Chapter 4: Functions of Two Variables

Transcription

1 Chapter 4 Functions o Two Variables Applied Calculus 39 Chapter 4: Functions o Two Variables PreCalculus Idea -- Topographical Maps I ou ve ever hiked, ou have probabl seen a topographical map. Here is part o a topographic map o Stowe, Vermont, USA (courtes o United States Geological Surve and Points with the same elevation are connected with curves, so ou can read not onl our eastwest and our north-south location, but also our elevation. You ma have also seen weather maps that use the same principle points with the same temperature are connected with curves (isotherms), or points with the same atmospheric pressure are connected with curves (isobars). These maps let ou read not onl a place s location but also its temperature or atmospheric pressure. In this chapter, we ll use that same idea to make graphs o unctions o two variables. This chapter is (c) 03. It was remied b David Lippman rom Shana Calawa's remi o Contemporar Calculus b Dale Homan. It is licensed under the Creative Commons Attribution license.

2 Chapter 4 Functions o Two Variables Applied Calculus 40 Section : Functions o Two Variables Real lie is rarel as simple as one input one output. Man relationships depend on lots o variables. Eamples: I I put a deposit into an interest-bearing account and let it sit, the amount I have at the end o 3 ears depends on P (how much m initial deposit is), r (the annual interest rate), and n (the number o compoundings per ear). The air resistance on a wing in a wind tunnel depends on the shape o the wing, the speed o the wind, the wing s orientation (pitch, aw, and roll), plus a mriad o other things that I can t begin to describe. The amount o our television cable bill depends on which basic rate structure ou have chosen and how man pa-per-view movies ou ordered. Since the real world is so complicated, we want to etend our calculus ideas to unctions o several variables. Functions o Two Variables I,, 3,..., n are real numbers, then,, 3,..., n is called an n-tuple. This is an etension o ordered pairs and triples. A unction o n variables is a unction whose domain is some set o n-tuples and whose range is some set o real numbers. For much o what we do here, everthing would work the same i we were working with, 3, or 47 variables. Because we re tring to keep things a little bit simple, we ll concentrate on unctions o two variables. A Function o Two Variables A unction o two variables is a unction that is, to each input is associated eactl one output. The inputs are ordered pairs, (, ). The outputs are real numbers. The domain o a unction is the set o all possible inputs (ordered pairs); the range is the set o all possible outputs (real numbers). The unction can be written z = (,). Functions o two variables can be described numericall (a table), graphicall, algebraicall (a ormula), or in English. We will oten now call the amiliar = () a unction o one variable. Eample The cost o renting a car depends on how man das ou keep it and how ar ou drive. Represent this using a unction. Let d = the number o das ou rent the car, and m = the number o miles ou drive. Then the cost o the car rental C(d, m) is a unction o two variables. This chapter is (c) 03. It was remied b David Lippman rom Shana Calawa's remi o Contemporar Calculus b Dale Homan. It is licensed under the Creative Commons Attribution license.

3 Chapter 4 Functions o Two Variables Applied Calculus 4 Eample The demand or hot dog buns depends on the price or the hot dog buns and also on the price or hot dogs. Represent this as a unction. The demand q p, p B B D is a unction o two variables. (The demand or hot dogs also depends on the price o both dogs and buns). Formulas and Tables Just as in the case o unctions o one variable, we can displa a unction o two variables in a table. The two inputs are shown in the margin (top row, let column), and the outputs are shown in the interior cells. Eample 3 Here is a table that shows the cost C(d, m) in dollars or renting a car or d das and driving it m miles: d m a. What is the cost o renting a car or 3 das and driving it 00 miles? b. What is C(00, 4)? What is C(4, 00)? c. Suppose we rent the car or 3 das. Is C an increasing unction o miles? a. According to the table, renting the car or 3 das (row with d = 3) and driving it 00 miles (column with m = 00) will cost $50 (highlighted in aqua). b. Careul now the input is an ordered pair, so in C(00, 4), the 00 has to be a value o d and the 4 has to be a value o m. C(00, 4) would be the cost o renting a car or 00 das and driving it 4 miles. That cost is not in the table. (And that would be a prett sill wa to rent a car.) On the other hand, C(4, 00) is the cost o renting or 4 das and driving 00 miles the table sas that would cost $75. c. I we know that d is ied at 3, we re looking at C(3, m). This is now a unction o variable, just m. We can see the table that displas values o this unction b ocusing our attention on just the row where d = 3: d m Now we can see that i we rent or 3 das, the cost appears to be an increasing unction o the number o miles we drive, which shouldn't be surprising.

4 Chapter 4 Functions o Two Variables Applied Calculus 4 The idea o iing one variable and watching what happens to the unction as the other varies will come up again and again. It s hard to displa a unction o more than two variables in a table. But it s convenient to work with ormulas or unctions o two variables, or as man variables as ou like. Eample 4 The cost C(d,m) in dollars or renting a car or d das and driving it m miles is given b the ormula C( d, m) 40d. 5m a. What is the cost o renting a car or 3 das and driving it 00 miles? b. What is C(00, 4)? What is C(4, 00)? c. Suppose we rent the car or 3 das. Is C an increasing unction o miles? a. C 3, $ 50. This is the same value we got rom the table. The ormula will give us the same answers or an o the table values. b. C(00, 4) makes perect sense to the ormula (even i it doesn t make sense or actuall renting a car). So now we can get an answer. To rent the car or 00 das and drive it or 4 miles should cost $ C(4, 00) = $75, as beore. c. I we i d = 3, then C(d, m) becomes C(3, m) = 40(3) +.5m = 0 +.5m. Yes, this is an increasing unction o m; I can tell because it s linear and its slope is.5 > 0. Realit check the ormula that gives the cost or the rental car makes sense or all values o d and m. But that s not how the real cost works ou can t rent the car or a negative number o das or drive a negative number o miles. (That is, there are domain restrictions.) In addition, most car rental agreements don t compute a charge or ractions o das; the round up to the net whole number o das. Eample 5 Let,, z, w 35 w z. Evaluate 0,,,3 z. Remember that this is an ordered 4-tuple; make sure the numbers get substituted into the correct places: 0,,, Graphs The graph o a unction o two variables is a surace in three-dimensional space. Let's start b looking at the 3 dimensional rectangular coordinate sstem, how to locate points in three dimensions, and distance between points in three dimensions.

