Increments, differentials, and local approximations

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1 Increments, differentials, and local approximations Bro. David E. Brown, BYU Idaho Dept. of Mathematics. All rights reserved. Version 0.5, of November 16, 013 Draft version. This document is very much under construction! Contents 1 Introduction Real-valued functions of one variable.1 Increments and differentials Local linear approximations Local quadratic approximations Parametrized curves Increments and differentials of parameterized curves Local linear approximations to parameterized curves Local quadratic approximations to parameterized curves Surfaces: Real-valued functions of two variables Increments and differentials of surfaces Local linear approximations to surfaces Local quadratic approximations to surfaces Isn t there a better way to write this stuff? Local linear approximations to level curves Real-valued functions of three or more variables Three independent variables Local linear approximations to level surfaces Four independent variables Parametrized surfaces 0 7 Chain rules, differentials, and local approximations 0 8 Any number of variables 0 9 Error bounds 0 10 Applications 1 11 The technical stuff 1 1 Answers to the exercises 1 1

2 1 Introduction This document is an overview of increments, differentials, and local approximations, for students of multivariable calculus, differential equations, and partial differential equations. Someday, it will include graphics, enough examples and exercises, and solutions or answers to the same. Hopefully, it s more or less useful as it is. I don t intend to put many applications in this document, because your calculus, ODE s or PDE s course should have applications in it. Increments are often thought of as small changes in quantities. Certain ratios of increments are called difference quotients; taking an appropriate limit of a difference quotient can yield a derivative of some sort. Increments are used in constructing so-called discrete models of physical and other phenomena. Taking appropriate limits of appropriate parts of a discrete model can yield what s called a continuous model, in the form of a differential equation (whether ordinary or partial). Such a differential equation has differentials in it (of course!). Conversely, sometimes one can construct a continuous model using derivatives or differentials, instead of increments. Judicious use of local linear or quadratic approximations can yield simpler continuous models (with differentials in them) or discrete models (with increments in them), according to one s needs. Local approximations are useful in their own right. Linearizing a function gives information about rates of change including slopes or partial slopes and about its critical points. Local quadratic approximations give information about concavity and curvature. In many practical applications, messy or intractable mathematical models are replaced by local linear or quadratic approximations. The approximations are often much simpler to work with, while still yielding enough information about the phenomenon being modeled that the messy model almost becomes superfluous. We ll start with the one-variable case, then look at parametric curves, then functions of two and three variables, then parametric surfaces. By the time we re done, you should be able to predict how this goes for any number of input or output variables. If you find errors in this document, no matter how small they may seem, please bring them to my attention. Thanks! Real-valued functions of one variable For a function to be real-valued means that its outputs are real numbers (as opposed to whole numbers or complex numbers or fence posts, or something). To be a function of one variable means there is exactly one independent variable. In symbols, we re talking about y = f(x), only real numbers being used. We will let x 0 be a value of x; specifically, x = x 0 will be the center of attention. (You ll see what this means later.) Finally, at times we will use the fact that the point ( x 0, f(x 0 ) ) is on the graph of y = f(x)..1 Increments and differentials Recall the following definitions from your Calc I course. (You may have used different notation. We ll address this later.) Definition.1.1. Let y = f(x) be differentiable at x = x The increment of x is x = x x 0.. The increment of y is y = f(x) f(x 0 ). 3. The differential of x is dx = x. 4. The differential of y at x 0 is dy = f (x 0 ) dx. Note that dy is sometimes called df, in which case we think of it as the differential of f. 1 Recall that this means that the derivative f (x 0 ) exists. Page

3 In many instances, we think of x as being small. This is often useful, but it s a bad habit. There s no restriction on the size of x: it can be as large or as small as we like. In this document, it will always be small enough for the job at hand, a notion that will be illustrated later. We also often think of y as being small, though the size of y will depend on things like the physical context, the nature of the function f, and the size of x. Before you have to ask: Yes, dx really is the same thing as x. No, it s NOT the same dx that appears in the symbol dy dx, because this symbol is defined by dy dx = lim x 0 y x, so that dy dx isn t even a ratio. But when we write dx by itself, thinking of it as some (typically small) change in x, we haven t made dx go to 0; no limits are involved. So a loose dx by itself is just a change in x. In other words, it s a x, or even just x x 0. It s not the bottom half of a derivative. Editorial comment: Students sometimes wonder why we want df if we can use f, instead. Part of the answer is that you can t always use f! Sometimes, you don t even know which function f is. Sometimes, you do know, but you don t have a formula for it. Sometimes, you only have data, and maybe you don t have data that say what happens at x 0 or at some nearby x. Or if you re creating a mathematical model, then you don t know any of the values of f, so you can t possibly evaluate it at x 0, so you can t possibly find f. Another part of the answer is that using df incorporates information about the derivative of f, which means df has more precise info in it that f, in some sense. More precise is often more useful. Exercise.1.. Write down the increments of x and y, if y = ln x and x 0 = e. Exercise.1.3. Calculate the differentials of f(x) = cos nπ nπ L x and f(t) = ae( L ) t. Assume a, n, and L are nonzero constants, and that L is positive. (Hint: Don t forget the chain rule.) Sometimes, you have to write the differential of some function without knowing the formula for that function. The chain rule is usually your only recourse, at such a time. For example, for a solid object of constant mass, density depends on temperature (though only weakly for many commonly experienced solids), because as the temperature of the object changes, so does its volume. So if density is ρ, volume is V and temperature is T, you can write such things as ρ = ρ(v ) (that is, density is a function of volume), ρ = ρ(v (T )) (which says that density is a function of volume, which is itself a function of temperature), or even ρ = ρ(t ) (i.e., density is a function of temperature). Now, how to calculate the differential of ρ, as a function of T, at some given temperature T 0? What we need is dρ = dρ T. That s fine, but what s dρ? Well, the chain rule says T =T0 T =T0 dt dρ dt T =T0 = dρ dv dt V =V (T0) d dt V. T =T0 So the differential of ρ can be written as dρ = dρ dv V =V (T0) d dt V T. T =T0 Notice that dρ dρ dv is a function of volume, V, not T. So we can t evaluate dv at T 0 directly. By writing we re saying we re going to calculate the volume of the object when its temperature is T 0, V =V (T0) dρ dv and evaluate dρ dv at that volume. There is a theorem that relates increments to differentials and local linear approximations. It s called the increment theorem, and it is the following: Here s something fun and strange for you: If you define dy and dx as in Definition.1.1, and if you divide dy by dx, you get dy dx = dy dx = f (x 0 ) dx = f (x 0 ). So it s safe to pretend that the symbol df is a fraction, even though it s not! dx dx Page 3

