tangent lines to the two curves obtained by intersecting the surface with the planes x = a and y = b. In particular the tangent plane at the point ( a, b, f(a, b) ) contains this point, and should contain the two tangent lines through this point in the directions specified by the vectors obtained from these partial derivatives. In the plane y = b a tangent vector is t x = ( 1, 0, f x (a, b) ), and in the plane x = a a tangent vector is t y = ( 0, 1, f y (a, b) ). Both of these vectors lie in the plane, hence their vector product n = t x t y is a normal vector to the plane: i j k n = t x t y = 1 0 f x (a, b) 0 1 f y (a, b) = ( f x (a, b), f y (a, b), 1 ), and so the tangent plane has equation [ (x, y, z) ( a, b, f(a, b) )]. ( f x (a, b), f y (a, b), 1 ) = 0 f x (a, b)(x a) f y (a, b)(y b) + z f(a, b) = 0 z = f(a, b) + f x (a, b)(x a) + f y (a, b)(y b) This function of x and y is also known as the linearisation of z = f(x, y) at the point (a, b). Exercise 4.4 (A04 8(b)). Find the equation of the tangent plane and the normal line to the surface z = x 2 y 3 sin ( πx + π 2 y) at the point (2, 1). Find the intersection of this normal line with the (x, y)-plane. 64

7 Exercise 4.5 (S03 6(b)). Find the equation of the tangent plane to the surface z = f(x, y) = 8 3x2 y 2 at the point (1, 2, 1). Write down the linear approximation to f(x, y) at (1, 2) and use it to find an approximate value for f(1.05, 1.95). Exercise 4.6. Find the tangent planes to the surface z = f(x, y) = 3xy + x + y 2 at (1, 0) and at ( 1, 2). Find the line of intersection of these two planes. 65

8 Higher order derivatives Example 4.7. Find the tangent plane to the surface at the point (3, 1, e 6 + 1). Since z = e 2xy + tan x = 2ye2xy + π 16 z = e 2xy + tan π(x + y), we have 16 π(x + y) sec2, 16 π(x + y) 16 So when x = 3 and y = 1, z = e 6 + tan π 4 = e6 + 1 and y = 2xe2xy + π 16 π(x + y) sec2. 16 x = 2e6 + π 1 16 (1/ 2) = 2 2e6 + π 8, y = 6e6 + π 1 16 (1/ 2) = 2 6e6 + π 8 So the tangent plane has equation ( z = e e 6 + π ) ( (x 3) + 6e 6 + π ) (y 1). 8 8 Example 4.8 (S05 8(c)). Find the equation of the tangent plane to the surface with equation z = y cos(x y) at the point (2, 2, 2). Taking partial derivatives of z we get = y sin(x y) (x y) = y sin(x y), and x x y = cos(x y) y sin(x y) (x y) = cos(x y) + y sin(x y). y So at (2, 2, 2) we have = 2 sin0 = 0, and = cos0 + 2 sin0 = 1. Thus the tangent x y plane is z = (x 2) + 1 (y 2) z = y. 4.3 Higher order derivatives Suppose z = xsin y + x 2 y. Then x = siny + 2xy and y = xcos y + x2. Both of these partial derivatives are again functions of x and y, so we can differentiate both of them, either with respect to x, or with respect to y. This gives us a total of four second order partial derivatives: ( ) ( ) x 2 = x x y = x = 2y, x ) = cosy + 2x, ( y y x = y y 2 = y = cosy + 2x x ) = xsiny. ( y 66

9 Remark. The mixed partial derivatives in this case are equal: y x = 2 z. This is not x y something special about our particular example, but is true for all reasonably well-behaved functions. However it is possible to find functions for which this is not true (examples can be found in calculus text books). In the notation, the order of taking the derivatives is given by reading the variables in x the denominator from right to left. For example 2 z means calculate, then differentiate y x x the result with respect to y. When using the f x or f 1 notations, the convention is the other way the subscripts are read left to right. That is, f xy = (f x ) y, the derivative with respect y of the derivative with respect to x. However, in light of the remark above, these conventions can usually be ignored for most functions, since the results in either order will be the same. For example, if f(x, y) = x 2 + xy 2, then f x (x, y) = 2x + y 2, f y (x, y) = 2xy, and so f xx (x, y) = 2, f xy (x, y) = 2y = f yx (x, y), f yy (x, y) = 2x. Exercise 4.9. Compute all the second order partial derivatives of the function f(x, y) = sin(x + xy). Example Compute x, y and 2 z x 2 when z = x3 y + e x+y2 + y sin x. Since z = x 3 y + e x+y2 + y sin x then x = 3x2 y + e x+y2 + y cosx, and y = x3 + 2ye x+y2 + sin x x 2 = 6xy + ex+y2 y sin x, 67

