Review guide for midterm 2 in Math 233 March 30, 2009

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1 Review guide for midterm 2 in Math 2 March, 29 Midterm 2 covers material that begins approximately with the definition of partial derivatives in Chapter 4. and ends approximately with methods for calculating the double integral of a function f(x, y) over a domain described in the xy-plane. See the updated course web page for the exact material covered on this exam. This material begins with the definition of partial derivatives in Chapter 4. and ends with methods for calculating the double integral of a function f(x, y) over a domain described in the xy-plane. See sections 4. and 4.2 for material on continuity of functions. efinition (Partial erivatives) If f is a function of two variables, its partial derivatives are the functions f x and f y defined by f x (x, y) lim h f(x + h, y) f(x, y) h f(x, y + h) f(x, y) f y (x, y) lim. h h We have the following rule for calculating partial derivatives.. To find f x, regard y as a constant and differentiate f(x, y) with respect to x. 2. To find f y, regard x as a constant and differentiate f(x, y) with respect to y. Example 2 Calculate f x, f y for f(x, y) x 2 e xy + y 2. f x (x, y) 2xe xy + x 2 e xy y 2xe xy + x 2 ye xy f y (x, y) x 2 e xy x + 2y x e xy + 2y. efinition (Second Partial erivatives) For z f(x, y), we use the following notation: (f x ) x f xx ( ) f 2 f x x x 2 2 z x 2 (f x ) y f xy ( ) f 2 f y x y x 2 z y x (f y ) x f yx ( ) f 2 f x y x y 2 z x y (f y ) y f yy ( ) f 2 f y y y 2 2 z y 2 Example 4 Find the second partial derivatives of f(x, y) x + x 2 y 2y 2.

2 2 Note: Therefore, f x (x, y) x 2 + 2xy f y (x, y) x 2 y 2 4y. f xx 6x + 2y f xy 6xy 2 f yx 6xy 2 f yy 6x 2 y 4. Note that in the above example f xy f yx. This is no coincidence and fact follows from the next theorem that states that under weak conditions on f(x, y), taking partial derivatives is a commutative process. Theorem 5 (Clairaut s Theorem) Suppose f is defined on a disk that contains the point (a, b). If the functions f xy and f yx are both continuous on, then f xy (a, b) f yx (a, b). The next definition of tangent plane generalizes in a natural way the following equation of the tangent line of a function of variable: y y f (x )(x x ). efinition 6 (Tangent Plane) Suppose f has continuous partial derivatives. An equation of the tangent plane to the surface z f(x, y) at the point P (x, y, z ) is z z f x (x, y )(x x ) + f y (x, y )(y y ). Example 7 Find the tangent plane to the elliptic paraboloid z 2x 2 + y 2 at the point (,, ). Let f(x, y) 2x 2 + y 2. Then f x (x, y) 4x f y (x, y) 2y f x (, ) 4 f y (, ) 2. Then efinition 6 gives the equation of the tangent plane at (,, ) as or z 4(x ) + 2(y ) z 4x + 2y. The next definition of linear approximation generalizes the linear approximation L(x) of a function f(x) of variable at a point x a : L(x) f(x) + f (x)(x a). efinition 8 (Linear Approximation) The linear approximation of f(x, y) at (a, b) is L(x, y) f(a, b) + f x (a, b)(x a) + f y (a, b)(y b).