5 Chapter 4 Functions o Two Variables Applied Calculus 43 In the dimensional rectangular coordinate sstem we have two coordinate aes that meet at right angles at the origin, and it takes two numbers, an ordered pair (, ), to speci the rectangular coordinate location o a point in the plane ( dimensions). Each ordered pair (, ) speciies the location o eactl one point, and the location o each point is given b eactl one ordered pair (, ). The and values are the coordinates o the point (, ). The situation in three dimensions is ver similar. In the 3 dimensional rectangular coordinate sstem we have three coordinate aes that meet at right angles, and three numbers, an ordered triple (,, z), are needed to speci the location o a point. Each ordered triple (,, z) speciies the location o eactl one point, and the location o each point is given b eactl one ordered triple (,, z). The, and z values are the coordinates o the point (,, z). The igure below shows the location o the point (4,, 3). Tpicall we use a right-hand orientation. To see what this means, imagine our right hand in ront hand in ront o ou with the palm toward our ace, our thumb pointing up, ou inde inger straight out, and our net inger toward our ace (and the two bottom ingers bent into the palm. Then, in the right hand coordinate sstem, our thumb points along the positive z-ais, our inde inger along the positive -ais, and the other inger along the positive -ais. Other orientations o the aes are possible and valid (with appropriate labeling), but the right hand sstem is the most common orientation and is the one we will generall use. Each ordered triple (,, z) speciies the location o a single point, and this location point can be plotted b locating the point (,, 0) on the plane and then going up z units. (We could also get to the same (,, z) point b inding the point (, 0, z) on the z plane and then going units parallel to the ais, or b inding (0,, z) on the z plane and then going units parallel to the ais.)

6 Chapter 4 Functions o Two Variables Applied Calculus 44 Eample 6 Plot the locations o the points : P = (0, 3, 4), Q = (, 0, 4), R = (, 4, 0), S = (3,, ), and T(,, ). The points are shown to the right. Once we can locate points, we can begin to consider the graphs o various collections o points. B the graph o "z = " we mean the collection o all points (,, z) which have the orm "(,, )". Since no condition is imposed on the and variables, the take all possible values. The graph o z = is a plane parallel to the plane and units above the plane. Similarl, the graph o = 3 is a plane parallel to the z plane, and = 4 is a plane parallel to the z plane. (Note: The planes have been drawn as rectangles, but the actuall etend ininitel ar.) Distance Between Points In two dimensions we can think o the distance between points as the length o the hpothenuse o a right triangle, and that leads to the Pthagorean ormula: distance = +.

7 Chapter 4 Functions o Two Variables Applied Calculus 45 In three dimensions we can also think o the distance between points as the length o the hpothenuse o a right triangle, but in this situation the calculations appear more complicated. Fortunatel, the are straightorward: distance = base + height = ( + ) + z = + + z so distance = + + z. I P = (,, z ) and Q = (,, z ) are points in space, then the distance between P and Q is distance = + + z = ( ) + ( ) + (z z ). The 3 dimensional pattern is ver similar to the dimensional pattern with the additional piece z. Eample 7 Find the distances between points A = (,, 3) and B = (7, 5, 3) Dist(A, B) = ( 6) = = 8 = 9. In two dimensions, the set o points at a ied distance rom a given point is a circle, and we used the distance ormula to determine equations describing circles: the circle with center (, 3) and radius 5 is given b ( ) + ( 3) = 5 or =. The same ideas work or spheres in three dimensions.

8 Chapter 4 Functions o Two Variables Applied Calculus 46 Spheres The set o points (,, z) at a ied distance r rom a point (a, b, c) is a sphere with center (a, b, c) and radius r. The sphere is given b the equation ( a) + ( b) + (z c) = r. Eample 8 Write the equations o a sphere with center (, 3, 4) and radius 3 ( ) + (+3) + (z 4) = 3. Now suppose that we want to graph a surace. We can think o each input (,) as a location on the plane, and plot the point (,) units above that point. Graphing that can be challenging. We have a ew options:. Use a anc computer program to draw beautiul perspective drawings.. Tr to draw a perspective drawing b hand. This is ver challenging, and usuall not worth the eort. 3. Use level curves to draw contour diagrams, which is the approach we'll ocus on here. A contour diagram is like a topographical map points with the same elevation (outputs) are connected with curves. Each particular output is called a level, and these curves are called level curves or contours. The closer the curves are to each other, the steeper that section o the surace is. Topographical maps give hikers inormation about elevation, steep and shallow grades, peaks and valles. Contour diagrams give us the same kind o inormation about a unction.