10 Exercises Example Consider the following function of three variables: f(x, y, z) = x 2 y 3 + sin(x 2 + z) xy z. Find the partial derivatives f x, f y, f yx and f xyz. Since f(x, y, z) = x 2 y 3 + sin(x 2 + z) xy z then f x = 2xy 3 + 2xcos(x 2 + z) y z, f y = 3x 2 y 2 x z, f yx = (f y ) x = 6xy 2 1 z, f xyz = (f xy ) z = (f yx ) z = 1 z Exercises 1. Find the domains of the following functions. Sketch these domains in parts (i) and (iii). (i) f(x, y) = 8 3x 2 y 2 (ii) f(x, y) = sin(x y) + 1 x y 2 (iii) f(x, y) = ln(1 x 2 y 2 ) + x2 + 1 x + y (iv) f(x, y) = sin(x 2 + e xy ) 2. Find all the first order derivatives of the following functions: (i) f(x, y) = x 3 4xy 2 + y 4 (iii) f(x, y) = x 2 sin xy 3y 2 (ii) f(x, y) = x 2 e y 4y (iv) f(x, y, z) = 3xsiny + 4x 3 y 2 z 3. Find the indicated partial derivatives: (i) f(x, y) = x 3 4xy 2 + 3y : (ii) f(x, y) = x 4 3x 2 y 3 + 5y : f xx, f yy, f xy f xx, f xy, f xyy (iii) f(x, y, z) = e 2xy z2 y + xz sin y : f xx, f yy, f yyzz 4. Find the equation of the tangent plane and normal line to the following surfaces at the given point: (i) z = x 2 + y 2 1 at (2, 1, 4) (ii) z = e x2 y 2 at (0, 0, 1) (iii) z = sinxcos y at (0, π, 0) (iv) z = x 3 2xy at ( 2, 3, 4) (v) z = x 2 + y 2 at ( 3, 4, 5) (vi) z = 4x y at (1, 2, 2) 4.5 The Chain Rule For one-variable calculus if y = f(x), i.e. y is a function of x, and if x = x(t), i.e. x is a function of the variable t, then y can be viewed as a function of t, and has derivative dy dt = dy dx dx dt. Suppose instead that z = f(x, y) is a function of two variables, each of which is written in terms of the single variable t. Then we think of z just as a function of t, and differentiate. It depends on the partial derivatives with respect to x and y through the following formula: dz dt = dx x dt + dy y dt 68

11 As an example, suppose z = x 2 xy, and that x = sin t, y = t 2. Then dz dt = dx x dt + dy y dt = (2x y) d dt (sin t) + ( x) d dt (t2 ) = (2 sint t 2 )cost 2t sint. In this case we could avoid use of the chain rule, since direct substitution for x and y gives z = sin 2 t t 2 sin t. But substitution may not always be convenient, or even available. Exercise Suppose z = x 2 y + y 2, where x = cost and y = 1/t. Find dz dt. Exercise The pressure, volume and temperature of an ideal gas are related by the equation PV = 8.31T. Find the rate at which the pressure is changing when the temperature is 300K and increasing at a rate of 0.1Ks 1, and the volume is 100l and increasing at a rate of 0.2ls 1. Example The volume of a right circular cylinder of base radius r and height h is V = πr 2 h. If the radius is decreasing at a rate of 3cms 1 while the height is increasing at a rate of 2cms 1, what is the rate of the change of V when r = 40cm and h = 110cm? Since V = πr 2 h, V r we have, by the chain rule, For the problem we have r = 40, h = 110, dr dt = 2πrh and V h = πr2. So thinking of V as a function of t dv dt = V dr r dt + V dh h dt = 2πrhdr dh + πr2 dt dt. = 3 and dh dt = 2, and so dv dt = 2π ( 3) + π = 23200π cm 3 s 1. 69