4 Theorem.1.4. If f is differentiable at x 0, then for x sufficiently close to x 0, y dy, that is f(x) f(x 0 ) f (x 0 )(x x 0 ). What does for x sufficiently close to x 0 actually mean? The mathematician s answer may not be satisfactory to the practical mind: It means that if you divide the increment of f by x and let x go to 0, you get the derivative. Other peoples answers may be more appealing. If you re an engineer, things like x and y are tolerances within which you must operate. The tolerance of y depends on that of x, and design specifications typically tell how big these tolerances are allowed to be; if you re within tolerance, you re sufficiently close. If you re a scientist, these increments are related to error bars, which measure (in some sense) how much experimental error is involved in whatever you re doing. Sufficiently close means experiment agrees well with theory. If you re a statistician, they re deviations, sums of the squares of which you very likely want to minimize; sufficiently close is actually replaced by as close as we can get. And so on. In a differential equations or partial differential equations course, the phrase for x sufficiently close to 0 allows us to take a discrete model (which is based on increments) and turn it into a continuous model (which is based on differentials or derivatives), by letting x go to 0. Note that there are at least two more ways to write the increment theorem. For example: If y = f(x) is differentiable at x = x 0 then y f (x 0 ) x, for x sufficiently small. Exercise.1.5. Think of at least one more way to write the increment theorem. Exercise.1.6. Write out what the increment theorem says about the function y = sin x at x = 0.. Local linear approximations The easiest approximation anyone could use for any function is f(x) f(x 0 ). This is almost as bad an idea as it may sound. Say f(x) is 10, 000, 000x + 1, 000, 000x, for example, and you need to approximate this function close to x 0 = 0. Well, then f(0) = 0, and using f(x) f(0) = 0 as a local approximation doesn t make much sense, because f(x) isn t very close to 0 for very many values of x. We need a line that does a better job of following the function. A line with the same slope as the curve in question might have a chance of being a better approximation. In the example of the previous paragraph, the slope of f at x 0 = 0 is df dx = 0, 000, 000(0) + 1, 000, 000 = 1, 000, 000. x=0 A line passing through the origin with slope 1,000,000 will be more likely to follow the curve better for more values of x, than the line y = 0. This is the logic behind the local linear approximation. So start by supposing f (x 0 ) exists. It s the slope of a line. If we want this line to approximate the function f near x 0, it would be nice to have the point ( x 0, f(x 0 ) ) be on the line. No problem! Let s use the point-slope form of the line: y f(x 0 ) = f (x 0 )(x x 0 ), so f(x) f(x 0 ) + f (x 0 )(x x 0 ) should be the desired line, provided f (x 0 ) exists. So we were looking for a line that would intersect the graph of y = f(x) at x = x 0 and would have the same slope as the graph of y = f(x). Well, if you evaluate the local linear approximation at x = x 0, you get f(x 0 ). Try it: f(x 0 ) + f (x 0 )(x x 0 ), when evaluated at x = x 0, is f(x 0 ) + f (x 0 )(x 0 x 0 ) = f(x 0 ). This tells you that the local linear approximation intersects the graph of y = f(x) at x = x 0. Page 4

5 Likewise, if you calculate the derivative of the local linear approximation of y = f(x) at x = x 0, you ll get the right number: d dx (f(x 0) + f (x 0 )(x x 0 )) = d dx f(x 0) + d dx f (x 0 )(x x 0 ) = 0 + f (x 0 )(1) = f (x 0 ). So the local linear approximation has the desired slope. It looks like we have the approximation we were looking for. Let s formalize this approximation with a definition: Definition..1. Suppose the function f is differentiable at x = x 0. The local linear approximation to f at x = x 0 is f(x) f(x 0 ) + f (x 0 )(x x 0 ), for x sufficiently close to x 0. Example... Suppose f(x) = e x. The local linear approximation to f at x = 0 is e x f(x 0 ) + f (x 0 )(x x 0 ) = e 0 + e 0 (x 0) = 1 + x, for x sufficiently close to x 0 = 0. Example..3. Let g(x) = cos x, and take x = 0 again. The local linear approximation is for x sufficiently close to x 0 = 0. cos x g(x 0 ) + g (x 0 )(x x 0 ) = cos(0) ( sin(0) ) (x 0) = 1 0(x 0) = 1, Exercise..4. Convince yourself that the local linear approximation to y = e x at x = 0 is y 1 + x, for x sufficiently close to x 0 = 0. Caution: You will see several different names and symbols for local linear approximations: tangent line first-order Taylor polynomial local linearization linearization (For shame! The word local is very important here!) L(x) = f(x 0 ) + f (x 0 )(x x 0 ) P 1 (x) = f(c) + f (c)(x c) T 1 (x) = f(a) + f (a)(x a) y = f(x 0 ) + f (x 0 )(x x 0 ) Page 5

