Tangent Planes/Critical Points

Transcription

1 Tangent Planes/Critical Points Christopher Croke University of Pennsylvania Math 115 UPenn, Fall 2011

2 Problem: Find the tangent line to the curve of intersection of the surfaces xyz = 1 and x 2 + 2y 2 + 3z 2 = 6 at (1, 1, 1). (We use the fact that u v is perpendicular to both u and v).

3 Problem: Find the tangent line to the curve of intersection of the surfaces xyz = 1 and x 2 + 2y 2 + 3z 2 = 6 at (1, 1, 1). (We use the fact that u v is perpendicular to both u and v). We now want to estimate the change in f if we move in direction u a (small) distance ds.

4 Problem: Find the tangent line to the curve of intersection of the surfaces xyz = 1 and x 2 + 2y 2 + 3z 2 = 6 at (1, 1, 1). (We use the fact that u v is perpendicular to both u and v). We now want to estimate the change in f if we move in direction u a (small) distance ds. The change df should be {directional derivative} {increment}. I.e df = ( f (x0,y 0 ) u)ds.

5 Problem: Find the tangent line to the curve of intersection of the surfaces xyz = 1 and x 2 + 2y 2 + 3z 2 = 6 at (1, 1, 1). (We use the fact that u v is perpendicular to both u and v). We now want to estimate the change in f if we move in direction u a (small) distance ds. The change df should be {directional derivative} {increment}. I.e df = ( f (x0,y 0 ) u)ds. Problem: Estimate the change in f (x, y) = xy 2 if one moves.01 in the direction of (i.e. towards) (2, 3) starting from (1, 2).

6 Linearization The Linearization (or Linear approximation) of a function f (x, y) at a point (x 0, y 0 ) is the function; L(x, y) = f (x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ).

7 Linearization The Linearization (or Linear approximation) of a function f (x, y) at a point (x 0, y 0 ) is the function; L(x, y) = f (x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ). The point is that L(x, y) should be close to f (x, y) when (x, y) is close to (x 0, y 0 ). (In fact the definition of differentiable says precisely that these are close.)

8 Linearization The Linearization (or Linear approximation) of a function f (x, y) at a point (x 0, y 0 ) is the function; L(x, y) = f (x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ). The point is that L(x, y) should be close to f (x, y) when (x, y) is close to (x 0, y 0 ). (In fact the definition of differentiable says precisely that these are close.) Also called the Standard linear approximation of f at (x 0, y 0 ).

9 Linearization The Linearization (or Linear approximation) of a function f (x, y) at a point (x 0, y 0 ) is the function; L(x, y) = f (x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ). The point is that L(x, y) should be close to f (x, y) when (x, y) is close to (x 0, y 0 ). (In fact the definition of differentiable says precisely that these are close.) Also called the Standard linear approximation of f at (x 0, y 0 ). Problem: Find the linearization of f (x, y) = x 3 + 2xy y 2 at (1, 2).

10 Linearization The Linearization (or Linear approximation) of a function f (x, y) at a point (x 0, y 0 ) is the function; L(x, y) = f (x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ). The point is that L(x, y) should be close to f (x, y) when (x, y) is close to (x 0, y 0 ). (In fact the definition of differentiable says precisely that these are close.) Also called the Standard linear approximation of f at (x 0, y 0 ). Problem: Find the linearization of f (x, y) = x 3 + 2xy y 2 at (1, 2).

11 How good? How good is this linear approximation?

12 How good? How good is this linear approximation? Let f have continuous first and second partial derivatives in an open set containing a rectangle R centered at (x 0, y 0 ). Let M be an upper bound for f xx, f yy and f xy on R. The error E(x, y) = L(x, y) f (x, y) by using the linearization L of f at (x 0, y 0 ) instead of f satisfies Theorem E(x, y) 1 2 M( x x 0 + y y 0 ) 2.

13 How good? How good is this linear approximation? Let f have continuous first and second partial derivatives in an open set containing a rectangle R centered at (x 0, y 0 ). Let M be an upper bound for f xx, f yy and f xy on R. The error E(x, y) = L(x, y) f (x, y) by using the linearization L of f at (x 0, y 0 ) instead of f satisfies Theorem E(x, y) 1 2 M( x x 0 + y y 0 ) 2. Problem Find an upper bound for the error in our example on R: x and y What is the maximum relative error? The percentage error?

14 The Differential Definition The Total Differential of f (x, y) at a point (x 0, y 0 ) is df = f x (x 0, y 0 )dx + f y (x 0, y 0 )dy.

15 The Differential Definition The Total Differential of f (x, y) at a point (x 0, y 0 ) is df = f x (x 0, y 0 )dx + f y (x 0, y 0 )dy. It represents the change in the linearization as we move from (x 0, y 0 ) to a point (x 0 + dx, y 0 + dy) (nearby).

16 The Differential Definition The Total Differential of f (x, y) at a point (x 0, y 0 ) is df = f x (x 0, y 0 )dx + f y (x 0, y 0 )dy. It represents the change in the linearization as we move from (x 0, y 0 ) to a point (x 0 + dx, y 0 + dy) (nearby). Problem A cylindrical silo is full of wheat. The height is measured with an error of at most 2% and its radius with an error of at most 1%. Estimate the percentage error in calculating the total volume of wheat.

17 More variables Things look pretty much the same for more variables.

18 More variables Things look pretty much the same for more variables. For f (x, y, z) at a point P 0 = (x 0, y 0, z 0 ) we have Linearization L(x, y, z) = f (P 0 )+f x (P 0 )(x x 0 )+f y (P 0 )(y y 0 )+f z (P 0 )(z z 0 )

19 More variables Things look pretty much the same for more variables. For f (x, y, z) at a point P 0 = (x 0, y 0, z 0 ) we have Linearization L(x, y, z) = f (P 0 )+f x (P 0 )(x x 0 )+f y (P 0 )(y y 0 )+f z (P 0 )(z z 0 ) Error in Linearization E(x, y) 1 2 M( x x 0 + y y 0 + z z 0 ) 2.

