Multivariable Calculus

1. Multivariable Calculus This presentation is not complete. So if viewing this, remember to keep checking this presentation for updates. Chapter 17

2. FUNCTION OF TWO OR MORE VARIABLES The expression z = f(x, y) is a function of two variables if a unique value of z is obtained from each ordered pair of real numbers (x, y). The variables x and y are independent variables, and z is the dependent variable. The set of all ordered pairs of real numbers (x,y) such that f(x, y) exists is the domain of f; the set of all values of f(x, y) is the range. Similar definitions could be given for functions of three, four , or more independent variables. PARTIAL DERIVATIVES (INFORMAL DEFINITION) The partial derivative of fwith respect to x is the derivative of f obtained by treating x as a variable and y as a constant. The partial derivative of fwith respect to y is the derivative of f obtained by treating y as a variable and x as a constant.

3. PARTIAL DERIVATIVES (FORMAL DEFINITION) Let z = f(x, y) be a function of two independent variables. Let all indicated limits exist. Then the partial of fwith respect to x is And the partial derivative of fwith respect to y is If the indicated limits do not exist, then the partials derivatives do not exist. Notice these definitions are similar to those on the first test.

4. SECOND – ORDER PARTIALS DERIVATIVES For a function z = f (x, y), if the indicated partial derivatives exist then This is read as the second partial of z with respect to x the second time. The two’s are not exponents. This is read as the second partial of z with respect to x with respect to y. This is read as the second partial of z with respect to y with respect to x. This is read as the second partial of z with respect to y the second time. Notice when using the ∂ symbol it is read and found from right to left. However when using the subscript notation it is found and read from left to right.

5. Finding partials is very similar to doing implicit differentiation: zy contains all the terms that had dy/dx and zx contains all the terms that did not have dy/dx. Each variable is considered a constant when working with another variable. Find all the second partials for z = f(x, y) = 5x 3y 2– 4x 2y 4 + 7y 3– 8x + y – 10. First we must find the first partials of which there are two. zx = 15x 2y 2– 8xy 4 – 8 zy = 10x 3y – 16x 2y 3 + 21y 2 + 1 Study problems 21 – 24 and 33 – 36 page 989. Just practice finding all of the second partials. Next find all of the second partialsof which there are four. zxx = 30xy 2– 8y 4 zxy = 30x 2y– 32xy 3 zyx = 30x 2y– 32xy 3 z yy= 10x 3 – 48x 2y 2 + 42y Notice the zxy and zyx answers are the same. They should always be the same for any problems we work. Do not just copy the second answer. Find the answer and compare in case you made a mistake in finding the first one.

6. MAXIMA AND MINIMA RELATIVE MAXIMA AND MINIMA Let (a, b) be the center of a circular region contained in the xy-plane. Then, for a functionz = f(x, y) defined for every (x, y) in the region, f(a,b) is a relativemaximum if f(a,b) ≥f (x, y) for all points ( x, y ) in the circular region, and f ( a, b ) is a relative minimum if f(a,b) ≤f(x, y) for all points (x, y) in the circular region. LOCATION OF EXTREMA Let a functionz = f(x, y) have a relative maximum or relative minimum at the point (a,b). Let fx(a,b) and fy(a,b) both exist. Then fx(a,b) = 0 and fy(a,b) = 0.

7. TEST FOR RELATIVE EXTREMA For a functionz = f(x, y), let fxx, and fyy, and fxy all exist in a circular region contained in the xy-plane with center (a,b). Further, let fx(a,b) = 0 and fy(a,b) = 0. Define the number D by D = fxx(a,b) · fyy (a,b) – [fxy(a,b)]2. Then a. f(a,b) is a relative maximum if D > 0 and fxx(a,b) < 0; b. f(a,b) is a relative minimum if D > 0 and fxx(a,b) > 0; c. f(a,b) is a saddle point (neither a maximum nor a minimum,) if D < 0 d. If D = 0, the test gives no information. This test is similar to the second derivative test used in test three.

8. STEPS FOR FINDING RELATIVE EXTREMA Step 1. Find the first partialsfx(x, y) and fy(x, y). Step 2. Find the critical points (a,b) by setting the first partials equal to zero and solving the system of equations by either elimination or substitution. Step 3. Find the second partials fxx(x, y), f xy(x, y), and fyy(x, y). Remember fxy(x, y) is the same answer as fyx(x, y). Step 4. Evaluate the second partials with the critical point (a,b): fxx(a,b),fxy(a,b), and fyy(a,b). Step 5. Determine the value of D:D = fxx(a,b) · fyy(a,b) – [fxy(a,b)]2. Step 6. Identify the extrema using criteria on the previous slide. Remember to write the answer correctly. Study problems 1 – 16 page 1001

9. Find all points where the functions have any relative extrema. Identify any saddle points. Example 1: f(x,y) = 5xy– 7x2– y2 + 3x– 6y – 4 We are solving this system of equations by elimination. This is a procedure you were supposed to have learned in an algebra class. The equations were rearranged using rules of algebra. Step 1. fx(x,y) = 5y– 14x + 3 fy(x,y) = 5x – 2y – 6 Step 2. fx(x,y) = 5y– 14x + 3 = 0 fy(x,y) = 5x – 2y – 6 = 0 You will not necessarily be using 2 and –5. 14x– 5y = 3 multiply by 2 5x – 2y = 6 multiply by –5 Then substituting into either equation 5(–8) – 2y = 6 –40 – 6 = 2y–46 = 2y –23 = y 28x– 10y = 6 – 25x + 10y = – 30 3x = – 24x = – 8 Critical Point: (–8,–23)

10. Step 3. fxx(x,y) = –14fxy(x,y) = 5fyy(x,y) =–2 Step 4. fxx(–8,–23) = –14fxy(– 8,–23) = 5fyy(– 8,–23) = –2 Replace all of the x’s with – 8 and all of they’s with – 23 in step 3 and evaluate. Luckily there is nothing to evaluate. Step 5. D =fxx(–8,–23)·fyy(–8,–23)–[fxy(–8,–23)]2D = ( – 14 )( – 2 ) – [ 5 ] 2D = 3 Step 6. Because D > 0 in step 5 andfxx(–8,–23) < 0 in step 4, the critical point found in step 2 is a relative maximum. Answer: (–8,–23) is a relative maximum

11. Example 2: to be done later

12. DOUBLE INTEGRAL The double integral of f(x,y) over a rectangular region R is written And equals either VOLUME Letz = f(x,y) be a function that is never negative on the rectangular region R defined by c≤ x ≤ d, a ≤ y ≤ b. The volume of the solid under the graph of fand over the region R is

13. Step 1. Evaluate the inside integral first. The dy means to integrate the y variable only. Notice sign change.

14. Step 2. Substitute this result into the original problem. Step 3. Evaluate the remaining integral.