12 The Chain Rule Example The voltage V in an electrical circuit is slowly decreasing as the battery wears out. The resistance is slowly increasing as the resistor heats up. Use Ohm s Law V = IR to find how the current I is changing at the moment when R = 400Ω, I = 0.08A, dv dt = 0.01Vs 1 and dr dt = 0.03Ωs 1. From V = IR we get I = V, and so the chain rule gives R di dt = I V = 1 R dv dt + I dr R dt dv dt V dr R 2 dt = 1 R dv dt I dr R dt = ( 0.01) = ( ) = As 1. Now suppose that z = f(x, y), a function of the two variables x and y, and that each of these in turn depend on two variables s and t. Then, viewing z as a function of s and t, we have the two partial derivatives s = x x s + y y s and t = x x t + y y t As an example, suppose that z = xy y 2, and that x = e s+t and y = s t. Then Similarly, s = x x s + y y s = y s (es+t ) + (x 2y) s( s ) t = s ( t es+t + e s+t 2s ) 1 t t = 1 [ (s + 1)te s+t t 2 2s ] t = s t 3 [ t(t 1)e s+t + 2s ]. Exercise If z = e x sin y where x = st 2 and y = s 2 t, find s. Exercise 4.17 (A04 8(c)). Suppose that z = f(u, v) and that the variables u and v depend on x and y through u = x 2 y + y 2 and v = e x cos(πy). If = 4 and = 3 at the point u v (u, v) = (4, 1), find x and y at the corresponding point (x, y) = (0, 2). 70

13 Exercise If g(s, t) = f(s 2 t 2, t 2 s 2 ), and if f has partial derivatives with respect to both variables, show that g satisfies the equation t g s + s g t = 0. In general, if z is a function depending on the m variables x 1, x 2,..., x m, and each of these are defined in terms of the n variables y 1, y 2,..., y n, then we can think of z as a function of the y j, and take n different partial derivatives, which are given by = x 1 + x x m. y j x 1 y j x 2 y j x m y j The previous two formulae given are special cases of this general version of the chain rule. 4.6 Directional derivatives; the gradient operator Given a function z = f(x, y), we have defined the two partial derivatives and x y by making small changes in one variable while holding the other fixed. This amounts to moving a small distance in the (x, y)-plane parallel to one or other of the coordinate axes. This can be generalised by moving in any direction in the (x, y)-plane as specified by a unit vector c = (c, d) (so c 2 + d 2 = 1). The directional derivative of z = f(x, y) at the point r = (a, b) in the direction c is f(a + hc, b + hd) f(a, b) D c f(r) = lim. h 0 h 71

14 Directional derivatives; the gradient operator Particular examples are given by taking c = i = (1, 0) or c = j = (0, 1), the unit vectors in the coordinate directions, since from these we just recover the partial derivatives: f(a + h, b) f(a, b) D i f(r) = lim = f x (r), D j f(r) = f y (r). h 0 h More generally, having fixed a unit vector c and a point r, we are taking the one-variable function g(h) := f(r + hc) = f(a + hc, b + hd), differentiating it, and evaluating at h = 0. Using the chain rule we get d dh g(h) = f x(a + hc, b + hd) d dh (a + hc) + f y(a + hc, b + hd) d (b + hd) dh = cf x (r + hc) + df y (r + hc) for all h R, and so setting h = 0 we get D c f(r) = cf x (r) + df y (r) = (c, d).(f x (r), f y (r)). That is, D c f(r) can be calculated by finding the vector of partial derivatives, evaluating this at the point r, and taking the dot product of the result with the direction vector c. The vector (f x (r), f y (r)) is known as the gradient of f, and is denoted f(r), so that D c f(r) = c. f(r) and so For example, if we take f(x, y) = 4xy x 3 y 2, c = ( 3 5, 4 5 ) and r = (2, 3) then f x (x, y) = 4y 3x 2 y 2 f x (r) = 120, f y (x, y) = 4x 2x 3 y f y (r) = 56, D c f(r) = ( 3 5 5), ( 120, 56) = = Recall that for any two vectors a and b, a.b = a b cos θ, where θ is the angle between these two vectors. Applying this to our formula for directional derivatives, and noting that c = 1, we get D c f(r) = f(r) c cos θ = f(r) cos θ. But 1 cosθ 1, so the directional derivative is maximised if we take θ = 0, when cosθ = 1. That is, if we take c in the same direction as f(r). Alternatively, if we take c in the opposite direction, so that θ = π, and hence cosθ = 1, then D c f(r) = f(r). Exercise Find the directional derivative of f(x, y) = x 3 3xy + 4y 2 in the direction given by the unit vector c at an angle π/6 to the x-axis. What is D c f(1, 2)? 72