9 Chapter 4 Functions o Two Variables Applied Calculus 47 This is a contour diagram o the same surace shown in above. The level curves are graphs in the -plane o curves (, ) = c or various constants c. Each o the squares corresponds to one o the bumps on the surace. I the contours are positive, as highlighted below, the bump is above the -plane. I the contours are negative, the bump etends below the -plane. Everwhere on the crisscrossed pattern o diagonal lines, the height o the surace is 0, so the surace is on the -plane. This is a eature that I couldn t see when I looked at the perspective drawing.

10 Chapter 4 Functions o Two Variables Applied Calculus 48 To better understand contour diagrams, suppose we had a table o elevation data. We could graph this b plotting the height at each point and connecting the dots with smooth curves, which would result in the something like the graph shown. I we "slice" the surace above with the plane z = 8, the points where the plane cuts the surace are those points where the elevation o the surace is 8 units above the plane. The igure below shows the surace being sliced b the planes z = 8 and z = 4. Slicing the surace at dierent elevations and sketching the curves where the plane intersects the surace results in the second graph below.

11 Chapter 4 Functions o Two Variables Applied Calculus 49 I we move all o those curves to the plane (or, equivalentl, view them rom directl overhead), the result is a dimensional graph o the level curves o the original surace. This is the contour diagram. Eample 9 Create a contour diagram or our car rental eample with cost unction Draw curves or when the cost is 0, 00, 00, 300, and 400. C( d, m) 40d. 5m. Set will set C( d, m) 40d. 5m c or c = 0, 00, 00, 300, and 400 and draw the curves in the dm-plane. The irst coordinate o the ordered pair is d, so the d-ais will be horizontal; the m-ais will be vertical. Remember that the domain or this unction is reall just where d 0 and m 0, so we will onl draw the curves in the irst quadrant. When c = 0: C( d, m) 40d.5m 0.5m 40d 40 m d 67 d.5 This is the equation o a line, with slope about 67, passing through the origin. Because o the domain restrictions, the curve I will draw or this level is simpl the origin. Putting this back into the car rental contet, the onl point where I pa $0 or renting the car is when I rent the car or 0 das and drive it 0 miles that is, i I don t rent it at all. When c = 00: C( d, m) 40d.5m 00.5m 40d 00 m 40.5 d d 667 This is the equation o a line, with slope about 67, and d-intercept o about 667. This section o this line that lies in the irst quadrant is shown with 00 labeling it.

12 Chapter 4 Functions o Two Variables Applied Calculus 50 Putting this into contet, an point on that line represents a (d, m) combination o das and miles that will make the cost eactl $00. So, or eample i I rent the car or 0 das and drive it 667 miles, it will cost me $00. I I rent the car or.5 das and don t drive an miles, it will cost me $00. We continue or c = 00, 300, and 400 and sketch the curves in the plane, resulting in the contour diagram shown to the right. Eample 0 The contour diagram or the cost C(d,m) in dollars or renting a car or d das and driving it m miles is shown in the previous eample. Use the diagram to answer the ollowing questions. a. What is the cost o renting a car or 3 das and driving it 00 miles? b. What is C(00, 4)? What is C(4, 00)? c. Suppose we rent the car or 3 das. Is C an increasing unction o miles? a. The point (3, 00) is between contours on this graph, so I can t get an eact answer or C(3, 00). (But it s tpical or a graph that we would have to estimate). It looks to me as i (3, 00) is halwa between the 00 and the 00 contours, so I will estimate that C(3, 00) is about $50. Estimates rom the graph are necessaril ver rough. The graph onl shows a little inormation (in this wa, a contour diagram is like a table), so I have to etrapolate in between. But or most graphs, I don t actuall know what happens between the contours. All I know or sure is that the output at (3, 00) is between the two levels I see. For this car rental eample, I also know a ormula, and m table showed this particular input, so I have other was to get a better answer. b. I can t ind (00, 4) on this diagram, so I can t make an estimate o C(00, 4) rom this graph. (4, 00) lies between the contours or 00 and 00. It looks closer to 00, so I ll estimate that C(4, 00) is about $80.

13 Chapter 4 Functions o Two Variables Applied Calculus 5 c. I we i d = 3, we get a vertical line. What happens as m increases on this vertical line? As m increases, the unction values shown on the contours increase C appears to be an increasing unction o miles. Eample Here is a contour diagram or a unction g(,). Use the diagram to answer the ollowing questions: a. What is g(3, 5)? b. What is the highest point shown on the diagram? What is the lowest point shown? c. I ou start at (3, 5) and head in the positive direction, do ou go uphill or downhill irst? a. g(3, 5) is 0.6. I can tell because the point is right on one o the contours.

14 Chapter 4 Functions o Two Variables Applied Calculus 5 b. The highest contour shown is 0.9, and there would be a contour or.0 i the surace had ever got that high. However, the height seems to be increasing as we move in toward the center, so I m guessing that the g gets to nearl in the center. The lowest contour is 0.. But again, I will guess that the height continues to decrease, so I think g is nearl 0 around the outside. c. Starting at the point (3, 5, 0.6) on the surace and traveling to the right along the horizontal line shown in Fig. 9, ou would cross the contour or 0.7 net. So the unction increases irst (we go uphill), and then decreases again. Note one more time we don t reall know what happens between the contours. All we can do is estimate rom the inormation in the graph. Eample Here is a contour diagram or a unction F(,). a) Describe the shape o the surace. b) Suppose ou travel along the surace in the positive -direction, starting on the surace at the point above (or below) the point (, ) = (-, ). Describe our journe. a) The surace is bump, with regularl spaced oval bumps. Notice that some o the bumps go up (positive contours), but others go down. Between the bumps, there are horizontal lines that are completel level, with an elevation o 0. b) It looks as i F(-,) is about 3. As I head in the positive -direction along the line shown below, I irst go uphill, nearl to 4, then I start going downhill. As I keep going north, I keep descending, going into the dip, until nearl -4. I m starting to go uphill again just as I leave the graph.