6 and so on. That last one is hazardous, because of the = sign. After all, the function to be approximated is often given as y = stuff. Example: It s easy to be confused by seeing y = e x and y = 1 + x in the same sentence (like this one!). How can y = e x, if y = 1 + x? Easy: The two y s are not the same. Well, if they re not the same, let s avoid confusion, either by using different symbols for them, or by not saying = when we mean. I prefer to write f(x) f(x 0 ) + f (x 0 )(x x 0 ). That way, instead of having y = e x and y = 1 + x in the same sentence, I end up with e x 1 + x. It also helps me remember that I m only approximating f(x). However, sometimes you need a name for the quantity f(x 0 ) + f (x 0 )(x x 0 ). I like the name L(x), as in L(x) = f(x 0 ) + f (x 0 )(x x 0 ). (You see, L is for linear. ) Now, let s be careful about using = and correctly: If L(x) = f(x 0 ) + f (x 0 )(x x 0 ) and if f(x) f(x 0 ) + f (x 0 )(x x 0 ), then f(x) L(x). Recall that the increment theorem says that if x is sufficiently close to x 0, then f df, or in other words, f(x) f(x 0 ) f (x 0 )(x x 0 ). Note that this is equivalent to the statement that f(x) f(x 0 )+f (x 0 )(x x 0 ). Therefore, the increment theorem can be understood to imply that if f is differentiable at x 0, then you may use the local linear approximation to f at x 0, if you like. Note that there are at least two more ways to write the local linear approximation of real-valued functions of one variable. We used x x 0 in the foregoing. We could just as easily have used x or dx. This gives the following: The local linear approximation to y = f(x) at x = x 0 is or or f(x) f(x 0 ) + f (x 0 )(x x 0 ) f(x) f(x 0 ) + f (x 0 ) x f(x) f(x 0 ) + f (x 0 ) dx, for x sufficiently close to x 0. Finally, note that the local linear approximation can also be thought of as f(x) f(x 0 ) + df. Exercise..5. Calculate the local linear approximation to h(x) = ln x at x = 1. Exercise..6. Calculate the local linear approximation to u(x) = cosh x at x = 1..3 Local quadratic approximations The local linear approximation to a function is well and good, but it only takes into account the slope of the graph of the function at the given point. Can we get a better approximation by also somehow including concavity? Yes, because our approximating line (or whatever it ends up being) would not have only the same slope as f, but also the same concavity as f (well, at x = x 0, anyway). And how would we go about including concavity? Well, concavity is measured by f, so we ought to make use of f. Shall we try to use a line again? The problem is that the graph of any line is straight, while the graph of f need not be. Let s use a curve, to get a better approximation to f. To keep things simple, let s use a parabola, the graph of a quadratic polynomial. Specifically, let s use the parabola tangent 3 to the graph of f at x = x 0, having the same concavity at x = x 0 as the graph of f has. How do we create such a parabola? As an approximation to the given function, our parabola probably ought to include the terms f(x 0 ) and f (x 0 )(x x 0 ) that the local linear approximation contains. (We re including additional information, about f, not replacing any information.) We ll also need a quadratic term, 3 Please excuse the barbarism. Page 6

7 and this term ought to include information about f (x 0 ). Obviously, we can only include such information if f (x 0 ) exists. Hmm... If we want a parabola, we ll need an x term. But I think we ought to use (x x 0 ) instead. You see, the x 0 in x x 0 represents shifting the coordinate system x 0 units to the left. Since (x x 0 ) is part of the local linear approximation, we ve already shifted the coordinate system left x 0 units. Our new quadratic term had better include this shift, as well. So, we ll use (x x 0 ) in our formula, instead of x. Our first guess at the new approximating curve is therefore f(x) f(x 0 ) + f (x 0 )(x x 0 ) + af (x 0 )(x x 0 ). The a is there to be an adjustable constant, so we can match the concavity of this new curve with that of the graph of f. Speaking of which, the concavity of the parabola we re building is d ( f(x0 dx ) + f (x 0 )(x x 0 ) + af (x 0 )(x x 0 ) ) x=x0 = d dx (f (x 0 ) + af (x 0 )(x x 0 )) x=x0 = (af (x 0 )) x=x0 = af (x 0 ). The concavity of f at x = x 0 is given by f (x 0 ). Setting the two concavities equal to each other yields f (x 0 ) = af (x 0 ), so a = 1. Our new approximation seems to be f(x) f(x 0 ) + f (x 0 )(x x 0 ) + 1 f (x 0 )(x x 0 ). Note that if x = x 0 then our approximation says that f(x 0 ) f(x 0 ) + f (x 0 )(x 0 x 0 ) + 1 f (x 0 )(x 0 x 0 ) = f(x 0 ). This says that the new approximation intersects the graph of y = f(x) at x = x 0, which would not be the case if we had used x instead of (x x 0 ). Also, note that our new approximation is, in fact, a quadratic polynomial, and therefore has a parabola as its graph. 4 We have guaranteed that the concavity of this approximation is the same as that of f at x = x 0. How about the slope? Well, the slope of f(x 0 ) + f (x 0 )(x x 0 ) + 1 f (x 0 )(x x 0 ) at x = x 0 is d dx ( f(x 0 ) + f (x 0 )(x x 0 ) + 1 ) f (x 0 )(x x 0 ) = [f x=x0 (x 0 ) + f (x 0 )(x x 0 )] x=x0 = f (x 0 ) + f (x 0 )(x 0 x 0 ) = f (x 0 ), which is the same as the slope of the graph of f, at x = x 0. Please note that using x x 0 and (x x 0 ) was important in ensuring that our parabola has the right concavity and the right slope. It also helped guarantee that the point (x 0, f(x 0 )) was on the graph of the approximation. It seems that powers of (x x 0 ) are a good choice for constructing local approximations to curves. Let s dignify our new approximation with an official definition: Definition.3.1. Suppose the function f is twice differentiable at x = x 0. The local quadratic approximation to f at x = x 0 is for x sufficiently close to x 0. f(x) f(x 0 ) + f (x 0 )(x x 0 ) + 1 f (x 0 )(x x 0 ), 4 Granted, if f (x 0 ) = 0, then the parabola is really a line, but I choose to call this line a degenerate parabola. Pretty lame, but it helps keep the technical jargon from mushrooming. Page 7

8 Cool! Let s calculate some. Example.3.. The local quadratic approximation to f(x) = e x at x = 0 is e x f(0) + f (0)(x 0) + 1 f (0)(x 0) = e 0 + e 0 (x 0) + 1 e0 (x 0) = 1 + x + 1 x. Nice. Notice that the two terms of the local linear approximation are the first two terms of the local quadratic approximation, so that we re building on the local linear approximation, as desired. Example.3.3. The local quadratic approximation to g(x) = cos x at x = π (for variety) is cos(x) f(π) + f (π)(x π) + 1 f (π)(x π) = cos π (sin π)(x π) 1 (cos π)(x π) = 1 0(x π) 1 ( 1)(x π) = (x π). Again, the first two terms (including the invisible 0(x π) ) of the local quadratic approximation are the first two terms of the local linear approximation. Exercise.3.4. Calculate the local quadratic approximation of h(x) = ln x at x = 1. The local quadratic approximation to the graph of a function is actually the parabola that comes closest to being the graph of the function, at the x-value x 0. In other words, it s the best parabolic approximation to the graph which is to say it s the best quadratic approximation to the graph at x = x Parametrized curves Suppose you have a position function #» r (t) that is twice differentiable. The graph of such a function is a curve possibly in the plane, possibly in space. Such a function has an increment, a differential, and local linear and quadratic approximations. And now you know what this section is about. 3.1 Increments and differentials of parameterized curves We will treat parameterized curves as though they were vector-valued functions. Increments in this context are just differences in values, as in the y = f(x) case. But the increment of some function #» r is a difference of two vectors, which have components, so there s more to write. The differential of a vector-valued function isn t hard to invent, either. Recall that the differential of y = f(x) is dy = f (x) dx. In like manner, the differential of #» r (t) is d #» r = #» r (t) dt. 5 The word best here has a technical meaning not discussed in the current edition of this document. Page 8

