lesson 15: gradients and level curves

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Section 11.6Gradients and Level Curves

Math 21a

March 10, 2008

Announcements◮ No Sophie session tonight. Problem sessions today:

◮ Lin Cong, 7:30 in SC 103b◮ Eleanor Birrell, 3:00pm in SC 310

◮ Office hours Tuesday, Wednesday 2–4pm SC 323◮ Midterm I, tomorrow, 7–9pm in SC Hall D

..Image: Flickr user Other Neither

. . . . . .

Announcements

◮ No Sophie session tonight. Problem sessions today:◮ Lin Cong, 7:30 in SC 103b◮ Eleanor Birrell, 3:00pm in SC 310

◮ Office hours Tuesday, Wednesday 2–4pm SC 323◮ Midterm I, tomorrow, 7–9pm in SC Hall D

. . . . . .

Outline

Definition of the gradient

Plotting the gradient

Gradients and Level Curves

Directional Derivatives

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Another kind of derivative expression

Let f be a function of two variables. The total differential of f isthe expression

df =∂f∂x

dx +∂f∂y

But what is this? One way to think about it is as a vector.

DefinitionLet f be a function of two (or three variables). The gradient of f at

(x, y) is the vector⟨

∂f∂x

,∂f∂y

⟩(add on the last partial if it’s 3D).

. . . . . .

Another kind of derivative expression

Let f be a function of two variables. The total differential of f isthe expression

df =∂f∂x

dx +∂f∂y

But what is this? One way to think about it is as a vector.

DefinitionLet f be a function of two (or three variables). The gradient of f at

(x, y) is the vector⟨

∂f∂x

,∂f∂y

⟩(add on the last partial if it’s 3D).

. . . . . .

Examples

ExampleFind the gradient of f(x, y) = 2x + y.

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

⟩= ⟨2, 1⟩

ExampleFind the gradient of f(x, y) = x2 + y2.

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

⟩= ⟨2x, 2y⟩

. . . . . .

Examples

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

⟩= ⟨2, 1⟩

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

⟩= ⟨2x, 2y⟩

. . . . . .

Examples

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

⟩= ⟨2, 1⟩

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

⟩= ⟨2x, 2y⟩

. . . . . .

Examples

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

⟩= ⟨2, 1⟩

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

⟩= ⟨2x, 2y⟩

. . . . . .

Examples

ExampleFind the gradient of f(x, y) = x2 − y2.

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

⟩= ⟨2x,−2y⟩

ExampleFind the gradient of f(x, y) = x3/4y1/4.

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

x−1/4y1/4,14

x3/4y−3/4⟩

. . . . . .

Examples

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

⟩= ⟨2x,−2y⟩

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

x−1/4y1/4,14

x3/4y−3/4⟩

. . . . . .

Examples

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

⟩= ⟨2x,−2y⟩

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

x−1/4y1/4,14

x3/4y−3/4⟩

. . . . . .

Examples

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

⟩= ⟨2x,−2y⟩

Solution

∇f(x, y) =

⟨∂f∂x

,∂f∂y

x−1/4y1/4,14

x3/4y−3/4⟩

. . . . . .

Examples

ExampleA three-variable one: Find the gradient of f(x, y, z) = e4x sin(2y + 3z).

Solution

∇f(x, y, z) =

⟨∂f∂x

,∂f∂y

,∂f∂z

⟨4e4x sin(2y + 3z), 2e4x cos(2y + 3z), 3e4x cos(2y + 3z)

. . . . . .

Examples

ExampleA three-variable one: Find the gradient of f(x, y, z) = e4x sin(2y + 3z).

Solution

∇f(x, y, z) =

⟨∂f∂x

,∂f∂y

,∂f∂z

⟨4e4x sin(2y + 3z), 2e4x cos(2y + 3z), 3e4x cos(2y + 3z)

. . . . . .

Outline

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ExamplePlot the gradient of f(x, y) = 2x + y.

. . . . . .

ExamplePlot the gradient of f(x, y) = 2x + y.

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ExamplePlot the gradient of f(x, y) = x2 + y2.

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ExamplePlot the gradient of f(x, y) = x2 + y2.

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ExamplePlot the gradient of f(x, y) = x2 − y2.

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ExamplePlot the gradient of f(x, y) = x2 − y2.

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ExamplePlot the gradient of f(x, y) = x3/4y1/4.

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ExamplePlot the gradient of f(x, y) = x3/4y1/4.

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Outline

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TheoremOn the graph of z = f(x, y), ∇f points in the direction in which f grows thefastest.

TheoremThe gradient ∇f is normal to the level curves f = c.

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TheoremOn the graph of z = f(x, y), ∇f points in the direction in which f grows thefastest.

TheoremThe gradient ∇f is normal to the level curves f = c.

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Tangent planes

ExampleFind the equations for the the tangent plane and normal line to thesurface x2 − y2 + z2 = 1 at the point (2, 3,

√6).

SolutionLet F(x, y, z) = x2 − y2 + z2. Then

∇F(2, 3,√

6) = ⟨2x,−2y, 2z⟩|(2,3,√

6) = (4,−6, 2√

So the tangent plane has equation

4(x − 2) − 6(y − 3) + 2√

6(z −√

6) = 0

and the normal line has parametric equations

x = 2 + 4t, y = 3 − 6t, z =√

6 + 2√

. . . . . .

Tangent planes

ExampleFind the equations for the the tangent plane and normal line to thesurface x2 − y2 + z2 = 1 at the point (2, 3,

√6).

SolutionLet F(x, y, z) = x2 − y2 + z2. Then

∇F(2, 3,√

6) = ⟨2x,−2y, 2z⟩|(2,3,√

6) = (4,−6, 2√

So the tangent plane has equation

4(x − 2) − 6(y − 3) + 2√

6(z −√

6) = 0

and the normal line has parametric equations

x = 2 + 4t, y = 3 − 6t, z =√

6 + 2√

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Outline

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DefinitionLet f be a function defined near a point P(x0, y0), and u = ⟨a, b⟩ a unitvector. The directional derivative of f at (x0, y0) in thedirection ̌ is defined by

Duf(x0, y0) = limh→0

f(x0 + ha, y0 + hb) − f(x0, y0)

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FactWe have

Duf(x0, y0) = ∇f(x0, y0) · u

Proof.Use the chain rule.

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FactWe have

Duf(x0, y0) = ∇f(x0, y0) · u

Proof.Use the chain rule.

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Theorem

◮ On the contour plot of f, ∇f points “uphill”, i.e., in the direction ofgreatest increase of f.

◮ The length |∇f| is the amount of increase in that direction.

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Who cares?

The gradient and the differential of a function contain the sameamount of information: a list of the partial derivatives. The gradienthas a geometric significance which will be useful to visualize things aswe get into two of the biggest topics in multivariable differentialcalculus:

◮ Unconstrained optimization of functions of several variables◮ Constrained optimization of functions of several variables

lesson 15: gradients and level curves

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