15 Chapter 4 Functions o Two Variables Applied Calculus 53 What happens i ou have a unction o more than two variables? Its graph will be a hpersurace. For eample, the graph o a unction o our variables will be a hper-surace in 5- dimensional space. This is hard (impossible or most o us) to visualize. Even the contours are hard to visualize instead o curves in the plane, the re hper-suraces in 4-dimensional space. So i ou have more than two variables, the graph isn t usuall ver useul. Functions o Two Real-Lie Variables Complementar goods and substitute goods The demand or some pairs o goods have a relationship, where the quantit demanded or one product depends somehow on the prices or both. Two goods are complementar i an increase in the price o either decreases the demand or both. Eamples: The demand or cars depends on both the price or cars and the price o gasoline. The demand or hot dog buns depends on both the price or the buns and the price or the hot dogs. Two goods are substitutes i an increase in the price o one increases the demand or the other. Eample: The demand or Brand A depends on its price and also on the price o its main competitor Brand B. I the Brand B raises its price, consumers will switch brands substitute and demand or Brand A will increase. Think brands o sot drinks, detergent, or paper towels. A traditional eample is coee and tea the idea is that consumers are simpl looking or a hot drink and the ll bu whatever is cheaper.

16 Chapter 4 Functions o Two Variables Applied Calculus 54 But this has alwas seemed ish to me I ve never met an coee- or tea-drinkers who would happil switch. These demand unctions are unctions o two variables. Eample 3 The demand unctions or two products are given below. p, p, q, and q are the prices (in dollars) and quantities or products and. q 00 3p p q 50 p p Are these two products complementar goods or substitute goods? What is the quantit demanded or each when the price or product is $0 per item and the price or product is $30 per item? These products are complementar an increase in either price decreases both demands. You can see that because the coeicients are both negative in each demand unction. When p = 0 and p = 30, we have q q 50 0 units are demanded or product and 70 units are demanded or product when the price or product is $0 per item and the price or product is $30 per item Cobb-Douglas Production unction Production unctions are used to model the total output o a irm or a variet o inputs (doesn t this sound like a unction o several variables?). One eample is a Cobb-Douglas Production unction: P AL K In this unction, P is the total production, A is a constant, α and β are constants between 0 and, L is the labor orce, and K is the capital ependiture. (And the units must be massaged well.) You can read more about Cobb-Douglas Production unctions at You can read about other kinds o production unctions at

17 Chapter 4 Functions o Two Variables Applied Calculus Eercises. F,. Find 0,4 a) F b) F 4,0 c) F,4 d) F 4, e) F 800,800 ) F, g) F,. g s, t st. Find a) g,9 b) g 9, c) g,t d) g s,9 e) g w, z 3. Let,, z, w z. Evaluate,,3,4 zw. 4. Let,, z, w w 0 z. Evaluate,,3,4 5. Here is a table showing the unction A t, r. t r a) Find A,.05 b) Find A.05,. A t,.06 an increasing or decreasing unction o t? c) Is d) Is A 3,r an increasing or decreasing unction o r? 6. Here is a table showing values or the unction t h H, t h a) Is H t,50 an increasing or decreasing unction o t? b) Is H 4, h an increasing or decreasing unction o h? c) Fill in the blanks: The maimum value shown on this table is, d) Fill in the blanks: The minimum value shown on this table is, H. H.

18 Chapter 4 Functions o Two Variables Applied Calculus 56 In problems 7 0, plot the given points. 7. A = (0,3,4), B = (,4,0), C = (,3,4), D = (, 4,) 8. E = (4,3,0), F = (3,0,), G = (0,4,), H = (3,3,) 9. P = (,3, 4), Q = (,,3), R = (4,, ), S = (,,3) 0. T = (,3, 4), U = (,0, 3), V = (,0,0), W = ( 3,, ) In problems 4, calculate the distances between the given points. A = (5,3,4), B = (3,4,4). A = (6,,), B = (3,,) 3. A = (3,4,), B = (,6, ) 4. A = (,5,0), B = (,3,) In problems 5 8, graph the given planes. 5. = and z = 6. = 4 and = 7. = and = 0 8. = and z = 0 In problems 9, the center and radius o a sphere are given. Find an equation or the sphere. 9. Center = (4, 3, 5), radius = 3 0. Center = (0, 3, 6), radius =. Center = (5,, 0), radius = 5. Center = (,, 3), radius = 4 In problems 3 4, the equation o a sphere is given. Find the center and radius o the sphere. 3. ( 3) + (+4) + (z ) = 6 4. (+) + + (z 4) = 5 For problems 5 through 30. Match the contour diagram to the computer-generated, perspective drawing (a through ) it matches. Briel eplain our answer.

19 Chapter 4 Functions o Two Variables Applied Calculus a. b. c. d. e.. For problems 3 through 36. Match the contour diagram to the equation (a through ) it matches. Briel eplain our answer.

