WEEK #18: Directional Derivatives and the Gradient; the Chain Rule

Transcription

1 WEEK #18: Directional Derivatives and the Gradient; the Chain Rule Goals: To introduce the directional derivative and the gradient vector. To learn how to compute the gradient vector and how it relates to the directional derivative. To explore how the gradient vector relates to level sets. To study the chain rule for functions of several variables. Textbook reading for Week #18: Study Sections 14.4, 14.5, and

2 2 Directional Derivatives Now we can return to the study of rates of change of a function f(x, y) whose domain is all or part of IR 2 (in other words, functions of two real variables, x and y). In our new terms, The partial derivative f x is the rate of change of f in the direction of the unit vector i (towards larger x values) The partial derivative f y is the rate of change of f in the direction of the unit vector j (towards larger y values)

3 Week 18 Directional Derivatives and the Gradient; the Chain Rule 3 On the surface below, find a point that has f x < 0 and f y > 0.

4 4 Suppose we now want to find the rate of change in an arbitrary direction. Any direction can be specified by a vector u of length 1. Vectors of length 1 are called unit vectors. Given a unit vector u, we want to find the rate of change of f(x, y) if we move away from (x, y) in the direction of u. From the same point on the graph, indicate a direction where the slope would be steeper than f x. Indicate another direction where the slope would be close to zero.

5 Week 18 Directional Derivatives and the Gradient; the Chain Rule 5 Directional Derivative Let u = (u 1 i + u 2 j) =< u 1, u 2 > with u u 2 2 = 1, so that u = 1. Then, at the point (a,b) in the domain of f, the rate of change of f in the direction of u is f(a + hu 1, b + hu 2 ) f(a,b) lim. h 0 h This is called the directional derivative of f at the point (a, b) in the direction of u and it is denoted by D u f(a, b) or f u (a,b) Note: This formula only applies if u is a unit vector.

6 6 The following diagram illustrates the geometric relationships involved in calculating the directional derivative of f(x, y) at (a,b) in the direction given by a unit vector u = (u 1, u 2 ). Part of a surface. The tangent plane to the surface at a point.

7 Week 18 Directional Derivatives and the Gradient; the Chain Rule 7 Using the tangent plane as our surface now, we take a step of length 1 in the direction u =< u 1, u 2 >. This produces a change in height of z Since the change in height, z comes from a step of length 1, we can interpret the answer as a slope, z length of step in xy plane = D uf(a, b)

8 8 Example: Write the meaning of D if(a, b) in a sentence. What would be another way to write D i f(a,b), using the partial derivatives we already know?

9 Week 18 Directional Derivatives and the Gradient; the Chain Rule 9 Example - The following is a contour diagram for a function f(x, y). A = (a, b) is a point in the domain of f. On the diagram, mark three directions u, v and w (at A), chosen so that D u f(a,b) > 0 D v f(a,b) < 0 D w f(a,b) = 0.

10 10 Computing D u f(a, b) How can we go about computing values for D u f(a,b) in a systematic way? Keep in mind the ingredients of our calculation: f(x, y) is a function of two variables, (a,b) is a point in the domain of f, u =< u 1,u 2 > with u u2 2 = 1 is a unit vector.

11 Week 18 Directional Derivatives and the Gradient; the Chain Rule 11 Then f(a + hu 1, b + hu 2 ) f(a, b) D u f(a, b) = lim h 0 h f(x, y) f(a,b) = lim (where x = a + hu 1 and y = b + hu 2 ) h 0 h Use local linearity to find an alternate expression for f(x, y) f(a, b)

12 12 Use that alternate expression to express the directional derivative in terms of partial derivatives.

13 Week 18 Directional Derivatives and the Gradient; the Chain Rule 13 Computing the Directional Derivative If u =< u 1, u 2 > is a unit vector ( u = 1), then D u f(a, b) = f x (a,b)u 1 + f y (a,b)u 2 NOTE: we don t define directional derivatives for non-unit vectors. To find the slope in the direction of a non-unit length vector, v, you must normalize it before computing the directional derivative. If v = (v 1,v 2 ) is not a unit vector, first find u = 1 v v = 1 v v2 v, 2 then compute D u f(a,b) This formula allows us to compute the slope in any direction simply by knowing the partial derivatives.

14 14 Example: Let f(x, y) = x 2 xy 2 and let u = ( 3 5, 4 5 ). Calculate D uf(2, 2). First: is u a unit vector? f x (x,y) = f y (x,y) = f x (2, 2) = f y (2, 2) = u 1 = u 2 = D u f(2, 2) =

15 Week 18 Directional Derivatives and the Gradient; the Chain Rule 15 Now compute the slope in the direction opposite of u. What do you notice about the slope?

16 16 Example: Find the slope of the surface f(x, y) = x 2 y 2 at (x,y) = (2, 3) if we were to move directly towards the origin.

17 Week 18 Directional Derivatives and the Gradient; the Chain Rule 17 Gradient Vector Note that the formula for directional derivatives could be written as a dot product if we so desired: D u f(x, y) = f x (a,b)u 1 + f y (a,b)u 2 = < f x (a,b), f y (a,b) > < }{{}} u {{ 1 u 2 > } new vector u This is the first appearance of an important vector function called the gradient of f. While f assigns a number to each point in its domain, the gradient of f assigns a vector to each point in the domain of f, provided both partial derivatives of f exist at that point. The gradient is denoted by either grad f or f.