9 Exercise Convince yourself that d #» r = #» r (t) dt is exactly analogous to dy = f (x) dx. Here s the official vocabulary list, assuming #» r has three components: Definition Suppose #» r = #» x(t) r (t) = y(t) is differentiable at t = t 0 (which implies that ẋ(t 0 ), ẏ(t 0 ), z(t) and ż(t 0 ) exist). Let x 0 = x(t 0 ), y 0 = y(t 0 ), and z 0 = z(t 0 ). 1. The increment of t is t = t t 0.. The increments of x, y, and z are x = x x 0, y = y y 0, and z = z z 0, respectively. 3. The increment of #» r is #» r = #» r (t) #» r (t 0 ). 4. The differential of t is dt = t. 5. The differentials of x, y, and z at t 0 are dx = ẋ(t 0 ) dt, dy = ẏ(t 0 ) dt, and dz = ż(t 0 ) dt, respectively. 6. The differential of #» r at t 0 is d #» r = #» r (t0 ) dt. Exercise The above list of definitions assumes #» r has three components. What changes would we have to make to the list if we were to assume #» r has only two components? Exercise Use the appropriate definitions in the above list to show that d #» dx r = dy. dz Exercise If you divide the increment of #» r by the increment of t and let the increment of t go to zero, what do you get? (Despite its appearance, this is more of a vocabulary exercise than a math exercise.) Exercise If #» t π r = ln t, calculate the increment and differential of #» r at t = 0. t + 1 It s customary to think of t as being small, but it doesn t have to be. The fact that #» r is differentiable at t 0 is an excuse for pretending to believe that #» r is also small, at least, when t is small. The truth is that #» r can be large, but as t t 0, #» r has to go to zero. So you may as well pretend it s small, when t is small. Of course, our curve might be parameterized in terms of arc length instead of time. We have the option of writing out the definitions of the increments and differentials if arc length is the parameter. But all we d have to do is take Definition 3.1. and replace all the t s with s s. Exercise Please do so. (WARNING: Stay awake and please pay attention to detail.) Exercise Use the appropriate chain rule to determine how d #» r (t 0 ) is related to d #» r (s 0 ). There is a version of the increment theorem for parameterized curves. Theorem If #» r (t) is differentiable at t = t 0, then for t sufficiently close to t 0, #» r d #» r ; that is, #» r (t) #» r (t0 ) #» r (t0 )(t t 0 ). Exercise Evaluate the increment and differential you calculated for Exercise 3.1.6, at t = 0.01 and compare the two. Do you feel the differential approximates the increment very well? Exercise Think of at least one more way to write the increment theorem. (Hint: Mimic what you did for Exercise.1.5.) Exercise Write out what the increment theorem says about the function #» cos t r (t) = sin t at t = 0. t (Hint: Mimic what you did for Exercise.1.6.) Page 9

10 3. Local linear approximations to parameterized curves Our next task is to invent the local linear approximation to r(t) #» at a given point. There are a couple of ways to do this. One is to start by thinking about what the phrase local linear approximation ought to mean, for a parameterized curve. The other is to imitate the local linear approximation to y = f(x) at a given point. I propose we do the former, for the sake of practice. We can use the latter as a check on our work. So: The local linear approximation is a line that intersects the graph of #» r (t) at the point #» r (t 0 ) and comes as close as possible to being the curve. Hmm... To ensure our local linear approximation intersects the graph of #» r (t) at #» r (t 0 ), we need #» r (t 0 ) to be on the graph of the local linear approximation. That can t be too hard to manage. What about the as close as possible to being the curve part? Perhaps we can make the local linear approximation point in the same direction as the curve does? What a weird thing to say! 6 But it can be done: Just use the unit tangent vector as the direction vector of the local linear approximation. The unit tangent vector points in the direction of motion along the curve, so we might as well use it for the direction vector of the line. But do we really need a unit vector, with its potentially messy denominator? Nope. We just need an equation for the local linear approximation. Any (nonzero) vector parallel to the unit tangent vector will serve. Perhaps the most convenient one to use is #» r (t0 ), the good old velocity vector. At this point, we ought to be able to write down the local linear approximation just as we have written other parameterized lines, being sure to use the vector #» r (t0 ) as the direction vector and the vector #» r (t 0 ) as the location vector. Doing so yields something like L(t) #» = t #» r (t0 ) + #» r (t 0 ). This equation looks fine: This line is parallel to the direction of motion along the curve at t = t 0, and if we let t = 0, we get L(0) #» = #» r (t 0 ), so that this line intersects the curve at the right place. But there s a cultural issue here. People want the point #» r (t 0 ) to be L(t #» 0 ), not L(0). #» The easiest way I know of to make this happen is to shift the coördinate system by t 0 units. If we do, we ll get #» L(t) = (t t 0 ) #» r (t0 ) + #» r (t 0 ). I find it convenient to change the order in which the parts of this equation are written. If I write it this way: #» L(t) = #» r (t 0 ) + #» r (t0 )(t t 0 ) then I can instantly see that this looks a lot like the equation for the local linear approximation of y = f(x): L(x) = f(x 0 ) + f (x 0 )(x x 0 ). This is the check on our work that we were going to do. Exercise Convince yourself that #» r (t) #» r (t 0 ) + #» r (t0 )(t t 0 ) is exactly analogous to f(x) f(x 0 ) + f (x 0 )(x x 0 ). One last detail: If #» r (t0 ) = #» 0, then there can be a cusp or corner in the graph of the curve at t = t 0, even though #» r is differentiable there. Cusps and corners make it really hard for there to be a local linear approximation. So, as usual, we will require that #» r be smooth at t = t 0. Let s celebrate, by formalizing our new local linear approximation. Definition 3... Suppose #» r (t) is smooth at t = t 0. The local linear approximation to #» r (t) at t = t 0 is #» r (t) #» r (t0 ) + #» r (t0 )(t t 0 ), (1) for t sufficiently close to t 0. This local linear approximation is the line that comes closest to being the curve, at the t-value t 0. In that sense, it s the best linear approximation to the curve, at t = t 0. Example Suppose #» cos t r (t) = 3 sin t and t 0 = π/6. Then #» r (t 0 ) = #» r (π/6) = 3/ 3/ and t 3 π 3 /16 #» r = sin t 3 cos t, so that #» r (t0 ) = #» r (π/6) = 3t 6 Uh, doesn t the curve go both ways? And the line, as well? 1/ 3 3/ π /1. Page 10