20 Chapter 4 Functions o Two Variables Applied Calculus a., b., c., 5 d., 5 e., 0.0.,

21 Chapter 4 Functions o Two Variables Applied Calculus The contour diagram shown is or a unction M(, ). Use the diagram to answer the ollowing: a) Estimate M(, 3) b) Estimate M(3, ) c) Is M(, 3) an increasing or decreasing unction o? d) Is M(3, ) an increasing or decreasing unction o? e) Find a value o c so that M(c, ) is a constant unction o. 38. The contour diagram shown is or a unction G(, ). Use the diagram to answer the ollowing: a) Estimate G(, 3) b) Suppose ou travel north (in the direction o increasing ) along the surace, starting above (, 3). Describe our journe. c) Suppose ou travel east (in the direction o increasing ) along the surace, starting above (, 3). Describe our journe.

22 Chapter 4 Functions o Two Variables Applied Calculus The demand unctions or two products are given below. p, p, q,and q are the prices (in dollars) and quantities or products and. Are these two products complementar goods or substitute goods q 00 3p p q 50 p p 40. The demand unctions or two products are given below. p, p, q,and q are the prices (in dollars) and quantities or products and. Are these two products complementar goods or substitute goods q 350 p p q 5 p p 4. Consider the Cobb-Douglas Production unction: P( L, K) L K o production when 9 units o labor and units o capital are invested. 4. Consider the Cobb-Douglas Production unction: P( L, K) 6L K production when 4 units o labor and 8 units o capital are invested.. Find the total units. Find the total units o

23 Chapter 4 Functions o Two Variables Applied Calculus 6 Section : Calculus o Functions o Two Variables Now that ou have some amiliarit with unctions o two variables, it s time to start appling calculus to help us solve problems with them. In Chapter, we learned about the derivative or unctions o two variables. Derivatives told us about the shape o the unction, and let us ind local ma and min we want to be able to do the same thing with a unction o two variables. First let s think. Imagine a surace, the graph o a unction o two variables. Imagine that the surace is smooth and has some hills and some valles. Concentrate on one point on our surace. What do we want the derivative to tell us? It ought to tell us how quickl the height o the surace changes as we move. Wait, which direction do we want to move? This is the reason that derivatives are more complicated or unctions o several variables there are so man directions we could move rom an point. It turns out that our idea o iing one variable and watching what happens to the unction as the other changes is the ke to etending the idea o derivatives to more than one variable. Partial Derivatives Partial Derivatives: Suppose that z = (, ) is a unction o two variables. The partial derivative o with respect to is the derivative o the unction (,) where we think o as the onl variable and act as i is a constant. The partial derivative o with respect to is the derivative o the unction (,) where we think o as the onl variable and act as i is a constant. The with respect to or with respect to part is reall important ou have to know and tell which variable ou are thinking o as THE variable. Geometricall the partial derivative with respect to gives the slope o the curve as ou travel along a cross-section, a curve on the surace parallel to the -ais. The partial derivative with respect to gives the slope o the cross-section parallel to the -ais. Notation or the Partial Derivative: The partial derivative o = () with respect to is written as,, or z simpl z The Leibniz notation is, or d d z We use an adaptation o the notation to mean ind the partial derivative o (,) with d respect to :, This chapter is (c) 03. It was remied b David Lippman rom Shana Calawa's remi o Contemporar Calculus b Dale Homan. It is licensed under the Creative Commons Attribution license.

24 Chapter 4 Functions o Two Variables Applied Calculus 6 To estimate a partial derivative rom a table or contour diagram: The partial derivative with respect to can be approimated b looking at an average rate o change, or the slope o a secant line, over a ver tin interval in the -direction (holding constant). The tinier the interval, the closer this is to the true partial derivative. To compute a partial derivative rom a ormula: I (,) is given as a ormula, ou can ind the partial derivative with respect to algebraicall b taking the ordinar derivative thinking o as the onl variable (holding ied). O course, everthing here works the same wa i we re tring to ind the partial derivative with respect to just think o as our onl variable and act as i is constant. The idea o a partial derivative works perectl well or a unction o several variables ou ocus on one variable to be THE variable and act as i all the other variables are constants. Eample Here is a contour diagram or a unction g(,). Use the diagram to answer the ollowing questions: 3,5 g 3,5 a. Estimate g and b. Where on this diagram is g greatest? Where is g greatest? g means we're thinking o as the onl variable, so we ll hold ied at = 5. That means we ll be looking along the horizontal line = 5. To estimate g, we need two unction values. (3, 5) lies on the contour line, so we know that g(3, 5) = 0.6. The net point as we move to the right is g(4.,5) = 0.7. a. 3,5 Now we can ind the average rate o change: g Average rate o change = (change in output) / (change in input) We can do the same thing b going to the net point we can read to the let, which is g(.4,5) = g Then the average rate o change is g 3,5 given the inormation we have, or ou could Either o these would be a ine estimate o take their average. We can estimate that 3,5. 5 g.

25 Chapter 4 Functions o Two Variables Applied Calculus 63 Estimate 3,5 g the same wa, but moving on the vertical line. Using the net point up, we g get the average rate o change. 5. Using the net point down, we get g Taking their average, we estimate g 3, b. g means is m onl variable, and we're thinking o as a constant. So we're thinking about moving across the diagram on horizontal lines. g will be greatest when the contour lines g are closest together, when the surace is steepest then the denominator in will be small, so g will be big. Scanning the graph, we can see that the contour lines are closest together when we head to the let or to the right rom about (0.5, 8) and (9, 8). So 8) and (9, 8). For g, I want to look at vertical lines. ). g is greatest at about (0.5, g is greatest at about (5, 3.8) and (5, Eample Cold temperatures eel colder when the wind is blowing. Windchill is the perceived temperature, and it depends on both the actual temperature and the wind speed a unction o two variables! You can read more about windchill at Below is a table (courtes o the National Weather Service) that shows the perceived temperature or various temperatures and windspeeds. Note that the also include the ormula, but or this eample we'll use the inormation in the table. a. What is the perceived temperature when the actual temperature is 5 F and the wind is blowing at 5 miles per hour? b. Suppose the actual temperature is 5 F. Use inormation rom the table to describe how the perceived temperature would change i the wind speed increased rom 5 miles per hour?

