Math 241: Multivariable Calculus, Lecture 11 Implicit Differentiation, Directional Derivative and Gradient

Math 241: Multivariable calculus, Lecture 11 Implicit Differentiation, Directional Derivative and Gradient. Section 14.5, 14.6

go.illinois.edu/math241fa17

Friday, Sept 22nd, 2017

Exam I Monday Evening, No class on Wednesday

go.illinois.edu/math241fa17. • y = s + t, 2 2 • z = es +t .

Math 241: Problem of the day

Use the chain rule to calculate ∂w (1, 0), ∂t

2 y where w = x + z2 , • x = x(s, t) = s cos(t),

go.illinois.edu/math241fa17. 2 2 • z = es +t .

Math 241: Problem of the day

Use the chain rule to calculate ∂w (1, 0), ∂t

2 y where w = x + z2 , • x = x(s, t) = s cos(t), • y = s + t,

go.illinois.edu/math241fa17. Math 241: Problem of the day

Use the chain rule to calculate ∂w (1, 0), ∂t

2 y where w = x + z2 , • x = x(s, t) = s cos(t), • y = s + t, 2 2 • z = es +t .

go.illinois.edu/math241fa17. n ∂f ∂ X ∂f ∂xi = f (x1(t1,..., tk ),..., xn(t1,..., tk )) = . ∂tj ∂tj ∂xi ∂tj i=1

Chain rule

For a function f (x1,..., xn) and functions x1 = x1(t1,..., tk ),..., xn = xn(t1,..., tk ) the chain rule says:

go.illinois.edu/math241fa17. Chain rule

For a function f (x1,..., xn) and functions x1 = x1(t1,..., tk ),..., xn = xn(t1,..., tk ) the chain rule says:

n ∂f ∂ X ∂f ∂xi = f (x1(t1,..., tk ),..., xn(t1,..., tk )) = . ∂tj ∂tj ∂xi ∂tj i=1

go.illinois.edu/math241fa17. Example 0.1 Suppose that x, y are related by

x3 − y 3 − 6xy = 0.

dy Assume y = y(x) and find dx .

Usually you can solve for y, so that y = y(x) satisfying F (x, y(x)) = 0. Now take the derivative wrt x on both sides and use the chain rule: ∂F dx ∂F dy dy ∂F + = 0 =⇒ = − ∂x . ∂x dx ∂y dx dx ∂F ∂y

Implicit Differentiation.

Suppose x, y are related by an equation F (x, y) = 0

go.illinois.edu/math241fa17. Example 0.1 Suppose that x, y are related by

x3 − y 3 − 6xy = 0.

dy Assume y = y(x) and find dx .

Now take the derivative wrt x on both sides and use the chain rule: ∂F dx ∂F dy dy ∂F + = 0 =⇒ = − ∂x . ∂x dx ∂y dx dx ∂F ∂y

Implicit Differentiation.

Suppose x, y are related by an equation F (x, y) = 0 Usually you can solve for y, so that y = y(x) satisfying F (x, y(x)) = 0.

go.illinois.edu/math241fa17. Example 0.1 Suppose that x, y are related by

x3 − y 3 − 6xy = 0.

dy Assume y = y(x) and find dx .

Implicit Differentiation.

Suppose x, y are related by an equation F (x, y) = 0 Usually you can solve for y, so that y = y(x) satisfying F (x, y(x)) = 0. Now take the derivative wrt x on both sides and use the chain rule: ∂F dx ∂F dy dy ∂F + = 0 =⇒ = − ∂x . ∂x dx ∂y dx dx ∂F ∂y

go.illinois.edu/math241fa17. Implicit Differentiation.

Suppose x, y are related by an equation F (x, y) = 0 Usually you can solve for y, so that y = y(x) satisfying F (x, y(x)) = 0. Now take the derivative wrt x on both sides and use the chain rule: ∂F dx ∂F dy dy ∂F + = 0 =⇒ = − ∂x . ∂x dx ∂y dx dx ∂F ∂y

Example 0.1 Suppose that x, y are related by

x3 − y 3 − 6xy = 0.

dy Assume y = y(x) and find dx . go.illinois.edu/math241fa17. Usually you can solve z in terms of x, y, giving a function z = z(x, y). So z has partial derivatives with respect to x, y. Take partial derivative first with respect to x and use the chain rule: ∂F ∂x ∂F ∂y ∂F ∂z + + = 0 ∂x ∂x ∂y ∂x ∂z ∂x

∂x ∂y Now ∂x = 1, ∂x = 0, so that we find

∂F ∂z ∂F ∂z = − ∂x , same way: = − ∂y ∂x ∂F ∂y ∂F ∂z ∂z

Implicit Differentiation for more variables

Now assume that x, y, z are related by

F (x, y, z) = 0.

go.illinois.edu/math241fa17. So z has partial derivatives with respect to x, y. Take partial derivative first with respect to x and use the chain rule: ∂F ∂x ∂F ∂y ∂F ∂z + + = 0 ∂x ∂x ∂y ∂x ∂z ∂x

∂x ∂y Now ∂x = 1, ∂x = 0, so that we find

∂F ∂z ∂F ∂z = − ∂x , same way: = − ∂y ∂x ∂F ∂y ∂F ∂z ∂z

Implicit Differentiation for more variables

Now assume that x, y, z are related by

F (x, y, z) = 0.