3 Example 9 Use the linear approximation L(x, y) to f(x, y) xe xy at (, 2) to estimate f(.,.8). Hence, f x (x, y) e xy + xye xy f y (x, y) x 2 e xy f x (, 2) e 2 + 2e 2 f y (, 2) e 2. L(x, y) f(, 2) + f x (, 2)(x ) + f y (, 2)(y 2). L(.,.8) e 2 + (e 2 + 2e 2 )(.) + e 2 (.2). efinition If z f(x, y), then f is differentiable at (a, b) if the change z of z can be expressed in the form where ε and ε 2 as ( x, y) (, ). z f x (a, b) x + f y (a, b) y + ε x + ε 2 y, The next theorem gives a simple condition for f(x, y) to satisfy in order to be differentiable. Theorem If the partial derivatives f x and f y exist near (a, b) and are continuous at (a, b), then f is differentiable at (a, b). efinition 2 (Total ifferential) For z f(x, y), dz f x (x, y) dx + f y (x, y) dy dx + x y dy. Example The base radius and height of a right circular cone are measured as cm and 25 cm, respectively, with a possible error in measurement of as much as. cm. in each. Use differentials to estimate the maximum error in the calculated volume of the cone. The volume V of a cone with base radius r and height h is V πr 2 h. So the differential of V is dv V r dr + V h dh 2πrh dr + πr2 dh. Since each error is at most. cm, we have r., h.. To find the largest error in the volume we take the largest error in the measurement of r and of h. Therefore, we take dr. and dh. along with r, h 25. This gives the estimate dv 5π π (.) + (.) 2π. Thus, the maximum error in the calculated volume is about 2π cm 6 cm.

4 4 Example 4 The dimensions of a rectangular box are measured to be 75 cm, 6 cm, and 4 cm, and each measurement is correct to within.2 cm. Use differentials to estimate the largest possible error when the volume of the box is calculated from these measurements. If the dimensions of the box are x, y, and z, its volume is V xyz and so dv V x dx + V y dy + V dz yz dx + xz dy + xy dz. We are given that x.2, y.2, and z.2. To find the largest error in the volume, we use dx.2, dy.2, and dz.2 together with x 75, y 6, and z 4: V dv (6)(4)(.2) + (75)(4)(.2) + (75)(6)(.2) 98. Thus, an error of only.2 cm in measuring each dimension could lead to an error of as much as 98 cm in the calculated volume! This may seem like a large error, but it s only about % of the volume of the box. Theorem 5 (Chain Rule Case ) Suppose that z f(x, y) is a differentiable function of x and y, where x g(t) and y h(t) are both differentiable functions of t. Then z is a differentiable function of t and dz dt f dx x dt + f dy y dt. Theorem 6 (Chain Rule Case 2) Suppose that z f(x, y) is a differentiable function of x and y, where x g(s, t) and y h(s, t) are both differentiable functions of s and t. Then z is a differentiable function of t and s x x s + y y s t x x t + y y t. Theorem 7 (The Chain Rule (General Version)) Suppose that u is a differentiable function of n variables x, x 2,..., x n and each x j is a differentiable function of the m variables t, t 2,..., t m. Then u is a function of t, t 2,..., t m and for each i, 2,..., m. u t i u x x t i + u x 2 x 2 t i u x n x n t i Example 8 If z x 2 y + xy 4, where x sin 2t and y cos t, find dz dt when t. The Chain Rule gives dz dt f dx x dt + f dy y dt (2xy + y4 )(2 cos 2t) + (x 2 + 2xy )( sin t). It s not necessary to substitute the expressions for x and y in terms of t. observe that when t we have x sin and y cos. Therefore, We simply dz ( + )(2 cos ) + ( + )( sin ) 6. dt t

5 5 Example 9 The pressure P (in kilopascals), volume V (in liters), and temperature T (in kelvins) of a mole of an ideal gas are related by the equation P V 8.T. Find the rate at which the pressure is changing when the temperature is K and increasing at a rate of. K/s and the volume is L and increasing at a rate of.2 L/s. If t represents the time elapsed in seconds, then at the given instant we have T, dt/dt., V, dv/dt.2. Since Case of the Chain Rule gives dp dt P dt T dt + P dv V P 8. T V, dt 8. V dt dt + 8.T dv V 2 dt 8. 8.() (.) + 2 (.2).455. The pressure is decreasing at a rate of about.42 kpa/s. Example 2 If z e sin y, where x st 2 and y s 2 t, find s and t. Applying Case 2 of the Chain Rule, we get dz ds x x s + y y s (ex sin y)(t 2 ) + (e x cos y)(2st) t 2 e st2 sin(s 2 t) + 2ste st2 cos(s 2 t), dz dt x x t + y y t (ex sin y)(2st) + (e x cos y)(s 2 ) 2sre st2 sin(s 2 t) + s 2 e st2 cos(s 2 t). Example 2 If u x 4 + y 2 z, where x rse t, y rs 2 e t, and z r 2 s sin t, find the value of u/ s when r 2, s, t. We have u s u x x s + u y y s + u s (4x y)(re t ) + (x 4 + 2zy )(2rse t ) + (y 2 z 2 )(r 2 sin t). When r 2, s, and t, we have x 2, y 2, and z, so u s (64)(2) + (6)(4) + ()() 92. Theorem 22 (Implicit ifferentiation) Suppose that z is given implicitly as a function z f(x, y) by an equation F (x, y, z), i.e., F (x, y, f(x, y)) for all (x, y) in the domain of f(x, y). Then: F F x x y y. F F