15 Exercise If f(x, y) = xe y, find the rate of change at the point P with position vector (2, 0) in the direction from P to the point Q with position vector ( 1 2, 2). In which direction does f have the maximum rate of change? What is this maximum rate of change? Example 4.21 (S05 8(b)). Find the directional derivative of the function f(x, y) = 5xy 2 4x 3 y at the point P = (1, 2) in the direction of the vector (5, 12). What is the maximum rate of decrease of the function at P, and in which direction does this occur? Since = 169 = 13 2, a unit vector in the required direction is c = 1 13 (5, 12). Also, f(x, y) = 5xy 2 4x 3 y f = (5y 2 12x 2 y, 10xy 4x 3 ) and so the directional derivative in the direction of c at P is 1 1 (5, 12). f(1, 2) = (5, 12).(20 24, 20 4) = ( ) = The maximum rate of decrease occurs in the direction of f(1, 2) = (4, 16), and is f(1, 2) = ( 4) = Tangent planes revisited Consider the equation x 2 + y 2 + z 2 = 1. This can be written as r 2 = 1, which is equivalent to r = 1, where r = (x, y, z) is the position of a general point in space. Thus a point satisfies the equation (S) precisely if it is distance 1 from the origin, that is, if it is on the sphere of radius 1 whose centre is the origin. (S) 73

16 Directional derivatives; the gradient operator It is impossible to rearrange this equation to get a single function z = f(x, y), since we have the two possibilities: z = 1 x 2 y 2 and z = 1 x 2 y 2, and note that these only make sense whenever x 2 + y 2 1, that is when the point (x, y) lies in the disc of radius 1 centred on (0, 0) in the (x, y)-plane. Define a function F : R 3 R by F(r) = F(x, y, z) = x 2 + y 2 + z 2. The equation of the sphere is given by F(r) = 1, that is, setting F equal to a constant value. Surfaces generated this way are called the level surfaces of the function F, and provide a more general way of describing surfaces than equations of the form z = f(x, y). Note that if we define F(r) := z f(x, y) then our first type of surface is nothing but the level surface F(r) = 0. Consider now the problem of finding the equation of a tangent plane to such a surface F(r) = k at the point a. To do this imagine a point moving around on the surface. At time t its position vector is r(t) = ( x(t), y(t), z(t) ), and suppose that at time t = 0 it passes through a. That is r(0) = ( x(0), y(0), z(0) ) = a and F ( r(t) ) = k for all t, the second equation following since the point is constrained to lie on the surface. Since the function t F ( r(t) ) is constant, its derivative is 0. But we can also evaluate this using the chain rule, which gives ( )dx(t) ( )dy(t) ( )dz(t) 0 = F x r(t) + F y r(t) + F z r(t) dt dt dt ( ( ) ( ) ( ) ) ( dx(t) = F x r(t), Fy r(t), Fz r(t)., dy(t), dz(t) dt dt dt = F ( r(t) ). dr(t) dt where F is the gradient vector of F made up of its partial derivatives, which is the 3D analogue of f considered above. Putting t = 0 in this equation shows that F(a) is orthogonal to dr(0), which is the tangent vector to the path taken by the point when passing dt through a (i.e. its velocity, when thinking in terms of its motion). Since this is true for any path passing through a, it follows that the vector F(a) must be orthogonal to the surface at this point, and so provides the normal vector for the tangent plane. Hence all points r on the tangent plane at a satisfy the equation (r a). F(a) = 0. ) F dr(t) dt 74