26 Chapter 4 Functions o Two Variables Applied Calculus 64 a. Reading the table, we see that the perceived temperature is 3 F. b. This is a question about a partial derivative. We re holding the temperature (T) ied at 5 F, and asking what happens as wind speed (V) increases rom 5 miles per hour. We re thinking o V as the onl variable, so we want WindChill V = W V when T = 5 and V = 5. We ll ind the average rate o change b looking in the column where T = 5 and letting V increase, and use that to approimate the partial derivative. W 3 W V 0.4 V 0 5 What are the units? W is measured in F and V is measured in mph, so the units here are F/mph. And that lets us describe what happens: The perceived temperature would decrease b about.4 F or each mph increase in wind speed. Eample 3 Find and at the points (0, 0) and (, ) i, 4 4 To ind, take the ordinar derivative o with respect to, acting as i is constant:, 4 Note that the derivative o the 4 term with respect to is zero it s a constant. Similarl,, 4 8. Now we can evaluate these at the points:

27 Chapter 4 Functions o Two Variables Applied Calculus 65 0,0 0 and 0,0 0 both lat at (0,0)., and, 4 ; this tells us that the cross sections parallel to the - and - aes are ; this tells us that above the point (, ), the surace decreases i ou move to more positive values and increases i ou move to more positive values. Eample 4 Find and i, ln 3 e means is our onl variable, we re thinking o as a constant. Then we ll just ind the ordinar derivative. From s point o view, this is an eponential unction, divided b a constant, with a constant added. The constant pulls out in ront, the derivative o the eponential unction is the same thing, and we need to use the chain rule, so we multipl b the derivative o that eponent (which is just ): e 3 means that we re thinking o as the variable, acting as i is constant. From s point o view, is a quotient plus a product we ll need the quotient rule and the product rule: 3 e e 3 3 ln Eample 5 Find i z,, z, w 35 w z z z means we act as i z is our onl variable, so we ll act as i all the other variables (, and w) are constants and take the ordinar derivative. z,, z, w z z

28 Chapter 4 Functions o Two Variables Applied Calculus 66 Using Partial Derivatives to Estimate Function Values We can use the partial derivatives to estimate values o a unction. The geometr is similar to the tangent line approimation in one variable. Recall the one-variable case: i is close enough to a known point a, then a ' a a. In two variables, we do the same thing in both directions at once: Approimating Function Values with Partial Derivatives To approimate the value o (, ), ind some point (a, b) where. (, ) and (a, b) are close that is, and a are close and and b are close.. You know the eact values o (a, b) and both partial derivatives there. Then, a, b a, b a a, b b Notice that the total change in is being approimated b adding the approimate changes coming rom the and directions. Another wa to look at the same ormula: How close is close? It depends on the shape o the graph o. In general, the closer the better. Eample 6 Use partial derivatives to estimate the value o, 4 4 at (0.9,.) Note that the point (0.9,.) is close to an eas point, (, ). In act, we alread worked out the partial derivatives at (, ):, 4 ;,., 4 8 ;, 4. We also know that,. So 0.9, Note that it would have been possible in this case to simpl compute the eact answer; 0.9, Our estimate is not perect, but it s prett close.

29 Chapter 4 Functions o Two Variables Applied Calculus 67 Eample 7 Here is a contour diagram or a unction g(,). Use partial derivatives to estimate the value o g(3., 4.7). This is the same diagram rom beore, so we alread estimated the value o the unction and the partial derivatives at the nearb point (3,5). g(3, 5) is 0.6, our estimate o g 3,5. 5, and our estimate o g 3, So 3., g Note that in this case, we have no wa to know how close our estimate is to the actual value.

30 Chapter 4 Functions o Two Variables Applied Calculus Eercises For problems through 6, ind and or the unction given. 5,. 4 5, 3. e 6, 4. e, , 6., 7. 6, 8. 6 ln, , 0. e 4, 4. e 5,. 6, 3. 7, e , 5., , 4

31 Chapter 4 Functions o Two Variables Applied Calculus Here is a table showing the unction A t, r t r a. Estimate A t,.05. b. Estimate A r,.05 c. Use our answers to parts a and b to estimate the value o A.5,.054, which shows the interest earned i 000 dollars is deposited in an account earning r annual interest, compounded continuousl, and let there or t ears. How close are our estimates rom parts a, b, and c? rt d. The values in the table came rom A t, r 000 e 8. Here is a table showing values or the unction t h H, t h H a. Estimate the value o dt at (3, 50). H b. Estimate the value o dh at (3, 50). c. Use our answers to parts a and b to estimate the value o H.6,56. d. The values in the table came rom Ht, h h 5t 4.9t, which gives the height in meters above the ground ater t seconds o an object that is thrown upward rom an initial height o h meters with an initial velocit o 5 meters per second. How close are our estimates rom parts a, b, and c?, 9. Given the unction a. Calculate, 4,, 4, and, 4 b. Use our answers rom part a to estimate.9, Given the unction, ln 0,,5, and,5 b. Use our answers rom part a to estimate.8, 4.8 a. Calculate,5

