Daily WeBWorK, #1. This means the two planes normal vectors must be multiples of each other.

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1 Daily WeBWorK, #1 Consider the ellipsoid x 2 + 3y 2 + z 2 = 11. Find all the points where the tangent plane to this ellipsoid is parallel to the plane 2x + 3y + 2z = 0. In order for the plane tangent to the ellipsoid at a point (a, b, c) to be parallel to the plane 2x + 3y + 2z = 0, the two planes normal vectors must be parallel. This means the two planes normal vectors must be multiples of each other. A vector normal to the plane 2x + 3y + 2z = 0 is < 2, 3, 2 >. Goal: Find all points (a, b, c) where a vector normal to the tangent to the ellipsoid at the point (a, b, c) is k< 2, 3, 2 > for some value of k. Math 236-Multi (Sklensky) In-Class Work April 5, / 32

2 Daily WeBWorK, #1, continued Consider the ellipsoid x 2 + 3y 2 + z 2 = 11. Find all the points where the tangent plane to this ellipsoid is parallel to the plane 2x + 3y + 2z = 0. Goal: Find all points (a, b, c) where a vector normal to the tangent to the ellipsoid at the point (a, b, c) is k< 2, 3, 2 > for some value of k. To find a vector normal to the ellipsoid at the point (a, b, c): Define the function f (x, y, z) = x 2 + 3y 2 + z Our specific ellipsoid is the level surface for this function with f (x, y, z) = 0. Showed last time: gradients are normal to level surfaces. Thus the gradient f (a, b, c) will be normal to the tangent plane to the ellipsoid at the point (a, b, c). f (x, y, z) =< 2x, 6y, 2z > Thus at a point (a, b, c), < 2a, 6b, 2c > is normal to the tangent plane to the ellipsoid. Math 236-Multi (Sklensky) In-Class Work April 5, / 32

3 Daily WeBWorK, #1, continued Consider the ellipsoid x 2 + 3y 2 + z 2 = 11. Find all the points where the tangent plane to this ellipsoid is parallel to the plane 2x + 3y + 2z = 0. Goal: Find all points (a, b, c) where a vector normal to the tangent to the ellipsoid at the point (a, b, c) is k< 2, 3, 2 > for some value of k. At a point (a, b, c), < 2a, 6b, 2c > is normal to the tangent plane to the ellipsoid. We want to find all points (a, b, c) where < 2a, 6b, 2c >= k < 2, 3, 2 > for some k. 2a = 2k a = k 6b = 3k b = k/2 2c = 2k c = k Thus all points (k, k/2, k) which also lie on the ellipsoid will have their tangent planes parallel to the given plane. Math 236-Multi (Sklensky) In-Class Work April 5, / 32

4 Daily WeBWorK, #1, continued Consider the ellipsoid x 2 + 3y 2 + z 2 = 11. Find all the points where the tangent plane to this ellipsoid is parallel to the plane 2x + 3y + 2z = 0. Goal: Find all points (a, b, c) where a vector normal to the tangent to the ellipsoid at the point (a, b, c) is k< 2, 3, 2 > for some value of k. At a point (a, b, c), < 2a, 6b, 2c > is normal to the tangent plane to the ellipsoid. All points (k, k/2, k) which also lie on the ellipsoid will have their tangent planes parallel to the given plane. ( ) 2 Find all values of k for which (k) 2 k (k) 2 = 11. Simplifying and solving for k, we find that k 2 = 4, so k = ±2 Thus at (2, 1, 2) and at ( 2, 1, 2), the plane tangent to the ellipsoid is parallel to 2x + 3y + 2z = 0. Math 236-Multi (Sklensky) In-Class Work April 5, / 32

5 Wrapping up the gradient: As your book says: always think of gradients as vector-valued functions whose values specify the direction of maximum increase of a function, and provide normal vectors to the level curves, in two dimensions, and to the level surfaces in three dimensions. Remember that this also gives us a way to find planes tangent to surfaces Math 236-Multi (Sklensky) In-Class Work April 5, / 32