11 Substitute these values into (1) to get #» r (t) 3/ 1/ 3/ + 3 3/ (t π/6) π 3 /16 π /1 which may feel a little odd. If you like, you can write #» r (t) 3/ 1/ 3/ + (t π/6) 3 3/. π 3 /16 π /1 I ve never found it helpful to try to simplify this sort of expression. Recall that the increment theorem says that if #» r (t) is differentiable at t = t 0, then for t sufficiently close to t 0, #» r (t) #» r (t 0 ) #» r (t0 )(t t 0 ). It follows that #» r (t) #» r (t0 ) + #» r (t0 )(t t 0 ). We conclude that the increment theorem implies that if #» r (t) is differentiable at t 0, then you may use the local linear approximation at t = t 0, if you like. 3.3 Local quadratic approximations to parameterized curves Our next task is to invent the local quadratic approximation to #» r (t). The local quadratic approximation ought to be the parabola that comes closest to being the curve, at the given point. How to construct it? We had pretty good luck back when it was y = f(x). We started with the local linear approximation and added a quadratic term. You may recall we decided to use (x x 0 ) rather than just x, because in the linear term (the x x 0 term), we had shifted the coördinate system. So we probably ought to use (t t 0 ) in this new approximation, as opposed to t. What benefit can we get from using a parabola that we can t get from using a line? A line is straight, and therefore cannot have much to do with the concavity of the curve. Suppose we include information about the concavity, then. The calculus tool for measuring concavity is the second derivative. So we ll use #» r (t0 ). You may feel it weird to use a vector to measure concavity, where we used to use a number in Calc I. But this is no weirder than using the vector #» r (t 0 ) to measure slope, i.e., direction. (Well, maybe a little weirder.) With #» r (t0 ) in hand, we ll going to guess that the local quadratic approximation has the form #» r (t) #» r (t0 ) + #» r (t0 )(t t 0 ) + a #» r (t0 )(t t 0 ). The a is there so we can adjust the quadratic term to match the second derivative of this new approximation with that of the given curve. Speaking of which, let s differentiate twice, to try to find the right value of a: d ( #»r (t0 dt ) + #» r (t0 )(t t 0 ) + a #» r (t0 )(t t 0 ) ) = d #» r (t0 dt ) + d #» r dt (t0 )(t t 0 ) + d dt a #» r (t0 )(t t 0 ) = 0 + d #» r dt (t0 ) + d dt a #» r (t0 )(t t 0 ) = a #» r (t0 ) = a #» r (t0 ). Page 11

12 If we want this to be the same as the second derivative of the curve, which is #» r (t0 ), then a had better be 1/. Hey! That s what we got when we went through this exercise when it was y = f(x)! In fact, a side-by-side comparison of the two local quadratic approximations could be informative. Compare with #» r (t) #» r (t0 ) + #» r (t0 )(t t 0 ) + 1 #» r (t0 )(t t 0 ) f(x) f(x 0 ) + f (x 0 )(x x 0 ) + 1 f (x 0 )(x x 0 ). Nice match, yes? Definition Suppose the function #» r (t) is twice differentiable at t = t 0 and that neither #» r (t0 ) not #» r (t0 ) is #» 0. The local quadratic approximation to #» r at t = t 0 is for t sufficiently close to t 0. #» r (t) #» r (t0 ) + #» r (t0 )(t t 0 ) + 1 #» r (t0 )(t t 0 ), () The requirement that v #» (t0 ) #» 0 helps ensure that the graph of #» r is smooth enough at t = t 0 for there to be a parabola that does a good job of approximating this graph. If you re curious about this, come talk to me. (It s about things that are similar to cusps, but subtler than cusps.) Example Suppose #» cos t r (t) = 3 sin t and t 0 = π/6. Then t 3 #» r (t0 ) = 3/ 3/ ; π 3 /16 sin t 1/ #» r = 3 cos t, so that #» r (t0 ) = 3 3/ ; 3t π /1 cos t 3/ #» r = 3 sin t, so that #» r (t0 ) = 3/. 6t π Substitute these values into Equation (3.3.1) to get #» r (t) 3/ 1/ 3/ + 3 3/4 3/ (t π/6) + 3/4 (t π/6), π 3 /16 π /1 π/ which may feel a little odd. If you like, you may write #» r (t) 3/ 1/ 3/ + (t π/6) 3 3/4 3/ + (t π/6) 3/4. π 3 /16 π /1 π/ Again, I ve never found it helpful to multiply out the (t π/6) through or try to simplify these expressions in any way. Exercise Calculate [ the increments, ] differentials, local linear approximation, and local quadratic approximation to #» cos r (t) = 3 t sin 3 at t = π/3. t Exercise Repeat the previous exercise, locating everything at t = 0, instead of t = π/3. Exercise Make a nice graph of the curve that appears in the previous two exercises. (Be sure to include t = 0 and t = π/3 in the range of t-values that you use.) Explain the results you got for the total differential and the local linear approximation in the previous exercise. and Page 1