32 Chapter 4 Functions o Two Variables Applied Calculus 70 In problems - 6, use the contour plot shown to estimate the desired value.., 5. 5, 3. 5,5 4. 0,0 5., ,

33 Chapter 4 Functions o Two Variables Applied Calculus 7 Section 3: Optimization The partial derivatives tell us something about where a surace has local maima and minima. Remember that even in the one-variable cases, there were critical points which were neither maima nor minima this is also true or unctions o man variables. In act, as ou might epect, the situation is even more complicated. Second Derivatives When ou ind a partial derivative o a unction o two variables, ou get another unction o two variables ou can take its partial derivatives, too. We ve done this beore, in the one-variable setting. In the one-variable setting, the second derivative gave inormation about how the graph was curved. In the two-variable setting, the second partial derivatives give some inormation about how the surace is curved, as ou travel on cross-sections but that s not ver complete inormation about the entire surace. Imagine that ou have a surace that s ruled around a point, like what happens near a button on an overstued soa, or a pinched piece o abric, or the wrinkl skin near our thumb when ou make a ist. Right at that point, ever direction ou move, something dierent will happen it might increase, decrease, curve up, curve down A simple phrase like concave up or concave down can t describe all the things that can happen on a surace. Surprisingl enough, though, there is still a second derivative test that can help ou decide i a point is a local ma or min or neither, so we still do want to ind second derivatives. Second Partial Derivatives Suppose, is a unction o two variables. Then it has our second partial derivatives: ; ; ; and are called the mied (second) partial derivatives o Leibniz notation or the second partial derivatives is a bit conusing, and we won t use it as oten: ; ; ; Notice that the order o the variables or the mied partials goes rom right to let in the Leibniz notation instead o let to right. This chapter is (c) 03. It was remied b David Lippman rom Shana Calawa's remi o Contemporar Calculus b Dale Homan. It is licensed under the Creative Commons Attribution license.

34 Chapter 4 Functions o Two Variables Applied Calculus 7 Eample Find all our partial derivatives o, 4 4 We have to start b inding the (irst) partial derivatives:, 4, 4 8 Now we re read to take the second partial derivatives:, 4, 4 4, 4 8 4, You might have noticed that the two mied partial derivatives were equal in this last eample. It turns out that it s not a coincidence it s a theorem. Mied Partial Derivative Theorem I, Then,,. =, and are all continuous (no breaks in their graph) In act, as long as and all its appropriate partial derivatives are continuous, the mied partials are equal even i the are o higher order, and even i the unction has more than two variables. This theorem means that the conusing Leibniz notation or second derivatives is not a big problem in almost ever situation, the mied partials are equal, so it doesn t matter in which order we compute them. Eample Find e or, ln 3 We alread ound the irst partial derivatives in an earlier eample: e 3

35 Chapter 4 Functions o Two Variables Applied Calculus 73 3 e e 3 3 ln Now we need to ind the mied partial derivative the Theorem sas it doesn t matter whether we ind the partial derivative o e with respect to or the partial derivative o 3 rather do? 3 e e 3 3 ln with respect to. Which would ou It looks like it will be easier to compute the mied partial b inding the partial derivative o e with respect to it still looks mess, but it looks less mess: 3 e 3 3 e e 3 3 I ou d decided to do this the other wa, ou d end up in the same place. Eventuall. Local Maima, Minima, and Saddle Points Let s briel review ma-min problems in one variable. A local ma is a point on a curve that is higher than all the nearb points. A local min is lower than all the nearb points. We know that local ma or min can onl occur at critical points, where the derivative is zero or undeined. But we also know that not all critical points are ma or min, so we also need to test them, with the First Derivative or Second Derivative Test. The situation with a unction o two variables is much the same. Just as in the one-variable case, the irst step is to ind critical points, places where both the partial derivatives are either zero or undeined. Deinition: has a local maimum at (a, b) i (a, b) (, ) or all points (, ) near (a, b) has a local minimum at (a, b) i (a, b) (, ) or all points (, ) near (a, b) A critical point or a unction (, ) is a point (, ) (or (,, (, )) where both the ollowing are true: 0 or is undeined and 0 or is undeined Just as in the one-variable case, a local ma or min o can onl occur at a critical point.

36 Chapter 4 Functions o Two Variables Applied Calculus 74 And then, just as in the one-variable setting, not all critical points are local ma or min. For a unction o two variables, the critical point could be a local ma, local min, or a saddle point. A point on a surace is a local maimum i it s higher than all the points nearb; a point is a local minimum i it s lower than all the points nearb. A saddle point is a point on a surace that is a minimum along some paths and a maimum along some others. It s called this because it s shaped a bit like a saddle ou might use to ride a horse. You can see a saddle point b making a ist between the knuckles o our inde and middle ingers, ou can see a place that is a minimum as ou go across our knuckles, but a maimum as ou go along our hand toward our ingers. Here is a picture o a saddle point rom a ew dierent angles. This is the surace, 5 3 0, and there is a saddle point above the origin. The lines show what the surace looks like above the - and -aes. Notice how the point above the origin, where the lines cross, is a local minimum in one direction, but a local maimum in the other direction. Second Derivative Test Just as in the one-variable case, we ll need a wa to test critical points to see whether the are local ma or min. There is a second derivative test or unctions o two variables that can help but, just as in the one-variable case, it won t alwas give an answer.