6 Recall: Critical points for a function f (x, y): If f (x, y) is a smooth function, so the partials f x (x, y) and f y (x, y) exist everywhere, then the point ( a, b, f (a, b) ) can only be a local maximum, local minimum, or saddle point if f (x, y) is flat at ( a, b, f (a, b) ) in all directions that is, only if D u f (a, b) = 0 for all unit vectors u R 2 Thus if f x (a, b) 0 or f y (a, b) 0, the point ( a, b, f (a, b) ) can not be a local maximum, local minimum, or saddle point. We therefore call all inputs (a, b) such that f (a, b) = 0 critical points. Note: This does not mean that every critical point is a local maximum, local minimum, or saddle point. Math 236-Multi (Sklensky) In-Class Work April 5, / 32

7 Recall from Calc 1: For g(x), if g (a) = 0 (so that g is flat at x = a), then the 2nd Derivative Test gives g (a) > 0 = g is concave up at x = a = x = a is a minimum g (a) < 0 = g is concave down at x = a = x = a is a maximum g (a) = 0 = the test is inconclusive Is there an analogous result for f (x, y)? Math 236-Multi (Sklensky) In-Class Work April 5, / 32

8 Question: Suppose f (x, y) is a smooth function and f x (x 0, y 0 ) = f y (x 0, y 0 ) = 0 f xx (x 0, y 0 ) > 0, f yy (x 0, y 0 ) > 0 That is, in both the x and y directions, f is concave up Must f have a local min at the point (x 0, y 0 )? Math 236-Multi (Sklensky) In-Class Work April 5, / 32

9 Question: Suppose f (x, y) is a smooth function and f x (x 0, y 0 ) = f y (x 0, y 0 ) = 0 f xx (x 0, y 0 ) > 0, f yy (x 0, y 0 ) > 0 That is, in both the x and y directions, f is concave up Must f have a local min at the point (x 0, y 0 )? No! Math 236-Multi (Sklensky) In-Class Work April 5, / 32

10 Example: Let f (x, y) = x 2 + 3xy + 2y 2 9x 11y. f x (x, y) = 2x + 3y 9 = f x ( 3, 5) = = 0 f y (x, y) = 3x + 4y 11 = f y ( 3, 5) = = 0 f xx (x, y) = 2 = f xx ( 3, 5) > 0 f yy (x, y) = 4 = f yy ( 3, 5) > 0 Math 236-Multi (Sklensky) In-Class Work April 5, / 32

11 Example: Let f (x, y) = x 2 + 3xy + 2y 2 9x 11y. f x (x, y) = 2x + 3y 9 = f x ( 3, 5) = = 0 f y (x, y) = 3x + 4y 11 = f y ( 3, 5) = = 0 f xx (x, y) = 2 = f xx ( 3, 5) > 0 f yy (x, y) = 4 = f yy ( 3, 5) > 0 Thus at ( 3, 5), f is flat in both the x and y directions, and is concave up in both the x and y directions... Math 236-Multi (Sklensky) In-Class Work April 5, / 32

12 Example: Let f (x, y) = x 2 + 3xy + 2y 2 9x 11y. f x (x, y) = 2x + 3y 9 = f x ( 3, 5) = = 0 f y (x, y) = 3x + 4y 11 = f y ( 3, 5) = = 0 f xx (x, y) = 2 = f xx ( 3, 5) > 0 f yy (x, y) = 4 = f yy ( 3, 5) > 0 Thus at ( 3, 5), f is flat in both the x and y directions, and is concave up in both the x and y directions... but f does not have a minimum there: Math 236-Multi (Sklensky) In-Class Work April 5, / 32

13 Recall: As long as f xy (x, y) and f yx (x, y) are continuous, they are equal: f xy (x, y) = f yx (x, y) Math 236-Multi (Sklensky) In-Class Work April 5, / 32

14 In Class Work 1. Locate all critical points and classify them using the 2nd Derivatives Test: (a) f (x, y) = x 3 3xy + y 3 (b) f (x, y) = y 2 + x 2 y + x 2 2y 2. Let f (x, y) = 4xy x 3 2y 2 (a) Find and classify all critical points of f. (b) Find the maximum value of f (x, y) on the unit disk {(x, y) x 2 + y 2 1}. Math 236-Multi (Sklensky) In-Class Work April 5, / 32