6 6 Example 2 Find x and y if x + y + z + 6xyz. Let F (x, y, z) x + y + z + 6xyz. Then, from Theorem 22, we have x F x F z x2 + 6yz z 2 + 6xy x2 + 2yz z 2 + 2xy y F y F z y2 + 6xz z 2 + 6xy y2 + 2xz z 2 + 2xy. efinition 24 (irectional erivative) The directional derivative of f at (x, y ) in the direction of a unit vector u a, b is if this limit exists. u f(x, y ) lim h f(x + ha, y + hb) f(x, y ) h efinition 25 (irectional erivative) The directional derivative of f at (x, y, z ) in the direction of a unit vector u a, b, c is if this limit exists. u f(x, y, z ) lim h f(x + ha, y + hb, z + hc) f(x, y, z ) h efinition 26 (Gradient) If f is a function of two variables x and y, then the gradient of f is the vector function f defined by f(x, y) f x (x, y), f y (x, y) f x i + f y j. efinition 27 (Gradient) For f(x, y, z), a function of three variables, f f x, f y, f z f x i + f y j + f k. The next two theorems give a simple rule for calculating the directional derivative of a function in 2 or variables in terms of the gradient of the function. Theorem 28 If f is a differentiable function of x and y, then f has a directional derivative in the direction of any unit vector u a, b and u f(x, y) f x (x, y)a + f y (x, y)b. Theorem 29 If f is a differentiable function of x, y, and z, then f has a directional derivative in the direction of any unit vector u a, b, c and u f(x, y, z) f(x, y, z) u. By the above two theorems, we have for any unit vector u, Thus, the next theorem holds. u f f u f u cos(θ) f cos(θ).

7 7 Theorem Suppose f is a differentiable function of two or three variables. The maximum value of the directional derivative u f(x) is f(x) and it occurs when u has the same direction as the gradient vector f(x). Example Find the directional derivative of the function f(x, y) x 2 y 4y at the point (2, ) in the direction of the vector v 2i + 5j. We first compute the gradient vector at (2, ): f(x, y) 2xy i + (x 2 y 2 4)j f(2, ) 4i + 8j. Note that v is not a unit vector, but since v 29, the unit vector in the direction of v is u v v 2 i + 5 j Therefore, by Theorem 28, we have u f(2, ) f(2, ) u ( 4i + 8j) ( 2 29 i j) Theorem 2 Suppose S is a surface determined as F (x, y, z) k constant. Then F is everywhere normal or orthogonal to S. In particular, if P (x, y, z ) S, then the equation of the tangent plane to S at p is: F x (x, y, z )(x x ) + F y (x, y, z )(y y ) + F z (x, y, z )(z z ) () Example Find the equations of the tangent plane and normal line at the point ( 2,, ) to the ellipsoid x y2 + z2 9. The ellipsoid is the level surface (with k ) of the function Therefore we have F (x, y, z) x2 4 + y2 + z2 9. F x (x, y, z) x 2 F y (x, y, z) 2y F z (x, y, z) 2z 9 F x ( 2,, ) F y ( 2,, ) 2 F z ( 2,, ) 2. as Then Equation in Theorem 2 gives the equation of the tangent plane at ( 2, 2, ) (x + 2) + 2(y ) 2 (z + ), which simplifies to x 6y + 2z 8. Since F ( 2,, ), 2, 2, the vector equation of the normal line is: L(t) 2,, + t, 2, 2.