37 Chapter 4 Functions o Two Variables Applied Calculus 75 The Second Derivative Test or Functions o Two Variables: Find all critical points o (,). D, and evaluate it at each critical point. Compute (a) I D > 0, then has a local ma or min at the critical point. To see which, look at the sign o : I 0, then has a local minimum at the critical point. I 0, then has a local maimum at the critical point. (b) I D < 0 then has a saddle point at the critical point. (c) I D = 0, there could be a local ma, local min, or neither. Eample 3 Find all local maima, minima, and saddle points or the unction 3 3, First we need the partial derivatives: 3 6 and 3 6 Critical points are the places where both o these are zero (neither is ever undeined): when = 0 or when = when = 0 or when =. Putting these together, we get our critical points: (0, 0), (, 0), (0, ), and (, ). Now to classi them, we ll use the Second Derivative Test. We ll need all the second partial derivatives: 6 6, 6 6, 0 Then D. Now look at each critical point in turn: At (0, 0): D ; there is a saddle point at (0, 0). 0 At (, 0): D , and a local maimum at (, 0). D and 6 0 ; there is a local minimum at (0, ). At (0, ): At (, ): 66 0 D ; there is another saddle point at (, ). ; there is

38 Chapter 4 Functions o Two Variables Applied Calculus 76 Eample 4 Find all local maima, minima, and saddle points or the unction 3 3 z Solution: We ll need all the partial derivatives and second partial derivatives, so let s compute them all irst: 7 4; 4; z z z 54 ; z ; z 4 z Now to ind the critical points: We need both z and z to be zero (neither is ever undeined), so we need to solve this set o equations simultaneousl: z z 4 0 Perhaps it s been a while since ou solved sstems o equations. Just remember the substitution method solve one equation or one variable and substitute into the other equation: solve 4 0 or, then substitute into the other equation Now we have just one equation in one variable to solve. Factoring out a gives , so = 0 or , giving Plugging back in to the equation 4 4 (0, 0) and,. 9 3 to ind gives us the two critical points: 4 Now to test them. Compute D Evaluate it at the two critical points, and see: At (0,0): D = 6 < 0, so there is a saddle point at (0, 0).. At , : D = 48 > 0, and 0, so there is a local minimum at,

39 Chapter 4 Functions o Two Variables Applied Calculus 77 Applied Optimization Eample 5 A compan makes two products. The demand equations or the two products are given below. p, p, q,and q are the prices and quantities or products and. q 00 3p p q 50 p p Find the price the compan should charge or each product in order to maimize total revenue. What is that maimum revenue? Revenue is still price quantit. I we re selling two products, the total revenue will be the sum o the revenues rom the two products: R( p, p ) p q p q p 00 3p p p 50 p p R( p, p ) 00 p 3p p p 50 p p This is a unction o two variables, the two prices, and we need to optimize it just as in the previous eamples. First we ind critical points. The notation here gets a bit hard to look at, but hang in there this is the same stu we ve done beore. R p 00 6 p and R p 50 p 4 p p Solving these simultaneousl gives the one critical point p 5,5 p., To conirm that this gives maimum revenue, we need to use the Second Derivative Test. Find all the second derivatives: R 6, R 4, and R R p p p p So p p p p R p p D and 0, so this reall is a local maimum. To maimize revenue, the compan should charge $5 per unit or both products. This will ield a maimum revenue o $4375.

40 Chapter 4 Functions o Two Variables Applied Calculus Eercises For problems through 6, ind.,, and or the unction given. Conirm that., , , , e 5., ln , Find the critical points o, 5 and use the Second Derivative Test to classi them. I the test ails, sa the test ails. 8. Find the critical points o, 6 4 and use the Second Derivative Test to classi them. I the test ails, sa the test ails. 9. Find the critical points o, 4ln 4 and use the Second Derivative Test to classi them. I the test ails, sa the test ails. 0. Find the critical points o, 6 3 and use the Second Derivative Test to classi them. I the test ails, sa the test ails The origin is a critical point or the unction,, and D = 0 there. That is, the Second Derivative Test ails. Use what ou know about shapes o unctions to decide i there is a local minimum, local maimum, or saddle point or this unction at (0, 0).. The origin is a critical point or the unction, 5, and D = 0 there. That is, the Second Derivative Test ails. Use what ou know about shapes o unctions to decide i there is a local minimum, local maimum, or saddle point or this unction at (0, 0). For problems 3 through 8, ind all local maima, minima, and saddle points or the unction.

41 Chapter 4 Functions o Two Variables Applied Calculus 79 3., , , 3 6., , e, or 0 and 0. 8., ln 9. The demand unctions or two products are given below. p, p, q,and q are the prices (in dollars) and quantities or products and. q 00 3p p q 50 p p a. Are these two products complementar goods or substitute goods? b. What is the quantit demanded or each when the price or product is $0 per item and the price or product is $30 per item? c. Write a unction R p, p that epresses the total revenue rom these two products. d. Find the price and quantit or each product that maimizes the total revenue. 0. The demand unctions or two products are given below. p, p, q,and q are the prices (in dollars) and quantities or products and. q 350 p p q 5 p p a. Are these two products complementar goods or substitute goods? b. What is the quantit demanded or each when the price or product is $0 per item and the price or product is $30 per item? c. Write a unction R p, p that epresses the total revenue rom these two products. d. Find the price and quantit or each product that maimizes the total revenue., p q, p, where p, p, q, and q are the prices (in dollars) and quantities or products and. Consider q q q q the our partial derivatives,,, and. Tell the sign o each o these partial p p p p derivatives i a. the products are complementar goods. b. the products are substitute goods.. Suppose the demand unctions or two products are q p and gp