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 {directional derivative} × {increment}. I.e

df = (∇f |(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 {directional derivative} × {increment}. I.e

df = (∇f |(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 {directional derivative} × {increment}. I.e

df = (∇f |(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 |fxx |,|fyy | and |fxy | 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 |E(x, y)| ≤ M(|x − x | + |y − y |)2. 2 0 0 Problem Find an upper bound for the error in our example on R: |x − 1| ≤ 0.1 and |y − 2| ≤ 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: |x − 1| ≤ 0.1 and |y − 2| ≤ 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 |fxx |,|fyy | and |fxy | 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 |E(x, y)| ≤ M(|x − x | + |y − y |)2. 2 0 0

Christopher Croke Calculus 115 How good?

E(x, y) = L(x, y) − f (x, y)

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 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 |E(x, y)| ≤ M(|x − x | + |y − y | + |z − z |)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 |E(x, y)| ≤ M(|x − x | + |y − y | + |z − z |)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. 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 |E(x, y)| ≤ M(|x − x | + |y − y | + |z − z |)2. 2 0 0 0

Christopher Croke Calculus 115 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)

Error in Linearization 1 |E(x, y)| ≤ M(|x − x | + |y − y | + |z − z |)2. 2 0 0 0

Diﬀerential

df = fx (P0)dx + fy (P0)dy + fz (P0)dz

Christopher Croke Calculus 115 The tangent plane of such a point will be horizontal.

Finding Maxima and Minima

For a function of two variables what does a relative maximum or relative minimum look like?

Christopher Croke Calculus 115 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.

Christopher Croke Calculus 115 First Derivative Test: If f (x, y) has either a relative maximum or minimum at at point (a, b) then ∂f ∂f (a, b) = 0 and (a, b) = 0. ∂x ∂y In other words ∇f (a, b) = 0.

What does that say about the partial derivatives?

Christopher Croke Calculus 115 In other words ∇f (a, b) = 0.

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 ∂f (a, b) = 0 and (a, b) = 0. ∂x ∂y

Christopher Croke Calculus 115 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 ∂f (a, b) = 0 and (a, b) = 0. ∂x ∂y In other words ∇f (a, b) = 0.

Christopher Croke Calculus 115 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 fx (a, b) = 0 and fy (a, b) = 0.

(a, b) is an interior point and fx (a, b) or fy (a, b) is not deﬁned. (a, b) is a boundary point. Points in the ﬁrst two category are called Critical Points of f .

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.

Finding Maxima and Minima on a region