15 1. Locate all critical points and classify them using the 2nd Derivatives Test: (a) f (x, y) = x 3 3xy + y 3 f x = 0 = 3x 2 3y = 0 = y = x 2 f y = 0 = 3x + 3y 2 = 0 = x = y 2 Putting those two results together, y = x 2 = (y 2 ) 2 = y(y 3 1) = 0 = y = 0, y = 1 Thus our critical points are (0, 0) and (1, 1). Math 236-Multi (Sklensky) In-Class Work April 5, / 32

16 1. Locate all critical points and classify them using the 2nd Derivatives Test: (a) f (x, y) = x 3 3xy + y 3 We found the only critical points are (0, 0) and (1, 1) To classify them, we will use the Second Derivatives Test. To use that, we need to find f xx (a, b) and the discriminant D(a, b): f xx (x, y) = x (3x 2 3y) = 6x D(x, y) = f xx(x, y) f xy (x, y) f xy (x, y) f yy (x, y) = 6x 3 3 6y = 36xy 9 Math 236-Multi (Sklensky) In-Class Work April 5, / 32

17 1. Locate all critical points and classify them using the 2nd Derivatives Test: (a) f (x, y) = x 3 3xy + y 3 We found the only critical points are (0, 0) and (1, 1) Also found f xx (x, y) = 6x and D(x, y) = 36xy 9 At (0, 0), f xx (0, 0) = (6)(0) = 0 and D(0, 0) = (36)(0)(0) 9 < 0. According to the Second Derivatives Test, because D < 0, f has a saddle point at the point (0, 0). Math 236-Multi (Sklensky) In-Class Work April 5, / 32

18 1. Locate all critical points and classify them using the 2nd Derivatives Test: (a) f (x, y) = x 3 3xy + y 3 We found the only critical points are (0, 0) and (1, 1) We found (0, 0) is a saddle point. f xx (x, y) = 6x and D(x, y) = 36xy 9 At (1, 1), f xx (1, 1) = (6)(1) > 0 and D(1, 1) = (36)(1)(1) 9 > 0. According to the Second Derivatives Test, because D > 0 and f xx > 0, f has a local minimum at (1, 1). Math 236-Multi (Sklensky) In-Class Work April 5, / 32

19 1. Locate all critical points and classify them using the 2nd Derivatives Test: (a) f (x, y) = x 3 3xy + y 3 We found the only critical points are (0, 0) and (1, 1) We found (0, 0) is a saddle point and (1, 1) is a local minimum. We can see these results on a contour plot of f (x, y): Math 236-Multi (Sklensky) In-Class Work April 5, / 32

20 1. Locate all critical points and classify them using the 2nd Derivatives Test: (b) f (x, y) = y 2 + x 2 y + x 2 2y f x = 0 = 2xy + 2x = 0 = 2x(y + 1) = 0 = x = 0 or y = 1 f y = 0 = 2y + x 2 2 = 0 = If x = 0, 2y = 0 = y = 1 = If y = 1, 2( 1) + x 2 2 = 0 = x 2 = 4 = x = ±2 Thus our critical points are (0, 1), (2,-1), ( 2, 1). Math 236-Multi (Sklensky) In-Class Work April 5, / 32

21 1. Locate all critical points and classify them using the 2nd Derivatives Test: (b) f (x, y) = y 2 + x 2 y + x 2 2y Our critical points are (0, 1), (2,-1), ( 2, 1). To classify them using the 2nd Derivatives Test, we need: f xx (x, y) = (2xy + 2x) = 2y + 2 x D(x, y) = f xx(x, y) f xy (x, y) f xy (x, y) f yy (x, y) = 2y + 2 2x 2 2x 2 = 4y + 4 4x Math 236-Multi (Sklensky) In-Class Work April 5, / 32

22 1. Locate all critical points and classify them using the 2nd Derivatives Test: (b) f (x, y) = y 2 + x 2 y + x 2 2y Our critical points are (0, 1), (2,-1), ( 2, 1). f xx (x, y) = 2y + 2 and D(x, y) = 4y + 4 4x 2 At (0, 1) f xx (0, 1) = 2(1) + 2 > 0 and D(0, 1) = 4(1) + 4 4(0) > 0. According to the 2nd Derivatives Test, because D > 0 and f xx > 0, f has a local minimum at (0, 1). Math 236-Multi (Sklensky) In-Class Work April 5, / 32