18 18 Gradient Vector Definition gradf = f = f x (x,y) i + f y (x,y) j =< f x (x,y), f y (x,y) > Alternate Directional Derivative Definition D u f(x, y) = (gradf) u

19 Week 18 Directional Derivatives and the Gradient; the Chain Rule 19 Example - Let f(x,y) = xe y grad f(x, y) = grad f(1, 0) = grad f(0, 1) = grad f(2, 3) = For each point in the domain of f where the partial derivatives are both defined, the gradient vector is also defined.

20 20 Example: Use the gradient-based definition of the directional derivative to determine the direction in which a surface has the largest positive slope. The direction of grad f(a,b) is the direction of maximum increase of the function f at the point (a,b).

21 Week 18 Directional Derivatives and the Gradient; the Chain Rule 21 Example: Consider the plane z = x + 2y + 3. At the point (x,y) = (1, 1), in which (x,y) direction should we move to move uphill the most quickly?

22 22 Support your answer, using the contour diagram for z = x + 2y + 3 shown below

23 Week 18 Directional Derivatives and the Gradient; the Chain Rule 23 Properties of the Gradient Vector Use the properties of the directional derivative and the dot product to justify the following conclusions: grad f(a,b) is perpendicular to the contour of f that passes through the point (a,b) grad f(a,b) gives the direction of maximum decrease of the function f at the point (a,b).

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25 24 grad f(a,b) (i.e. the length or magnitude of the gradient vector) is the maximum rate of change of f at (a,b).

26 Week 18 Directional Derivatives and the Gradient; the Chain Rule 25 Reminding ourselves of these properties of the gradient vector, consider the contour diagram for a function f(x, y) For each of the points. A, B, and C, draw a vector that points in the direction of the gradient vector at that point.

27 26 At which of the points is the gradient vector longest? At which of the points is the gradient vector shortest? Justify your answers.

28 Week 18 Directional Derivatives and the Gradient; the Chain Rule 27 Putting Gradients and Contours Together We said earlier that the gradient is perpendicular to the contour at the same point. However, that isn t very precise, given that the contours are curves themselves. It is better to say that contours, as curves, have tangent lines, and gradient is perpendicular to those tangent line. Consider the 2-variable function f(x, y) = xy 2. Write an equation for the contour (level curve), C, of f that passes through the point (2, 1).

29 28 Find grad f(2, 1). Find an equation for the tangent line at (2, 1) to the contour (level curve) C, through (2, 1)

30 Week 18 Directional Derivatives and the Gradient; the Chain Rule 29 On the axes below, sketch the level curve C indicate the vector grad f(2, 1), and draw the tangent line to the contour at (2, 1). 5 4 y x

31 30 Note The idea of directional derivative and gradient are new, and are easily confused at first. The following reminders can be useful to help you check that you are on the right track. D u f(a, b) is a number. It is a rate of change associated with a specific direction, chosen regardless of the surface. grad f(a,b) is a vector. Its direction is the direction of maximum increase of f at (a,b). Its length is a number which represent the rate of change in the gradient s direction.

32 2 4 8 Week 18 Directional Derivatives and the Gradient; the Chain Rule 31 For reference, here is a more detailed contour diagram of the function f(x, y) = xy 2, used in the previous question

33 32 Composing Functions of Several Variables To prepare for this topic, you should read Section 14.6 in the Hughes-Hallett. There are many ways to compose multi-variable functions. Suppose we have a function f(x, y) with output z. That is, z = f(x,y). Suppose we let x be the output of a function g(t) and y the output of a function h(s). Then, by substituting these functions in for x and y, we get a new function f(g(t), h(s)). This is a function of two variables t and s. If, on the other hand, the second function were also a function of t, as in y = h(t), we would end up with a function of a single variable, f(g(t), h(t)).

34 Week 18 Directional Derivatives and the Gradient; the Chain Rule 33 Example (Ideal Gas) The pressure P (in kilopascals), volume V (in litres), and temperature T (in K) of one mole of an ideal gas are related by the formula PV = nrt = (8.31)T if n = 1. Suppose the pressure is increasing at 0.05 kpa per second, and the temperature is increasing at 0.15 K per second. Identify the way in which functions are composed in this problem.

35 34 Now suppose we want to find the rate of change of volume at the moment when the pressure is 20 kpa and the temperature is 320 K. In other words, we are given dp dt, dt dv, P, and T, and are trying to find dt dt. The Chain Rule The Chain Rule for the form of composition dealt with above should answer the following question: If z = f((g(t), h(t)), how does the derivative dz dt relate to the derivatives of f,g, and h?

36 Week 18 Directional Derivatives and the Gradient; the Chain Rule 35 We start with our understanding of linear approximations: z f x (x,y) x + f y (x,y) y, x g (t) t, y h (t) t, x = g(t) and y = h(t). Combining these, we get z f x (g(t), h(t)) g (t) t + f y (g(t),h(t)) h (t) t. Dividing by t and taking limits to get derivatives, we obtain For short, dz dt = z dx x dt + z dy y dt

37 36 Now apply this Chain Rule to solve the ideal gas question on the preceding page.

38 Week 18 Directional Derivatives and the Gradient; the Chain Rule 37 Example: Consider a hillside defined by the function z = f(x, y) = xy 2 where z is in meters, x and y in kilometers. We are walking along a straight path, with x(t) = t and y(t) = 3t where t is measured in hours. Give the units of df dt, dx z, and dt x. How quickly are we moving up or downhill one hour into the hike (at t = 1)?

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40 38 Indicate the interpretation of your result on the contour diagram for f(x, y) = xy

41 Week 18 Directional Derivatives and the Gradient; the Chain Rule 39 Summary for Week #18 We introduced the directional derivative We expressed the directional derivative in terms of a new vector, the gradient vector We discovered many important properties of the gradient vector We studied the chain rule; and we saw that there are different ways to compose functions.