20 More variables Things look pretty much the same for more variables. For f (x, y, z) at a point P 0 = (x 0, y 0, z 0 ) we have Linearization L(x, y, z) = f (P 0 )+f x (P 0 )(x x 0 )+f y (P 0 )(y y 0 )+f z (P 0 )(z z 0 ) Error in Linearization E(x, y) 1 2 M( x x 0 + y y 0 + z z 0 ) 2. Differential df = f x (P 0 )dx + f y (P 0 )dy + f z (P 0 )dz

21 Finding Maxima and Minima For a function of two variables what does a relative maximum or relative minimum look like?

22 Finding Maxima and Minima For a function of two variables what does a relative maximum or relative minimum look like? The tangent plane of such a point will be horizontal. What does that say about the partial derivatives?

23 Finding Maxima and Minima For a function of two variables what does a relative maximum or relative minimum look like? The tangent plane of such a point will be horizontal. What does that say about the partial derivatives? First Derivative Test: If f (x, y) has either a relative maximum or minimum at at point (a, b) then f x (a, b) = 0 and f (a, b) = 0. y

24 Finding Maxima and Minima For a function of two variables what does a relative maximum or relative minimum look like? The tangent plane of such a point will be horizontal. What does that say about the partial derivatives? First Derivative Test: If f (x, y) has either a relative maximum or minimum at at point (a, b) then In other words f (a, b) = 0. f x (a, b) = 0 and f (a, b) = 0. y

25 Problem: The function f (x, y) = 3x 2 xy + 2y 2 + 3x + 2y + 4 has a relative minimum (graph it with Maple). Find it.

26 Problem: The function f (x, y) = 3x 2 xy + 2y 2 + 3x + 2y + 4 has a relative minimum (graph it with Maple). Find it. Finding Maxima and Minima on a region For a function f (x, y) defined on a region the maximum and minimum values of f on the region can only happen at a point (a, b) where one of: (a, b) is an interior point and f x (a, b) = 0 and f y (a, b) = 0.

27 Problem: The function f (x, y) = 3x 2 xy + 2y 2 + 3x + 2y + 4 has a relative minimum (graph it with Maple). Find it. Finding Maxima and Minima on a region For a function f (x, y) defined on a region the maximum and minimum values of f on the region can only happen at a point (a, b) where one of: (a, b) is an interior point and f x (a, b) = 0 and f y (a, b) = 0. (a, b) is an interior point and f x (a, b) or f y (a, b) is not defined.

28 Problem: The function f (x, y) = 3x 2 xy + 2y 2 + 3x + 2y + 4 has a relative minimum (graph it with Maple). Find it. Finding Maxima and Minima on a region For a function f (x, y) defined on a region the maximum and minimum values of f on the region can only happen at a point (a, b) where one of: (a, b) is an interior point and f x (a, b) = 0 and f y (a, b) = 0. (a, b) is an interior point and f x (a, b) or f y (a, b) is not defined. (a, b) is a boundary point.

29 Problem: The function f (x, y) = 3x 2 xy + 2y 2 + 3x + 2y + 4 has a relative minimum (graph it with Maple). Find it. Finding Maxima and Minima on a region For a function f (x, y) defined on a region the maximum and minimum values of f on the region can only happen at a point (a, b) where one of: (a, b) is an interior point and f x (a, b) = 0 and f y (a, b) = 0. (a, b) is an interior point and f x (a, b) or f y (a, b) is not defined. (a, b) is a boundary point. Points in the first two category are called Critical Points of f.

30 Problem: A cardboard box is to be built with a double thick bottom and a volume of 324 cubic inches. Find the dimensions of the cheapest such box.

31 Problem: A cardboard box is to be built with a double thick bottom and a volume of 324 cubic inches. Find the dimensions of the cheapest such box. So just like the one variable case to find maxes and mins we take the first derivatives and set to 0.

32 Problem: A cardboard box is to be built with a double thick bottom and a volume of 324 cubic inches. Find the dimensions of the cheapest such box. So just like the one variable case to find maxes and mins we take the first derivatives and set to 0. Is there a second derivative test?

33 Problem: A cardboard box is to be built with a double thick bottom and a volume of 324 cubic inches. Find the dimensions of the cheapest such box. So just like the one variable case to find maxes and mins we take the first derivatives and set to 0. Is there a second derivative test? Yes, but it is a little more complicated because there are saddle points.

34 Problem: A cardboard box is to be built with a double thick bottom and a volume of 324 cubic inches. Find the dimensions of the cheapest such box. So just like the one variable case to find maxes and mins we take the first derivatives and set to 0. Is there a second derivative test? Yes, but it is a little more complicated because there are saddle points.

35 Second derivative test: If (a, b) is a point such that f f x (a, b) = 0 and y (a, b) = 0 let D(a, b) 2 f 2 f x 2 y 2 ( 2 f ) 2. x y (D is called the discriminant.) Then If D(a, b) > 0 and 2 f > 0 then (a, b) is a relative minimum. x 2 If D(a, b) > 0 and 2 f < 0 then (a, b) is a relative maximum. x 2 If D(a, b) < 0 then (a, b) is a saddle point (hence neither a relative Max nor a relative min). If D(a, b) = 0 then we get no information.

36 Problem: Find all possible relative maxima and minima of f (x, y) = 3x 2 6xy + y 3 9y and determine the nature of each point.