13 4 Surfaces: Real-valued functions of two variables Recall that the graph of a function of the form z = f(x, y) is a surface. 7 If we want information about increments, differentials, and local linearizations of such functions, we should be able to get a fair amount of good out of applying Sections and 3 to surfaces. 4.1 Increments and differentials of surfaces Increments are just as simple as they can be: Subtract a pair of values of the same object, and you have an increment. For example, if we have z = f(x, y), we can see that x is still x x 0, though y is now y y 0 because y is now an independent variable. Finally, z = f(x, y) f(x 0, y 0 ), as you might guess. Differentials of independent variables are always the same as the increments of those variables, so dx = x and dy = y. But dz should be more complicated, shouldn t it? Based on previous experience, we can expect dz to have first derivatives in it. Careful! There are infinitely many first derivatives available. 8 Well, f(x) depends on x, and we use df/dx in the differential of f(x). Since f(x, y) depends on x and y, let s use f/ x and f/ y in the differential of f(x, y). The differentials in the previous section have things like dx or dt in them. How do such things figure in to the local linear approximation of f(x, y)? To answer this question, recall why they are used in the local linearization of y = f(x). The idea was to approximate the change in output value (the rise, or dz) by multiplying the slope ( rise over run, or f ) by the change in input values (the run, or dx). Now, for a function z = f(x, y), there are two kinds of input values we can change: x-values and y-values. So the total change in output (dz) can be viewed as the-change-in-output-due-to-a-change-in-the-x-values PLUS the-change-in-output-due-to-a-change-in-the-y-values. So we can approximate the change in output value (dz) by totaling (adding) the changes in output due to x (that is, f x dx) and those due to y (that is, f y dy). Well, that would mean dz = f x (x 0, y 0 ) dx + f y (x 0, y 0 ) dy. (3) This is the total differential of f(x, y) at (x 0, y 0 ). If you like, you can call f x (x 0, y 0 ) dx the partial differential of f, with respect to x, and likewise for f y (x 0, y 0 ) dy, but I don t know anyone that does. ( π Exercise Calculate the total differential of T (x, y) = sin x cosh y at 4 )., 1 (If you d like a concise listing of information about hyperbolic functions, you can find one in the Mathematics Resources section of my faculty website.) Exercise Calculate the total differential of φ(t, x) = ( sin λx ) e kλt. Assume λ and k are positive constants. There is an increment theorem for surfaces, just as there is for parameterized curves and for the y = f(x) case. We state this theorem as: Theorem If f(x, y) is differentiable at (x 0, y 0 ) then f df, or for (x, y) sufficiently close to (x 0, y 0 ). f(x, y) f(x 0, y 0 ) f x (x 0, y 0 ) dx + f y (x 0, y 0 ) dy, Exercise Calculate all the increments associated with 9 z = 1 ln x + 3 y at (e, π), and give the total differential, as well. 7 In a partial differential equations course, you don t often think in terms of surfaces, but you certainly can. 8 There are infinitely many if #» f exists, anyway, because then you can have all the directional derivatives. 9 Meaning the increments of the dependent and all the independent variables. Page 13

14 4. Local linear approximations to surfaces We will use a different method of inventing the local linear approximation in this section than we have before. The method will be absurdly simple; so simple, in fact, that we ll want a way to verify that we re actually getting the right thing. Hmm... What should a local linear approximation to a surface be like? Well, it probably should contain the point ( x 0, y 0, f(x 0, y 0 ) ) where we want the approximation located. It ought to be straight, in some sense. And it ought to be close to the surface, in some verifiable sense. I m concerned about the word straight. Straight sounds like line to me, but would a line be an adequate approximation to a surface? I propose replacing the word straight with the word flat. In Latin, flat is planus, as in plane. Wouldn t a plane be a more suitable approximation than a line? So: We re looking for a plane that contains the point ( x 0, y 0, f(x 0, y 0 ) ) and is close to the surface over some region of useful size. Recall that the increment theorem states that if f(x, y) is differentiable at (x 0, y 0 ) then f(x, y) f(x 0, y 0 ) f x (x 0, y 0 ) dx + f y (x 0, y 0 ) dy, for (x, y) sufficiently close to (x 0, y 0 ). If we add f(x 0, y 0 ) to both sides of this approximation and write dx as (x x 0 ) and dy as (y y 0 ), we get f(x, y) f(x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ), for (x, y) sufficiently close to (x 0, y 0 ). That s fine, but did we get a plane containing the point ( x 0, y 0, f(x 0, y 0 ) )? Is it close to the surface z = f(x, y) over a region of useful size? You can verify immediately that that replacing x with x 0 and y with y 0 yields f(x 0, y 0 ): f(x 0, y 0 ) + f x (x 0, y 0 )(x 0 x 0 ) + f y (x 0, y 0 )(y 0 y 0 ) = f x (x 0, y 0 )(0) + f y (x 0, y 0 )(0) = f(x 0, y 0 ). Verifying that we got a plane is Exercise 4..4, which you ll do after seeing an example or two. (You ll be convinced, by then, which will make this task easier for you.) But is the above approximation is close to the surface over a region of useful size? Short answer: Yep, because f(x, y) is differentiable. Some day, I ll include the details in the section on the technical stuff. For now, it s enough to observe that we derived our linear approximation from the increment theorem, which says the increment is close to the total differential when f(x, y) is differentiable. Definition If f(x, y) is differentiable at (x 0, y 0 ), the local linear approximation to f at (x 0, y 0 ) is for (x, y) sufficiently close to (x 0, y 0 ). f(x, y) f(x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ), Example 4... Try it: Write down the local linear approximation of z = xy + cos y at (, π/3). Solution: Apparently, x 0 = and y 0 = π/3, and f = xy + cos y. So f x = y and f y = x sin y. At (, π/3), these last three are f(, π/3) = π/3 + 1/ = ( 4π + 3)/6, f x (, π/3) = π/3, and f y (, π/3) = 3/ = (4 + 3)/, respectively. It must be that for (x, y) sufficiently close to (, π/3), z 4π π 3 (x ) ( y π ). (4) 3 (I ve never known it to be worth the trouble to multiply all that out and combine like terms.) Page 14