23 1. Locate all critical points and classify them using the 2nd Derivatives Test: (b) f (x, y) = y 2 + x 2 y + x 2 2y Our critical points are (0, 1), (2,-1), ( 2, 1). (0, 1) is a local minimum. f xx (x, y) = 2y + 2 and D(x, y) = 4y + 4 4x 2 At (2, 1) f xx (2, 1) = 2( 1)+2 = 0 and D(2, 1) = 4( 1)+4 4(2) 2 < 0. According to the 2nd Derivatives Test, because D < 0, f has a saddle point at (2, 1). Math 236-Multi (Sklensky) In-Class Work April 5, / 32

24 1. Locate all critical points and classify them using the 2nd Derivatives Test: (b) f (x, y) = y 2 + x 2 y + x 2 2y Our critical points are (0, 1), (2,-1), ( 2, 1). (0, 1) is a local minimum and (2, 1) is a saddle point. f xx (x, y) = 2y + 2 and D(x, y) = 4y + 4 4x 2 At ( 2, 1) f xx ( 2, 1) = 2( 1)+2 = 0 and D(2, 1) = 4( 1)+4 4( 2) 2 < 0 According to the 2nd Derivatives Test, because D < 0, f has a saddle point at ( 2, 1). Math 236-Multi (Sklensky) In-Class Work April 5, / 32

25 1. Locate all critical points and classify them using the 2nd Derivatives Test: (b) f (x, y) = y 2 + x 2 y + x 2 2y Our critical points are (0, 1), (2,-1), ( 2, 1). (0, 1) is a local minimum and both of (±2, 1) are saddle points. We can again see these results illustrated: Math 236-Multi (Sklensky) In-Class Work April 5, / 32

26 2. Let f (x, y) = 4xy x 3 2y 2 (a) Find and classify all critical points of f. f x = 0 = 4y 3x 2 = 0 = y = 3 4 x 2 f = 0 y = 4x 4y = 0 = y = x x = y = 3 4 x 2 = x (1 34 ) x = 0 = x = 0 or x = 4 3. x = 0 = y = 0 x = 4 3 = y = 4 3. Thus our critical points are (0, 0) and (4/3, 4/3). Math 236-Multi (Sklensky) In-Class Work April 5, / 32

27 2. Let f (x, y) = 4xy x 3 2y 2 (a) Find and classify all critical points of f. The critical points are (0, 0) and (4/3, 4/3). To classify them using the 2nd Derivatives Test, we need: f xx (x, y) = x (4y 3x 2 ) = 6x D(x, y) = f xx f xy = 6x = 24x 16 f xy f yy Math 236-Multi (Sklensky) In-Class Work April 5, / 32

28 2. Let f (x, y) = 4xy x 3 2y 2 (a) Find and classify all critical points of f. The critical points are (0, 0) and (4/3, 4/3). f xx (x, y) = 6x and D(x, y) = 24x 16 At (0, 0), f xx (0, 0) = ( 6)(0) and D(0, 0) = 24(0) 16 < 0. By the 2nd Derivatives Test, because D < 0, f has a saddle point at (0, 0). Math 236-Multi (Sklensky) In-Class Work April 5, / 32

29 2. Let f (x, y) = 4xy x 3 2y 2 (a) Find and classify all critical points of f. The critical points are (0, 0) and (4/3, 4/3). f xx (x, y) = 6x and D(x, y) = 24x 16 f has a saddle point at (0, 0). At (4/3, 4/3), f xx (4/3, 4/3) = ( 6)(4/3) < 0 and D(4/3, 4/3) = 24(4/3) 16 > 0 By the 2nd Derivatives Test, because D > 0 and f xx < 0, f has a local maximum. Math 236-Multi (Sklensky) In-Class Work April 5, / 32

30 2. Let f (x, y) = 4xy x 3 2y 2 (a) Find and classify all critical points of f. The critical points are (0, 0) and (4/3, 4/3). f xx (x, y) = 6x and D(x, y) = 24x 16 f has a saddle point at (0, 0). f has local maximum at (4/3, 4/3). To illustrate these points: Math 236-Multi (Sklensky) In-Class Work April 5, / 32