8 8 efinition 4 A function of two variables has a local maximum at (a, b) if f(x, y) f(a, b) when (x, y) is near (a, b). (This means that f(x, y) f(a, b) for all points (x, y) in some disk with center (a, b).) The number f(a, b) is called a local maximum value. If f(x, y) f(a, b) for all f(x, y) in the domain of f, then f has an absolute maximum at (a, b). If f(x, y) f(a, b) when (x, y) is near (a, b), then f(a, b) is a local minimum value. If f(x, y) f(a, b) for all (x, y) in the domain of f, then f has an absolute minimum at (a, b). The next theorem explains how to find local maxima and local minima for a function in two variables. Theorem 5 If f has a local maximum of minimum at (a, b) and the first-order partial derivatives of f exist there, then f x (a, b) and f y (a, b). efinition 6 A point (a, b) is called a critical point of f(x, y) if f x (a, b) f y (a, b). The next theorem gives a method for testing critical points of a function f(x, y) to see if they represent local minima, local maxima or saddle points (a critical point (a, b) is a saddle point if the Hessian defined in the next theorem is negative). Theorem 7 (Second erivative Test) Suppose the second partial derivatives of f are continuous on a disk with center (a, b), and suppose that f x (a, b) and f y (a, b) (that is, (a, b) is a critical point of f). Let (a, b) f xx (a, b)f yy (a, b) [f xy(a,b) ] 2. (a) If > and f xx (a, b) >, then f(a, b) is a local minimum. (b) If > and f xx (a, b) <, then f(a, b) is a local maximum. (c) If <, then f(a, b) is a saddle point. To remember the formula for it s helpful to write it as a determinant: f xx f xy f xxf yy (f xy ) 2. f yx f yy Example 8 Find the local maximum and minimum values and saddle points of f(x, y) x 4 + y 4 4xy +. We first locate the critical points: f x 4x 4y f y 4y 4x. Setting these partial derivatives equal to, we obtain the equations x y y x. To solve these equations we substitute y x from the first equation into the second one. This gives x 9 x x(x 8 ) x(x 4 )(x 4 + ) x(x 2 )(x 2 + )(x 4 + )

9 9 so there are three real roots: x,,. The three critical points are (, ), (, ), and (, ). Next we calculate the second partial derivatives and (x, y): f xx 2x 2 f xy 4 f yy 2y 2 (x, y) f xx f yy (f xy ) 2 44x 2 y 2 6. Since (, ) 6 <, it follows from case (c) of the Second erivative Test that the origin is a saddle point; that is, f has no local maximum or minimum at (, ). Since (, ) 28 > and f xx (, ) 2 >, we see from case (a) of the test that f(, ) is a local minimum. Similarly, we have (, ) 28 > and f xx (, ) 2 >, so f(, ) is also a local minimum. efinition 9 A subset R 2 is closed if it contains all of its boundary points. efinition 4 A subset R 2 is bounded plane. if it is contained within some disk in the Theorem 4 (Extreme Value Theorem for Functions of Two Variables) If f is continuous on a closed, bounded set in R 2, then f attains an absolute maximum value f(x, y ) and an absolute minimum value f(x 2, y 2 ) at some points (x, y ) and (x 2, y 2 ) in. To find the absolute maximum and minimum values of a continuous function f on a closed, bounded set :. Find the values of f at the critical points of f in. 2. Find the extreme values of f on the boundary of.. The largest of the values from steps and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. The next theorem is stated for a function f of three variables but there is a similar theorem for a function of two variables (see Example 4 below). Theorem 42 (Method of Lagrange Multipliers) To find the maximum and minimum values of f(x, y, z) subject to the constraint g(x, y, z) k (assuming that these extreme values exist and g on the surface g(x, y, z) k):. Find all values of x, y, z, and λ such that f(x, y, z) λ g(x, y, z) and g(x, y, z) k. 2. Evaluate f at all the points (x, y, z) that result from step. The largest of these values is the maximum value of f; the smallest is the minimum value of f. Example 4 Find the extreme values of the function f(x, y) x 2 + 2y 2 on the circle x 2 + y 2.