15 Exercise Convince yourself that the equation we ended up with in Example 4.. is the equation of a plane. Exercise Convince yourself that the equation in Definition 4..1 is the equation of a plane. We really ought to assemble all the relevant vocabulary for the two-variable case. Here it is: Definition Let z = f(x, y) be differentiable at (x 0, y 0 ). 1. The increments of x and y are x = x x 0 and y = y y 0, respectively.. The increment of z is z = f(x, y) f(x 0, y 0 ). 3. The differentials of x and y are dx = x and dy = y, respectively. 4. The total differential of z is dz = f x (x 0, y 0 ) dx + f y (x 0, y 0 ) dy. Note that dz is sometimes called df, in which case we think of it as the total differential of f. 5. The local linear approximation to z = f(x, y) at (x 0, y 0 ) is for (x, y) near (x 0, y 0 ). f(x, y) f(x 0, y 0 ) + f x (x 0, y 0 ) dx + f y (x 0, y 0 ) dy, Exercise Write down the increment of and the local linear approximation to z = xy 3xy + 4x at (1, ). Note that the local linear approximation can also be thought of as f(x, y) f(x 0, y 0 ) + df, which is what the increment theorem for z = f(x, y) says. Later on, we may want the convenience of writing the linear terms (the f x (x 0, y 0 )dx + f y (x 0, y 0 )dy part) as a dot product. To[ this ] end, make a column matrix (that is, a vector) out of dx and dy and call it dx d(x, y). That is, d(x, y) =. Then dy This turns the local linear approximation into f x (x 0, y 0 )dx + f y (x 0, y 0 )dy = #» f(x 0, y 0 ) d(x, y). f(x, y) f(x 0, y 0 ) + #» f d(x, y), for (x, y) near (x 0, y 0 ). That f #» d(x, y) sure looks like a chain rule, but it s not. Chain rules look like f #» (x,y) t, or something. The Jacobian matrix (x,y) t in the chain rule is a full-fledged derivative, not some differential. 4.3 Local quadratic approximations to surfaces The local quadratic approximation is missing from Definition 4..5, because it needs additional attention, being more complicated than any local approximation we ve seen so far. We expect from our previous experience that the local quadratic approximation to a surface ought to have a 1/ in front of its quadratic term. And it ought to involve second derivatives of f(x, y). But which ones? Both f x and f y have infinitely many first derivatives. 10 When we invented the total differential and the local linear approximation, we worked around this problem by just differentiating in both the x- and y-directions. This suggests differentiating both f x and f y with respect to both x and y, which is what we ll do. 10 Think about differentiating f y, for example, in any desired direction. Page 15

16 Indeed, if we were to think about how the first derivatives of f change as (x, y) varies near (x 0, y 0 ), we d eventually write down the differential of f x, or We d do the same for f y, writing d(f x ) (f x ) x (x 0, y 0 )dx + (f x ) y (x 0, y 0 )dy = f xx (x 0, y 0 )dx + f xy (x 0, y 0 )dy. d(f y ) (f y ) x (x 0, y 0 )dx + (f y ) y (x 0, y 0 )dy = f yx (x 0, y 0 )dx + f yy (x 0, y 0 )dy. So what? So think about a change in dz. That should be d(dz) = d ( f x (x 0, y 0 )dx + f y (x 0, y 0 )dy ) = d ( f x (x 0, y 0 )dx ) + d ( f y (x 0, y 0 )dy ). It would sure be convenient if d(f x (x 0, y 0 )dx) were the same thing as d(f x )(x 0, y 0 )dx. It is. 11 we can keep forging ahead, to get that d(dz) ought to be This means d(f x )(x 0, y 0 )dx + d(f x )(x 0, y 0 )dx = ( f xx (x 0, y 0 )dx + f xy (x 0, y 0 )dy ) dx + ( f yx (x 0, y 0 )dx + f yy (x 0, y 0 )dy ) dy = f xx (x 0, y 0 )(dx) + f xy (x 0, y 0 )dydx + f yx (x 0, y 0 )dxdy + f yy (x 0, y 0 )(dy). If the second partial derivatives of f are continuous at (x 0, y 0 ), then f xy (x 0, y 0 ) = f yx (x 0, y 0 ). This means we can rewrite the foregoing as f xx (x 0, y 0 )(dx) + f xy (x 0, y 0 )dxdy + f yy (x 0, y 0 )(dy). But we have seen that the quadratic term in other local quadratic approximations needs a coefficient of 1/. So multiply by 1/ to get 1 f xx(x 0, y 0 )(dx) + f xy (x 0, y 0 )dxdy + 1 f yy(x 0, y 0 )(dy), which is the quadratic term in the local quadratic approximation to z = f(x, y). Well, we already know what the local linear approximation to z = f(x, y) is. Let s tack on the quadratic term and see what we get: Definition If z = f(x, y) is twice differentiable at (x 0, y 0 ) then the local quadratic approximation to f at (x 0, y 0 ) is f(x, y) f(x 0, y 0 ) + f x (x 0, y 0 )dx + f y (x 0, y 0 )dy + 1 f xx(x 0, y 0 )(dx) + f xy (x 0, y 0 )dxdy + 1 f yy(x 0, y 0 )(dy), for (x, y) near (x 0, y 0 ). Exercise Write down the local quadratic approximation to z = xy 3xy + 4x at (1, ). 4.4 Isn t there a better way to write this stuff? If you want a better way to write this stuff, think about matrix multiplication. You know, like when we were discussing chain rules. For example, if f(x, y) = 3x y cos(x y), we d say something like Let u = x y and v = x y, so that f = 3u cos v. Then we d calculate #» f = [ 3 sin v ] u=x y,v=x y = [ 3 sin(x y) ] u=x y,v=x y and (u, v) (x, y) = [ ] xy x As opposed to being something like d(f x)(x 0, y 0 )dx + f x(x 0, y 0 )d(dx). Actually, it turns out that d(dx) = 0. If you want to know why, read the section on the technical stuff, at the end of this document. Page 16

17 and then ( f #» f ) (u, v) (x, y) = u=x y,v=x y (x, y) = [ 3 sin(x y) ] [ ] xy x = 1 1 (Sigh. Good times!) But here s the magic moment: The multiplication of [ 3 sin(x y) ] [ ] xy x by 1 1 consists of taking the dot product of the row known as f #» with each column of the Jacobian matrix. We d get [ 6xy + sin(x y) 3x sin(x y) ]. Point is, to multiply two matrices together, take the dot product of each row of the left-hand matrix with each column of the right-hand matrix. [ ] r x [ ] 1 3 Example a b s y a + b + 3c r + s + 3t x + y + 3z = a + 5b + 6c 4r + 5s + 6t 4x + 5y + 6z c t z Example [ ] ( )(3) + (1)(0) ( )( 4) + (1)(1) ( )(1) + (1)( 1) = (3)(3) + (0)(0) (3)( 4) + (0)(1) (3)(1) + (0)( 1) = (5)(3) + ( 7)(0) (5)( 4) + ( 7)(1) (5)(1) + ( 7)( 1) You can multiply more than two matrices together. Just start at the left and work your way right: Example [ a b ] [ [ ] 1 a = 3 4] [ a + 3b a + 4b ] [ ] a = (a + 3b)a + (a + 4b)b = a b b + 5ab + 4b Matrix multiplication is associative, so you can start from the right, instead, if you prefer: Example [ a b ] [ [ ] 1 a = 3 4] [ a b ] [ ] a + b = a(a + b) + b(3a + 4b) = a b 3a + 4b + 5ab + 4b Exercise Multiply together the matrices [ a b ] [ [ ] 1 a. 4 3] b Here s how matrix multiplication helps with local quadratic approximations. Suppose f = f(x, y) has continuous second partial derivatives. Define the Hessian matrix of f to be [ ] fxx f H f = xy. f yx (It s the matrix of second partial [ derivatives ] of f, just as the gradient is the vector of first partial derivatives dx of f.) Recall that d(x, y) =. Transpose this column to make a row out of it, dy 1 and call the result d(x, y) T ; that is, d(x, y) T = [ dx dy ]. Exercise Calculate the Hessian matrix of f(x, y) = x Arctan y y ln x and evaluate it at (1, 1), at (3, 0), and at (e 3, 3). (This should convince you that the Hessian matrix can change, as x and y change.) 1 Transposing a column turns it into a row; transposing a row gives you a column. Or, if your matrix has multiple rows or [ ] T 1 3 columns, then transposing it turns all the rows into columns and all the columns into rows: = f yy Page 17