17 For an example consider F(x, y, z) = x 2 + y 2 + z 2 as above. We have F = ( x (x2 + y 2 + z 2 ), y (x2 + y 2 + z 2 ), ) (x2 + y 2 + z 2 ) = 2(x, y, z), a vector parallel to the position vector of the point r, showing that the normal line to the sphere at any point can be continued back to the origin. In particular at the point (1, 0, 0) on the surface of the sphere, the tangent plane has equation [ (x, y, z) (1, 0, 0) ].(2, 0, 0) = 0 2(x 1) = 0 x = 1. This equation could not have been calculated with our earlier method. Finally, if we are given a surface by means of the equation z = f(x, y) then, rewriting this as F(x, y, z) := z f(x, y) = 0, we see that the normal vector at the point (x, y, z) = (a, b, f(a, b)) is ( f x (a, b), f y (a, b), 1), as shown previously. Exercise Find the tangent plane to the hyperboloid x 2 y 2 + 2z 2 = 1 at the point (3, 4, 2). At which points is the normal line to the surface parallel to the line through the points (3, 1, 0) and (5, 3, 6). 75

18 Directional derivatives; the gradient operator Exercise Show that (0, 1, 2) lies on both of the following surfaces: x 2 + 4y + z 2 = 0 and x 2 + y 2 + z 2 6z + 7 = 0. Show, moreover, that the surfaces are tangent to one another at this point. 76

19 4.7 Critical/stationary points A function z = f(x, y) has a local maximum at the point (a, b) if f(a, b) f(x, y) for all (x, y) close to (a, b). More precisely, z = f(x, y) has a local maximum at the point (a, b) if we can find some number r > 0 such that when we evaluate f(x, y) at any point in the disc with centre (a, b) and radius r, we have f(a, b) f(x, y). Similarly, z = f(x, y) has a local minimum at the point y (c, d) if f(c, d) f(x, y) for all points (x, y) close to (c, d) in the same sense, i.e. all points in some disc of positive radius centred on (c, d). r We shall make use of partial derivatives to locate candidates for such points. Note that if (a, b) is a local maximum or a local minimum, then the tangent plane at this x (a, b) point should be horizontal, which is equivalent to saying that the normal vector n to the surface/tangent plane should have 0 for its x- and y- components. Recall that n = ( f x (a, b), f y (a, b), 1 ). Consequently a point (a, b) in the domain of a function z = f(x, y) is called a critical or stationary point if f x (a, b) = f y (a, b) = 0. This is a necessary condition for there to be a local maximum or a local minimum at (a, b), but is not a sufficient condition. It tallies with the fact that if there was a local maximum in the surface at that point then x = a would be a local maximum in the curve z = g(x) = f(x, b), and y = b would be a local maximum in the curve z = h(y) = f(a, y) the curves obtained by intersecting the surface with the planes y = b and x = a respectively. 77

20 Critical/stationary points The second derivative test Recall that for a function y = f(x), if f (a) = 0 then we can check to see whether there is a local maximum or local minimum at x = a by calculating f (a) (providing these derivatives exist) and seeing if the result is positive or negative. Given a function z = f(x, y) of two variables, its discriminant is the function D f (a, b) = f xx (a, b)f yy (a, b) f xy (a, b) 2 = f xx(a, b) f xy (a, b) f yx (a, b) f yy (a, b) where we are assuming that f xy (a, b) = f yx (a, b). Theorem Suppose that z = f(x, y) has partial derivatives up to second order, and that (a, b) is a critical point of the function. (i) If D f (a, b) > 0 and f xx (a, b) > 0 then f has a local minimum at (a, b). (ii) If D f (a, b) > 0 and f xx (a, b) < 0 then f has a local maximum at (a, b). (iii) If D f (a, b) < 0 then f has a saddle point at (a, b). (iv) If D f (a, b) = 0 then no conclusion can be drawn. Remark. The matrix whose determinant is taken to form the discriminant is symmetric, and so, by an earlier theorem, we know that it has two real (possibly repeated) eigenvalues. The inequality D f (a, b) > 0 is equivalent to saying that the eigenvalues are either both positive or both negative. The inequality D f (a, b) < 0 is equivalent to saying that the eigenvalues are of different sign. If D f (a, b) = 0 then at least one eigenvalue is 0, which leads to the lack of conclusion. The theorem follows from the multidimensional version of Taylor s Theorem which implies that for points (x, y) close to (a, b) f(x, y) f(a, b) + f x (a, b)(x a) + f y (a, b)(y a) + [ x a y b ] [ ] [ ] f xx (a, b) f xy (a, b) x a. f yx (a, b) f yy (a, b) y b So if there is a critical point at (a, b) then f(x, y) f(a, b) + [ x a y b ][ ][ ] f xx (a, b) f xy (a, b) x a f yx (a, b) f yy (a, b) y b Local maxima and minima are relatively easy to visualise, and the final possibility (D f (a, b) = 0) being inconclusive can happen for reasons similar to the one variable case (e.g. there is a point of inflection when looking at the curve passing through ( a, b, f(a, b) ) in some direction). Case (iii) is a phenomenon that occurs for surfaces but not for curves. In one direction the curve obtained by intersecting the surface with a plane has a local minimum, in the orthogonal direction the corresponding curve has a local maximum. y x y x y x Figure 1: z = x 2 + y 2 Figure 2: z = x 2 y 2 Figure 3: z = x 2 y 2 78