10 We are asked for the extreme values of f subject to the constraint g(x, y) x 2 + y 2. Using Lagrange multipliers, we solve the equations f λ g, g(x, y), which can be written as or as f x λg x f y λg y g(x, y) 2x 2xλ (2) 4y 2yλ () x 2 + y 2. (4) From (2) we have x or λ. If x, then (4) gives y ±. If λ, then y from (), so then (4) gives x ±. Therefore, f has possible extreme values at the points (, ), (, ) (, ), and (, ). Evaluating f at these four points, we find that f(, ) 2 f(, ) 2 f(, ) f(, ). Therefore, the maximum value of f on the circle x 2 + y 2 is f(, ±) 2 and the minimum value is f(±, ). Example 44 Find the extreme values of f(x, y) x 2 + 2y 2 on the disk x 2 + y 2. We will compare the values of f at the critical points with values at the points on the boundary. Since f x 2x and f y 4y, the only critical point is (, ). We compare the value of f at that point with the extreme values on the boundary from Example 4: f(, ) f(±, ) f(, ±) 2. Therefore, the maximum value of f on the disk x 2 + y 2 is f(, ±) 2 and the minimum value is f(, ). We now start the second material for midterm 2 which concerns double integrals. For a positive, continuous function f(x, y) defined on a closed and bounded domain R 2, we denote by f(x, y) da, the volume under the graph of f(x, y) over. This volume for a rectangle R {(x, y) a x b, c y d} [a, b] [b, c] R 2 can be estimated by the following Midpoint Rule for ouble Integrals described in the next theorem. We also use this rule for defining the double integral when f(x, y) is not necessarily positive. Theorem 45 (Midpoint Rule for ouble Integrals) Let m, n be positive integers. Let x a < x < x 2 <... < x m b be a division of [a, b] into n intervals [x i, x i + ] of equal width x b a m. Similarly, let y c < y < y 2 <... < y n d be a division of [c, d] into m intervals [y j, y j+ ] of equal widths y d c n. Then: R f(x, y) da m i j n f(x i, y j ) A, where x i is the midpoint of [x i, x i ] and y j is the midpoint of [y j, y j ].

11 efinition 46 If f is a continuous function of two variables, then its average value on a domain R 2 is: f(x, y) da Area() da. efinition 47 The iterated integral of f(x, y) on a rectangle R [a, b] [c, d] is b d a c f(x, y) dy dx or d b c a f(x, y) dx dy. One calculates the integral b d a c f(x, y) dy dx by first calculating A(x) d c f(x, y) dy, holding x constant, and then calculating b a A(x) dx and similarly, for calculating the other integral. Example 48 Evaluate the iterated integral. x 2 y dy dx Regarding x as a constant, we obtain ] y2 ( ) ( ) x 2 y dy [x 2 y2 2 x 2 2 x y x2. Thus, the function A in the preceding discussion is given by A(x) 2 x2 in this example. We now integrate this function of x from to : x 2 y dy dx [ ] x 2 y dy Example 49 Evaluate the iterated integral. dx x 2 y dx dy. Here we first integrate with respect to x: x 2 y dx dy [ ] x 2 y dx dy [ ] x x y x ] 2 x2 dx x dy ] 2 9y dy 9 y Theorem 5 (Fubini s Theorem) If f is continuous on the rectangle R {(x, y) a x b, c y d}, then R f(x, y) da b d a c f(x, y) dy dx d b c a f(x, y) dx dy. More generally, this is true if we assume that f is bounded on R, f is discontinuous only on a finite number of smooth curves, and the iterated integrals exist. Example 5 Evaluate the double integral R (x y2 ) da, where R {(x, y) x 2, y 2}.

