Tangent Planes/Critical Points Christopher Croke University of Pennsylvania Math 115 Christopher Croke Calculus 115 MAPLE For a similar intersection the picture is: Problem: Find the tangent line to the curve of intersection of the surfaces xyz = 1 and x2 + 2y 2 + 3z2 = 6 at (1; 1; 1). (We use the fact that ~u × ~v is perpendicular to both ~u and ~v). Christopher Croke Calculus 115 Problem: Find the tangent line to the curve of intersection of the surfaces xyz = 1 and x2 + 2y 2 + 3z2 = 6 at (1; 1; 1). (We use the fact that ~u × ~v is perpendicular to both ~u and ~v). MAPLE For a similar intersection the picture is: Christopher Croke Calculus 115 The change df should be fdirectional derivativeg × fincrementg. I.e df = (rf j(x0;y0) · ~u)ds: Problem: Estimate the change in f (x; y) = xy 2 if one moves :01 in the direction of (i.e. towards) (1; 2) starting from (2; 3). We now want to estimate the change in f if we move in direction ~u a (small) distance ds. Christopher Croke Calculus 115 Problem: Estimate the change in f (x; y) = xy 2 if one moves :01 in the direction of (i.e. towards) (1; 2) starting from (2; 3). We now want to estimate the change in f if we move in direction ~u a (small) distance ds. The change df should be fdirectional derivativeg × fincrementg. I.e df = (rf j(x0;y0) · ~u)ds: Christopher Croke Calculus 115 We now want to estimate the change in f if we move in direction ~u a (small) distance ds. The change df should be fdirectional derivativeg × fincrementg. I.e df = (rf j(x0;y0) · ~u)ds: Problem: Estimate the change in f (x; y) = xy 2 if one moves :01 in the direction of (i.e. towards) (1; 2) starting from (2; 3). Christopher Croke Calculus 115 The point is that L(x; y) should be close to f (x; y) when (x; y) is close to (x0; y0). (In fact the definition of ”differentiable” says precisely that these are close.) Also called the Standard linear approximation of f at (x0; y0). Linearization The Linearization (or Linear approximation) of a function f (x; y) at a point (x0; y0) is the function; L(x; y) = f (x0; y0) + fx (x0; y0)(x − x0) + fy (x0; y0)(y − y0): Christopher Croke Calculus 115 Also called the Standard linear approximation of f at (x0; y0). Linearization The Linearization (or Linear approximation) of a function f (x; y) at a point (x0; y0) is the function; L(x; y) = f (x0; y0) + fx (x0; y0)(x − x0) + fy (x0; y0)(y − y0): The point is that L(x; y) should be close to f (x; y) when (x; y) is close to (x0; y0). (In fact the definition of ”differentiable” says precisely that these are close.) Christopher Croke Calculus 115 Linearization The Linearization (or Linear approximation) of a function f (x; y) at a point (x0; y0) is the function; L(x; y) = f (x0; y0) + fx (x0; y0)(x − x0) + fy (x0; y0)(y − y0): The point is that L(x; y) should be close to f (x; y) when (x; y) is close to (x0; y0). (In fact the definition of ”differentiable” says precisely that these are close.) Also called the Standard linear approximation of f at (x0; y0). Christopher Croke Calculus 115 Problem: Find the linearization of f (x; y) = x3 + 2xy − y 2 + 6 at (1; 2). As we zoom in linear approximation gets better. Christopher Croke Calculus 115 Problem: Find the linearization of f (x; y) = x3 + 2xy − y 2 + 6 at (1; 2). As we zoom in linear approximation gets better. Christopher Croke Calculus 115 Problem: Find the linearization of f (x; y) = x3 + 2xy − y 2 + 6 at (1; 2). As we zoom in linear approximation gets better. Christopher Croke Calculus 115 Problem: Find the linearization of f (x; y) = x3 + 2xy − y 2 + 6 at (1; 2). As we zoom in linear approximation gets better. Christopher Croke Calculus 115 Problem: Find the linearization of f (x; y) = x3 + 2xy − y 2 + 6 at (1; 2). As we zoom in linear approximation gets better. Christopher Croke Calculus 115 As