Math 32A Discussion Session Week 9 Notes November 28 and 30, 2017

Transcription

1 Math 3A Discussion Session Week 9 Notes November 8 an 30, 07 This week we ll explore some of the ieas from chapter 5, focusing mostly on the graient. We ll motivate this exploration with an example that we carry through the iscussion, but we ll see notably fewer worke examples this week than usual. The examples shoul become substantially easier once we have a mastery of the ieas involve. Throughout this iscussion, let s suppose we have a function f(x, y) which gives the elevation (in whatever units you like) at the point (x, y). For instance, Figure shows a possible graph of an elevation function, along with some level curves of that function. As you ve learne, the partial erivatives f x an f y are meant to give the slopes of z = f(x, y) in the x- an y-irections, respectively. That is, if we re staning at the point (a, b), our elevation is f(a, b) an we expect this elevation to increase by approximately f x (a, b) units if we move to (a +, b). But what if we move in some irection other than the x-irection or y-irection? You ve investigate this question in the context of local linear approximations; toay we want to think about it in a slightly ifferent way. (a) The graph z = f(x, y). (b) Some level curves. Figure : An elevation function. Suppose we rive aroun the region for which f(x, y) provies the elevation function an recor our elevation as we rive. For instance, Figure shows a path our car might follow, as well as the elevation we woul experience along this path. Notice that we begin with a relatively high elevation, escen to a sale point, continue escening into the valley surrouning the point (, ), an then ascen a bit; all of this elevation change is reflecte in the elevation profile on the right. If we write our path as r(t) = (x(t), y(t)), then the elevation we experience at time t shoul be given by f(r(t)) = f(x(t), y(t)). How quickly oes this elevation change with respect to time? This shoul be given by the erivative: t (f(r(t)) = (f(x(t), y(t))). t

2 (b) Our elevation as a function of time. (a) The path of our car. Figure : A possible path an elevation profile for our car. To evaluate this erivative it will help to remember the chain rule from single-variable calculus. Remember that if g an h are single-variable (ifferentiable) functions, then We can visualize the chain rule as follows: x (g(h(x))) = g (h(x))h (x). h x g h g. We want to know how a small change in the value of x affects the value of g(h(x)). When we change x we change h(x), an this change is measure by h (x). Changing h(x) then affects the value of g(h(x)), an this change is measure by g (h(x)). The total change is then the prouct g (h(x))h (x). Differentiating our elevation function will follow a similar pattern. We can visualize the function f(x(t), y(t)) as x x f x t f. y f y y The elevation f epens on the coorinates x an y, which in turn both epen on t. Changing the value of t will change the value of x(t), which will then affect the value of f(x(t), y(t)). But changing t will also change y(t) an this change will also affect the value of f(x(t), y(t)). Summing

3 Figure 3: Some level curves of f(x, y) = 50 x 8y an the path r(t) = cos t, sin t. these effects tells us that t (f(x(t), y(t))) = f x(x(t), y(t))x (t) + f y (x(t), y(t))y (t). Notice that the terms in this sum come in two types: there are erivatives of the elevation function f an erivatives of our path r(t). Moreover, we can separate these terms by writing the expression as a ot prouct: t (f(x(t), y(t))) = f x(x(t), y(t)), f y (x(t), y(t)) x (t), y (t). The first vector captures everything we nee to know (for this particular application) about erivatives of f, so we give it a name: the graient of f is a vector-value function of two variables efine by f(x, y) = f x (x, y), f y (x, y). Finally we have a very succinct way of expressing the time erivative of our elevation function: t (f(r(t))) = f(r(t)) r (t). To recap: we starte with an elevation function f(x, y) an a parametrization r(t). There are then two vectors base at r(t) that represent some sort of erivative. The graient f(r(t)) represents a erivative of f at r(t), an r (t), our velocity vector, is a erivative of r(t) at r (t). Dotting these vectors together gives the erivative of the function f(r(t)). Example. Consier the elevation function f(x, y) = 50 x 8y an the path r(t) = cos t, sin t, t [0, π]. Some level curves of f an the path r(t) are shown in Figure 3. How quickly is our elevation changing at time t = 3π/4? Ientify the critical points of the elevation function along the path. 3