31 2. Let f (x, y) = 4xy x 3 2y 2 (b) Find the max value of f (x, y) on the unit disk {(x, y) x 2 + y 2 1}. We know that f has a saddle point at (0, 0) and a local maximum at (4/3, 4/3), and that these are the only critical points. We can see that the maximum value will occur somewhere on the boundary of the unit disk. Math 236-Multi (Sklensky) In-Class Work April 5, / 32

32 2. Let f (x, y) = 4xy x 3 2y 2 (b) Find the max value of f (x, y) on the unit disk {(x, y) x 2 + y 2 1}. We know that the maximum value occurs on the boundary of the unit disk. How do we find where? Math 236-Multi (Sklensky) In-Class Work April 5, / 32

33 2. Let f (x, y) = 4xy x 3 2y 2 (b) Find the max value of f (x, y) on the unit disk {(x, y) x 2 + y 2 1}. We know that the maximum value occurs on the boundary of the unit disk. How do we find where? We know the boundary is parametrized by x = cos(t), y = sin(t), z = f (cos(t), sin(t)). Math 236-Multi (Sklensky) In-Class Work April 5, / 32

34 2. Let f (x, y) = 4xy x 3 2y 2 (b) Find the max value of f (x, y) on the unit disk {(x, y) x 2 + y 2 1}. We know that the maximum value occurs on the boundary of the unit disk. How do we find where? We know the boundary is parametrized by x = cos(t), y = sin(t), z = f (cos(t), sin(t)). = We want to maximize z = f (cos(t), sin(t)). Math 236-Multi (Sklensky) In-Class Work April 5, / 32

35 2. Let f (x, y) = 4xy x 3 2y 2 (b) Find the max value of f (x, y) on the unit disk {(x, y) x 2 + y 2 1}. We know that the maximum value occurs on the unit disk, so we want to maximize z = g(t) = f (cos(t), sin(t)). Find where g (t) = 0! First, need to find g (t). g (t) = f dx x dt + f dy y dt = ( 4y 3x 2)( sin(t) ) + ( 4x 4y )( cos(t) ) = ( 4 sin(t) 3 cos 2 (t) )( sin(t) ) + ( 4 cos(t) 4 sin(t) )( cos(t) ) = 4 sin 2 (t) + 3 cos 2 (t) sin(t) + 4 cos 2 (t) 4 cos(t) sin(t) = 4 cos 2 (t) 4 sin 2 (t) + 3 cos 2 (t) sin(t) 4 sin(t) cos(t) Math 236-Multi (Sklensky) In-Class Work April 5, / 32

36 2. Let f (x, y) = 4xy x 3 2y 2 (b) Find the max value of f (x, y) on the unit disk {(x, y) x 2 + y 2 1}. We know that the maximum value occurs on the unit disk, so we want to maximize z = g(t) = f (cos(t), sin(t)). We know we want to find where g (t) = 0, and we have found that g (t) = 4 cos 2 (t) 4 sin 2 (t) + 3 cos 2 (t) sin(t) 4 sin(t) cos(t). Using Maple, I can solve for where g (t) = 0, and I find that there are 4 possible values of t that could give us our absolute max on this region: t.92 t.68 t 2.06 t Math 236-Multi (Sklensky) In-Class Work April 5, / 32

37 2. Let f (x, y) = 4xy x 3 2y 2 (b) Find the max value of f (x, y) on the unit disk {(x, y) x 2 + y 2 1}. We know that the maximum value occurs on the unit disk, so we want to maximize z = g(t) = f (cos(t), sin(t)). We have found that the only possible values of t for which the maximum of f on the boundary could occur are: t.92 t.68 t 2.06 t Plugging each of these in (again using Maple), we find that f is largest at t But we wanted to know the x s and y s that gave us our max, so now I plug this value of t into x = cos(t), y = sin(t), and I find that f attains its absolute max on the unit disk at x.91, y.42. Math 236-Multi (Sklensky) In-Class Work April 5, / 32

38 2. Let f (x, y) = 4xy x 3 2y 2 (b) Find the max value of f (x, y) on the unit disk {(x, y) x 2 + y 2 1}. We have found that maximum value of f on the unit disk occurs at approximately (.91,.42). Math 236-Multi (Sklensky) In-Class Work April 5, / 32