12 2 Fubini s Theorem gives R (x y 2 ) da (x y 2 ) dy dx (x 7) dx x2 2 7x ] 2 2. [ xy y ] y2 y dx Example 52 Find the volume of the solid S that is bounded by the elliptic paraboloid x 2 + 2y 2 + z 6, the planes x 2 and y 2, and the three coordinate planes. We first observe that S is the solid that lies under the surface z 6 x 2 y 2 and above the square R [, 2] [, 2]. We are now in a position to evaluate the double integral using Fubini s Theorem. Therefore V R (6 x 2 2y 2 ) da [6x x 2y 2 x] x2 x dy ( 88 y 4y2 ) dy (6 x 2 2y 2 ) dx dy [ 88 4 ] 2 y 48. In general, for any continuous function f(x, y) on a closed and bounded domain R 2, the integral f(x, y) da is defined and it is equal to the area under the graph of f(x, y) on when the function is positive. There are two cases for, called type I and types II, where the integral f(x, y) da can be calculated in a straightforward manner. efinition 5 A plane region is said to be of type I, if it can be expressed as where g (x) and g 2 (x) are continuous. {(x, y) a x b, g (x) y g 2 (x)}, efinition 54 A plane region is said to be of type II, if it can be expressed as where h and h 2 are continuous. {(x, y) c y d, h (y) x h 2 (y)}, Theorem 55 If f is continuous on a type I region such that {(x, y) a x b, g (x) y g 2 (x)}, then f(x, y) da b g2 (x) a g (x) f(x, y) dy dx.

13 Theorem 56 f(x, y) da d h2 (y) c h (y) where is a type II region given by efinition 54. f(x, y) dx dy. Example 57 Evaluate (x + 2y) da, where is the region bounded by the parabolas y 2x 2 and y + x 2. The parabolas intersect when 2x 2 + x 2, that is x 2, so x ±. We note that the region, is a type I region but not a type II region and we can write {(x, y) x, 2x 2 y + x 2 }. Since the lower boundary is y 2x 2 and the upper boundary is y + x 2, efinition 5 gives (x + 2y) da +x 2 2x 2 (x + 2y) dy dx ( x 4 x + 2x 2 + x + ) dx x5 5 x x + x2 2 + x [ xy + y 2 ] y+x y2x 2 Example 58 Find the volume of the solid that lies under the paraboloid z x 2 + y 2 and above the region in the xy-plane bounded by the line y 2x and the parabola y x 2. We see that is a type I region and {(x, y) x 2, x 2 y 2x}. dx Therefore, the volume under z x 2 + y 2 and above is x V (x 2 + y 2 ) da (x 2 + y 2 ) dy dx ] y2x [x 2 y + y dx yx 2 ) ( x6 x4 + 4x x 2 [x 2 (2x) + (2x) x 2 x 2 (x2 ) ] ] 2 dx x7 2 x x Solution 2: We see that can also be written as a type II region: {(x, y) y 4, Therefore, another expression for V is V (x 2 + y 2 ) da 2 y x y}. 4 y (x 2 + y 2 ) dx dy 2 y dx

14 4 4 [ x + y2 x ] x y x 2 y dy y y y4 ( ) y 2 + y 5 y 2 24 y dy 2 ] Example 59 Evaluate the iterated integral x sin(y2 ) dy dx. If we try to evaluate the integral as it stands, we are faced with the task of first evaluating sin(y 2 ) dy. But it s impossible to do so in finite terms since sin(y 2 ) dy is not an elementary function. So we must change the order of integration. This is accomplished by first expressing the given iterated integral as a double integral. We have where x sin(y 2 ) dy dx sin(y 2 ) da, {(x, x, x y )}. We see that an alternative description of is {(x, y) y, x y}. This enables us to express the double integral as an iterated integral in the reverse order: x sin(y 2 ) dy dx sin(y 2 ) da y y sin(y 2 ) dy 2 cos(y2 ) sin(y 2 ) dx dy ] ( cos ). 2 [ x sin(y 2 ) ] xy x dy

Exam 2 Summary. 1. The domain of a function is the set of all possible inputes of the function and the range is the set of all outputs.

Exam 2 Summary. 1. The domain of a function is the set of all possible inputes of the function and the range is the set of all outputs. Exam 2 Summary Disclaimer: The exam 2 covers lectures 9-15, inclusive. This is mostly about limits, continuity and differentiation of functions of 2 and 3 variables, and some applications. The complete