21 For each of the above examples = 2x or = 2x, and = 2y or x x y y = 2y. Consequently in all cases the only critical point is the origin (x, y) = (0, 0). When z = x 2 + y 2, x 2 = 2, x y = 0, y 2 = 2 D f(0, 0) = 4. Since f xx (0, 0) = 2 > 0, the point (0, 0) is a local minimum for this function. When z = x 2 y 2, x 2 = 2, x y = 0, y 2 = 2 D f(0, 0) = 4. Since f xx (0, 0) = 2 < 0, the point (0, 0) is a local maximum for this function. When z = x 2 y 2, x 2 = 2, x y = 0, y 2 = 2 D f(0, 0) = 4, so the point (0, 0) is a saddle point for this function. Perhaps more instructively one should consider the intersections of each surface with the planes y = 0 and x = 0 which produces the curves z = ±x 2 and z = ±y 2. Exercise Locate and classify the stationary points of f(x, y) = x 3 2y 2 2y 4 + 3x 2 y. 79

22 Critical/stationary points Exercise 4.26 (S04 8(c)). Locate and classify the stationary points of the function z = 2x 2 + y 3 x 2 y 3y. 80

23 Example Locate and classify all the critical points of the function z = xsin y. z = xsin y, so = sin y and = xcosy. So to have = 0 we need siny = 0, x y x that is y = nπ for n = 0, ±1, ±2,... But note: cosnπ = ( 1) n 0 for all n, and we also need = xcosy = 0, where y cosy 0. So we must have x = 0. Thus the critical points of z are (0, nπ) for n = 0, ±1, ±2,... But now 2 z x 2 = 0 and 2 z x y = cosy, so that 2 z 2 ( z 2 ) 2 x 2 y 2 z = cos 2 y = 1 for x y x = 0 and y = nπ. Thus each of the critical points is a saddle point. 4.8 Exercises 1. Show that for any k 0 the function z = sin kxcoskct satisfies the wave equation c 2 2 z x 2 = t 2. Show that if f(u) is any function of one variable that is twice differentiable (i.e. f (u) and f (u) exist) then z = f(x ct) is also a solution of this equation. 2. Use the chain rule to calculate the indicated derivatives: (i) g (t) where g(t) = f ( x(t), y(t) ), f(x, y) = x 2 y siny, x = t 2 + 1, y = e t. (ii) g (t) where g(t) = f ( x(t), y(t) ), f(x, y) = x 2 + y 2, x = sin t, y = t (iii) u (iv) u and v, where z = sin(xy), x = u2 v, y = ve u. and v, where z = xy3, x = e u2, y = v sinu. 3. Find the directional derivatives of the given functions at the given point and in the given direction: (i) f(x, y) = xsin(xy) at (1, π 1 2 ) in the direction of c = 2 (1, 1). (ii) f(x, y) = x 2 + ln(x y) at (2e, e) in the direction of the vector a = (4, 7). (iii) g(x, y) = y 2 x 2 x at (4, 5) in the direction of the vector from P = (3, 7) to Q = ( 1, 9). (iv) f(x, y, z) = x + y 2 z xz 3 at (2, 0, 2) in the direction of the normal vector to the plane x 3y + 4z = Find the tangent plane to the following surfaces at the given point: (i) x 2 y 2 + z 2 = 13 at the point (4, 2, 1). (ii) xsin(yz) = 2 at the point (2, 2, π 4 ). (iii) xy e x yz = 31 at the point (16, 2, 8). 5. Locate and classify all of the critical points of the following functions: (i) z = e x2 (y 2 + 1) (ii) z = 4xy x 4 y (iii) z = y 2 + x 2 y + x 2 2y (iv) z = e x2 y 2 (v) z = x 2 4xy y 2 (vi) z = xye x2 y