4 (Solution) Our elevation at time t is given by f(r(t)). Because we have expressions for f(x, y) an r(t) we coul make appropriate substitutions an use the tools of single-variable calculus, but instea let s use the ieas evelope above. We have an r(3π/4) = /, /, so f(r(3π/4)) = We also have r (t) = sin t, cos t, so f(x, y) = 4x, 6y, 4, 3 =, 6. r (3π/4) = /,. Finally, t (f(r(t))) = f(r(3π/4)) r (3π/4) = ( )( / ) + ( 6 )( ) = 34. t=3π/4 Now the critical points of f(r(t)) will of course be the points where t (f(r(t))) vanishes. We have t (f(r(t))) = f(r(t)) r (t) = 4 cos t, 6( sin t) sin t, cos t = 4 cos t sin t 64 sin t cos t = 60 cos t sin t. So f(r(t)) will have a critical point any time either cos t or sin t is zero that is, at 0, π/, π, 3π/, an π. Notice that these values correspon to the points where our path is tangent to a level curve of f. Next, let s consier the following problem: we still have the elevation function f(x, y), an we re staning at the point (a, b) (which we assume is not a critical point for f). We plan to move away from (a, b) at unit spee an want to know in which irection we shoul travel so that (a) our elevation increases as quickly as possible; (b) our elevation ecreases as quickly as possible; (c) our elevation oes not change. If we move away from (a, b) along a straight line, then our path can be parametrize as r(t) = a, b + tu, t 0, where u is a unit vector. In this case r (0) = u an the conitions we consier above involve t (f(r(t))) = f(r(0)) r (0) = f(a, b) u. t=0 This is the slope along our path, which we re hoping to either maximize (in the case of (a)), minimize (in the case of (b)), or set to zero (in the case of (c)). (We assume that f(a, b) is not Inee, since we re moving at unit spee this is truly the slope of the tangent line to the curve (x(t), y(t), f(x(t), y(t)) in 3-space. 4

5 the zero vector.) If θ is the smaller angle between u (so that 0 θ π) an the vector f(a, b) then we can write f(a, b) u = f(a, b) u cos θ = f(a, b) cos θ, since u is a unit vector. This expression is maximize when cos θ =, which occurs when θ = 0. So if we want our elevation to increase as quickly as possible, we shoul move in the irection of f(a, b). Similarly, our elevation will ecrease as quickly as possible if we move in the irection of f(a, b), where we have θ = π an cos θ =. Finally, our elevation will not change (instantaneously) if we have θ = π/, for in this case cos θ = 0 an t (f(r(t))) = 0. In particular, a line perpenicular to the graient vector is tangent to a level curve. So we have the following observations:. At (a, b) the irection of greatest increase for f is f(a, b).. The irection of greatest ecrease is f(a, b). 3. The vector f(a, b) is perpenicular to the level curve passing through (a, b). Remember that in the previous example the critical points of our elevation function (where our slope was zero) were the points where our velocity vector was tangent to a level curve of the elevation function. This agrees with our newer observations. We actually have a name for the erivatives consiere above. Say we have a function f(x, y), a point P = (a, b) an a vector v base at P. We efine the irectional erivative of f at P in the irection of v to be D u f = f(p ) u, where u = v/ v is the unit vector in the irection of v. As we learne in our above iscussion, this number is the slope of the line tangent to z = f(x, y) passing through (a, b, f(a, b)) an moving in the irection of v. Example. Fin the irectional erivative of f(x, y, z) = x 3 + yz at the point P = (, 3, ) in the irection pointing to the origin. (Solution) The vector pointing from P to the origin is given by v =, 3,. The unit vector pointing in this irection is then u = v =, 3, We also have so f(p ) = 3,, 3. Then f = 3x, z, y, D u f = f(p ) u = 3,, +3, 3, =