Lesson 15: Gradients and level curves

Slide 1: . Section 11.6 Gradients 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 . . . . . . .

Slide 2: 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 . . . . . .

Slide 3: Outline Definition of the gradient Plotting the gradient Gradients and Level Curves Directional Derivatives . . . . . .

Slide 4: Another kind of derivative expression Let f be a function of two variables. The total differential of f is the expression ∂f ∂f df = dx + dy ∂x ∂y But what is this? One way to think about it is as a vector. . . . . . .

Slide 5: Another kind of derivative expression Let f be a function of two variables. The total differential of f is the expression ∂f ∂f df = dx + dy ∂x ∂y But what is this? One way to think about it is as a vector. Definition Let f be a function⟨ two (or three variables). The gradient of f at of ⟩ ∂f ∂f (x, y) is the vector , (add on the last partial if it’s 3D). ∂x ∂y . . . . . .

Slide 6: Examples Example Find the gradient of f(x, y) = 2x + y. . . . . . .

Slide 7: Examples Example Find the gradient of f(x, y) = 2x + y. Solution ⟨ ⟩ ∂f ∂f ∇f(x, y) = , = ⟨2 , 1 ⟩ ∂x ∂y . . . . . .

Slide 8: Examples Example Find the gradient of f(x, y) = 2x + y. Solution ⟨ ⟩ ∂f ∂f ∇f(x, y) = , = ⟨2 , 1 ⟩ ∂x ∂y Example Find the gradient of f(x, y) = x2 + y2 . . . . . . .

Slide 9: Examples Example Find the gradient of f(x, y) = 2x + y. Solution ⟨ ⟩ ∂f ∂f ∇f(x, y) = , = ⟨2 , 1 ⟩ ∂x ∂y Example Find the gradient of f(x, y) = x2 + y2 . Solution ⟨ ⟩ ∂f ∂f ∇f ( x , y ) = , = ⟨2x, 2y⟩ ∂x ∂y . . . . . .

Slide 10: Examples Example Find the gradient of f(x, y) = x2 − y2 . . . . . . .

Slide 11: Examples Example Find the gradient of f(x, y) = x2 − y2 . Solution ⟨ ⟩ ∂f ∂f ∇f(x, y) = , = ⟨2x, −2y⟩ ∂x ∂y . . . . . .

Slide 12: Examples Example Find the gradient of f(x, y) = x2 − y2 . Solution ⟨ ⟩ ∂f ∂f ∇f(x, y) = , = ⟨2x, −2y⟩ ∂x ∂y Example Find the gradient of f(x, y) = x3/4 y1/4 . . . . . . .

Slide 13: Examples Example Find the gradient of f(x, y) = x2 − y2 . Solution ⟨ ⟩ ∂f ∂f ∇f(x, y) = , = ⟨2x, −2y⟩ ∂x ∂y Example Find the gradient of f(x, y) = x3/4 y1/4 . Solution ⟨ ⟩ ⟨ ⟩ ∂f ∂f 3 −1/4 1/4 1 3/4 −3/4 ∇f(x, y) = , = x y , x y ∂x ∂y 4 4 . . . . . .

Slide 14: Examples Example A three-variable one: Find the gradient of f(x, y, z) = e4x sin(2y + 3z). . . . . . .

Slide 15: Examples Example A three-variable one: Find the gradient of f(x, y, z) = e4x sin(2y + 3z). Solution ⟨ ⟩ ⟨ ∂f ∂f ∂f ∇f ( x , y , z ) = , , = 4e4x sin(2y + 3z), 2e4x cos(2y + 3z), 3e4x co ∂x ∂y ∂z . . . . . .

Slide 16: Outline Definition of the gradient Plotting the gradient Gradients and Level Curves Directional Derivatives . . . . . .

Slide 17: Example Plot the gradient of f(x, y) = 2x + y. . . . . . .

Slide 18: Example Plot the gradient of f(x, y) = 2x + y. . . . . . .

Slide 19: Example Plot the gradient of f(x, y) = x2 + y2 . . . . . . .

Slide 20: Example Plot the gradient of f(x, y) = x2 + y2 . . . . . . .

Slide 21: Example Plot the gradient of f(x, y) = x2 − y2 . . . . . . .

Slide 22: Example Plot the gradient of f(x, y) = x2 − y2 . . . . . . .

Slide 23: Example Plot the gradient of f(x, y) = x3/4 y1/4 . . . . . . .

Slide 24: Example Plot the gradient of f(x, y) = x3/4 y1/4 . . . . . . .

Slide 25: Outline Definition of the gradient Plotting the gradient Gradients and Level Curves Directional Derivatives . . . . . .

Slide 26: Theorem On the graph of z = f(x, y), ∇f points in the direction in which f grows the fastest. . . . . . .

Slide 27: Theorem On the graph of z = f(x, y), ∇f points in the direction in which f grows the fastest. Theorem The gradient ∇f is normal to the level curves f = c. . . . . . .

Slide 28: Tangent planes Example Find the equations for the the tangent plane and normal line to the √ surface x2 − y2 + z2 = 1 at the point (2, 3, 6). . . . . . .

Slide 29: Tangent planes Example Find the equations for the the tangent plane and normal line to the √ surface x2 − y2 + z2 = 1 at the point (2, 3, 6). Solution Let F(x, y, z) = x2 − y2 + z2 . Then √ √ ∇F(2, 3, 6) = ⟨2x, −2y, 2z⟩|(2,3,√6) = (4, −6, 2 6) 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 6t . . . . . .

Slide 30: Outline Definition of the gradient Plotting the gradient Gradients and Level Curves Directional Derivatives . . . . . .

Slide 31: Definition Let f be a function defined near a point P(x0 , y0 ), and u = ⟨a, b⟩ a unit vector. The directional derivative of f at (x0 , y0 ) in the direction ̌ is defined by f(x0 + ha, y0 + hb) − f(x0 , y0 ) Du f(x0 , y0 ) = lim h→0 h . . . . . .

Slide 32: Fact We have D u f ( x 0 , y 0 ) = ∇f ( x 0 , y 0 ) · u . . . . . .

Slide 33: Fact We have D u f ( x 0 , y 0 ) = ∇f ( x 0 , y 0 ) · u Proof. Use the chain rule. . . . . . .

Slide 34: Theorem ◮ On the contour plot of f, ∇f points “uphill”, i.e., in the direction of greatest increase of f. ◮ The length |∇f| is the amount of increase in that direction. . . . . . .

Slide 35: Who cares? The gradient and the differential of a function contain the same amount of information: a list of the partial derivatives. The gradient has a geometric significance which will be useful to visualize things as we get into two of the biggest topics in multivariable differential calculus: ◮ Unconstrained optimization of functions of several variables ◮ Constrained optimization of functions of several variables . . . . . .