we zoom in linear approximation gets better. Problem: Find the linearization of f (x; y) = x3 + 2xy − y 2 + 6 at (1; 2). Christopher Croke Calculus 115 As we zoom in linear approximation gets better. Problem: Find the linearization of f (x; y) = x3 + 2xy − y 2 + 6 at (1; 2). Christopher Croke Calculus 115 Let f have continuous first and second partial derivatives in an open set containing a rectangle R centered at (x0; y0). Let M be an upper bound for jfxx j,jfyy j and jfxy j on R. The error E(x; y) = L(x; y) − f (x; y) by using the linearization L of f at (x0; y0) instead of f satisfies Theorem 1 jE(x; y)j ≤ M(jx − x j + jy − y j)2: 2 0 0 Problem Find an upper bound for the error in our example on R: jx − 1j ≤ 0:1 and jy − 2j ≤ 0:1. What is the maximum relative error? The percentage error? How good? How good is this linear approximation? Christopher Croke Calculus 115 Problem Find an upper bound for the error in our example on R: jx − 1j ≤ 0:1 and jy − 2j ≤ 0:1. What is the maximum relative error? The percentage error? 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 (x0; y0). Let M be an upper bound for jfxx j,jfyy j and jfxy j on R. The error E(x; y) = L(x; y) − f (x; y) by using the linearization L of f at (x0; y0) instead of f satisfies Theorem 1 jE(x; y)j ≤ M(jx − x j + jy − y j)2: 2 0 0 Christopher Croke Calculus 115 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 (x0; y0). Let M be an upper bound for jfxx j,jfyy j and jfxy j on R. The error E(x; y) = L(x; y) − f (x; y) by using the linearization L of f at (x0; y0) instead of f satisfies Theorem 1 jE(x; y)j ≤ M(jx − x j + jy − y j)2: 2 0 0 Problem Find an upper bound for the error in our example on R: jx − 1j ≤ 0:1 and jy − 2j ≤ 0:1. What is the maximum relative error? The percentage error? Christopher Croke Calculus 115 Definition The Total Differential of f (x; y) at a point (x0; y0) is df = fx (x0; y0)dx + fy (x0; y0)dy: The Differential Recall for a function of one variable: Christopher Croke Calculus 115 The Differential Recall for a function of one variable: Definition The Total Differential of f (x; y) at a point (x0; y0) is df = fx (x0; y0)dx + fy (x0; y0)dy: Christopher Croke Calculus 115 The Differential Recall for a function of one variable: Definition The Total Differential of f (x; y) at a point (x0; y0) is df = fx (x0; y0)dx + fy (x0; y0)dy: Christopher Croke Calculus 115 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. It represents the change in the linearization as we move from (x0; y0) to a point (x0 + dx; y0 + dy) (nearby). Christopher Croke Calculus 115 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. It represents the change in the linearization as we move from (x0; y0) to a point (x0 + dx; y0 + dy) (nearby). Christopher Croke Calculus 115 It represents the change in the linearization as we move from (x0; y0) to a point (x0 + dx; y0 + 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. Christopher Croke Calculus 115 For f (x; y; z) at a point P0 = (x0; y0; z0) we have Linearization L(x; y; z) = f (P0)+fx (P0)(x−x0)+fy (P0)(y−y0)+fz (P0)(z−z0) Error in Linearization 1 jE(x; y)j ≤ M(jx − x j + jy − y j + jz − z j)2: 2 0 0 0 Differential df = fx (P0)dx + fy (P0)dy + fz (P0)dz More variables Things look pretty much the same for more variables. Christopher Croke Calculus 115 Error in Linearization 1 jE(x; y)j ≤ M(jx − x j + jy − y j + jz − z j)2: 2 0 0 0 Differential df = fx (P0)dx + fy (P0)dy + fz (P0)dz More variables Things look pretty much the same for more variables. For f (x; y; z) at a point P0 = (x0; y0; z0) we have Linearization L(x; y; z) = f (P0)+fx (P0)(x−x0)+fy (P0)(y−y0)+fz (P0)(z−z0) Christopher Croke Calculus 115 Differential df = fx (P0)dx + fy (P0)dy + fz (P0)dz More variables Things look pretty much the same for more variables.