18 The quadratic term in the local quadratic approximation to f can now be written as 1 d(x, y)t H f d(x, y), when this matrix product is evaluated at (x 0, y 0 ). (Note the 1/, inherited from local quadratic approximations to curves.) Exercise If z = f(x, y) = sin x sin y and if (x 0, y 0 ) = (π/, π/), then calculate the quadratic term of the local quadratic approximation to f at (x 0, y 0 ), using the formula 1 d(x, y)t H f d(x, y). Exercise As a check on your work for the previous exercise, recalculate the quadratic term of the local quadratic approximation to f(x, y) = sin x sin y at (π/, π/), using the formula (You d better get the same answer!) 1 f xx(x 0, y 0 )(dx) + f xy (x 0, y 0 )dxdy + 1 f yy(x 0, y 0 )(dy). The previous two exercises suggest that 1 d(x, y)t H f (x 0, y 0 ) d(x, y) really is the quadratic term. Here s the calculation that proves it: 1 d(x, y)t H f d(x, y) = 1 [ ] [ ] [ ] f dx dy xx f xy dx f yx f yy dy = 1 [ fxx dx + f yx dy f xy dx + f yy dy ] [ ] dx dy = ( 1 (fxx dx + f yx dy ) dx + ( f xy dx + f yy dy ) ) dy When evaluated at (x 0, y 0 ), this becomes = 1 ( fxx (dx) + f xy dxdy + f yy (dy) ). 1 f xx(x 0, y 0 )(dx) + f xy (x 0, y 0 )dxdy + 1 f yy(x 0, y 0 )(dy), as advertised. So, the compact way of writing the local quadratic approximation to f at (x 0, y 0 ) is f(x, y) f(x 0, y 0 ) + #» f(x 0, y 0 ) d(x, y) + 1 d(x, y)t H f (x 0, y 0 ) d(x, y), for (x, y) near (x 0, y 0 ). (Some people like to write it as f(x, y) f(x 0, y 0 ) + #» f(x 0, y 0 ) d(x, y) + 1 H f (x 0, y 0 ) d(x, y), maybe to make it look a bit more like the one-variable version. But I think this is cheating just a little. 13 ) In its full glory, the local quadratic approximation is f(x, y) f(x 0, y 0 ) + f x (x 0, y 0 )dx + f y (x 0, y 0 )dy + 1 f xx(x 0, y 0 )(dx) + f xy (x 0, y 0 )dxdy + 1 f yy(x 0, y 0 )(dy) 13 A more elegant way to write the local quadratic approximation comes from realizing that the gradient is the first derivative of f and the Hessian is the second derivative. Then you can write Df(x 0, y 0 ) = #» f(x 0, y 0 ) and D f(x 0, y 0 ) = H f (x 0, y 0 ), and the local quadratic approximation becomes or in its cheating version, f(x, y) f(x 0, y 0 ) + Df(x 0, y 0 ) d(x, y) + 1 d(x, y)t D f(x 0, y 0 ) d(x, y), f(x, y) f(x 0, y 0 ) + Df(x 0, y 0 ) d(x, y) + 1 D (x 0, y 0 ) d(x, y). Page 18

19 = f(x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ) + 1 f xx(x 0, y 0 )(x x 0 ) + f xy (x 0, y 0 )(x x 0 )(y y 0 ) + 1 f yy(x 0, y 0 )(y y 0 ), for (x, y) near (x 0, y 0 ). Knowing how lazy I am, I bet you can guess how I prefer writing it! 4.5 Local linear approximations to level curves To be continued... 5 Real-valued functions of three or more variables It s not unusual to have three spatial variables and one temporal variable in a partial differential equation. Therefore, you may well benefit from figuring out how to calculate increments, differentials, and local linear approximations of functions of as many as four variables. But let s start with three variables. 5.1 Three independent variables For the next two exercises, suppose z = e kt sin λx + e kt cos λy and (x 0, y 0, t 0 ) = (π, π/4, 1). Assume k > 0 and λ > 0. Exercise What would be the increments of x, y, t, and z? Exercise What would be the differentials of x, y, and t? Exercise Write down formal definitions of the increments of x, y, t, and z and of the differentials of x, y, and t. The differential of z is more work, naturally, since it depends on x, y, and t. Take another look at Part 4 of Definition 4..5 and think about how you would change it to include the variable t. Exercise Write down a formal definition of dz, given that z = f(x, y, t) is differentiable. Exercise Calculate the total differential of z = e kt sin λx + e kt cos λy. Again, assume λ > 0. Now take another look at Part 5 of Definition 4..5 and think about how you would change it to include the variable t. Exercise Write down a formal definition of the local linear approximation of z = f(x, y, t) at (x 0, y 0, t 0 ). Exercise Calculate the local linear approximation of z = e kt sin λx + e kt cos λy at (x 0, y 0, t 0 ) = (π, π/4, 1). As before, assume k > 0 and λ > 0. Exercise Write down the increment theorem assuming f(x, y, t) is differentiable at (x 0, y 0, t 0 ). As you can see, having three variables really doesn t introduce anything new into the ideas of increment, differential, and local linear approximation. All that s new, really, is that there s more to write. Stripped down a bit more, to suppress the dependence on (x, y) and (x 0, y 0 ), we have Or if d #» [ x = x x0 y y 0 ], then we can write z z 0 + Dz d(x, y) + 1 D z d(x, y). z z 0 + Dz d #» x T + 1 D z d #» x. While this is a lot of notation to process, surely you can see that these abbreviated versions of the formula for local quadratic approximations could be handy to use. Page 19