Usually you can solve z in terms of x, y, giving a function z = z(x, y).

go.illinois.edu/math241fa17. Take partial derivative first with respect to x and use the chain rule: ∂F ∂x ∂F ∂y ∂F ∂z + + = 0 ∂x ∂x ∂y ∂x ∂z ∂x

∂x ∂y Now ∂x = 1, ∂x = 0, so that we find

∂F ∂z ∂F ∂z = − ∂x , same way: = − ∂y ∂x ∂F ∂y ∂F ∂z ∂z

Implicit Differentiation for more variables

Now assume that x, y, z are related by

F (x, y, z) = 0.

Usually you can solve z in terms of x, y, giving a function z = z(x, y). So z has partial derivatives with respect to x, y.

go.illinois.edu/math241fa17. ∂x ∂y Now ∂x = 1, ∂x = 0, so that we find

∂F ∂z ∂F ∂z = − ∂x , same way: = − ∂y ∂x ∂F ∂y ∂F ∂z ∂z

Implicit Differentiation for more variables

Now assume that x, y, z are related by

F (x, y, z) = 0.

Usually you can solve z in terms of x, y, giving a function z = z(x, y). So z has partial derivatives with respect to x, y. Take partial derivative first with respect to x and use the chain rule: ∂F ∂x ∂F ∂y ∂F ∂z + + = 0 ∂x ∂x ∂y ∂x ∂z ∂x

go.illinois.edu/math241fa17. ∂F ∂z ∂F ∂z = − ∂x , same way: = − ∂y ∂x ∂F ∂y ∂F ∂z ∂z

Implicit Differentiation for more variables

Now assume that x, y, z are related by

F (x, y, z) = 0.

Usually you can solve z in terms of x, y, giving a function z = z(x, y). So z has partial derivatives with respect to x, y. Take partial derivative first with respect to x and use the chain rule: ∂F ∂x ∂F ∂y ∂F ∂z + + = 0 ∂x ∂x ∂y ∂x ∂z ∂x

∂x ∂y Now ∂x = 1, ∂x = 0, so that we find

go.illinois.edu/math241fa17. ∂F ∂z same way: = − ∂y ∂y ∂F ∂z

Implicit Differentiation for more variables

Now assume that x, y, z are related by

F (x, y, z) = 0.

Usually you can solve z in terms of x, y, giving a function z = z(x, y). So z has partial derivatives with respect to x, y. Take partial derivative first with respect to x and use the chain rule: ∂F ∂x ∂F ∂y ∂F ∂z + + = 0 ∂x ∂x ∂y ∂x ∂z ∂x

∂x ∂y Now ∂x = 1, ∂x = 0, so that we find

∂z ∂F = − ∂x , ∂x ∂F ∂z

go.illinois.edu/math241fa17. Implicit Differentiation for more variables

Now assume that x, y, z are related by

F (x, y, z) = 0.

Usually you can solve z in terms of x, y, giving a function z = z(x, y). So z has partial derivatives with respect to x, y. Take partial derivative first with respect to x and use the chain rule: ∂F ∂x ∂F ∂y ∂F ∂z + + = 0 ∂x ∂x ∂y ∂x ∂z ∂x

∂x ∂y Now ∂x = 1, ∂x = 0, so that we find

∂F ∂z ∂F ∂z = − ∂x , same way: = − ∂y ∂x ∂F ∂y ∂F ∂z ∂z

go.illinois.edu/math241fa17. Example 0.2 Suppose that z is given implicitly as a function z = z(x, y) by the equation: x2 + 2y 2 + 3z2 = 1. √ Knowing that z(0, 0) = 1/ 3, find:

∂z ∂z , . ∂x ∂y

go.illinois.edu/math241fa17. 2 f : R → R and a UNIT vector u, how does f change in the direction u at the point (a, b)?

Define the directional derivative of f in the direction u at the point (a, b) to be

d Duf (a, b) = f (a + tu1, b + tu2) dt t=0

here u = hu1, u2i. (Evaluate the derivative at t = 0.)

Directional derivatives

go.illinois.edu/math241fa17. Define the directional derivative of f in the direction u at the point (a, b) to be

d Duf (a, b) = f (a + tu1, b + tu2) dt t=0

here u = hu1, u2i. (Evaluate the derivative at t = 0.)

Directional derivatives

2 f : R → R and a UNIT vector u, how does f change in the direction u at the point (a, b)?

go.illinois.edu/math241fa17. Directional derivatives

2 f : R → R and a UNIT vector u, how does f change in the direction u at the point (a, b)?

Define the directional derivative of f in the direction u at the point (a, b) to be

d Duf (a, b) = f (a + tu1, b + tu2) dt t=0 here u = hu1, u2i. (Evaluate the derivative at t = 0.)

go.illinois.edu/math241fa17. Example: What are Dif (a, b) and Djf (a, b)?

In general, for u = hu1, u2i, Duf (a, b) = ?

Duf (a, b) = fx (a, b)u1 + fy (a, b)u2

Define the gradient of f at (a, b) as the vector with partials as components: ∇f (a, b) = hfx (a, b), fy (a, b)i. Then the directional derivative is a dot product: Duf (a, b) = ∇f (a, b) · u.

3 2 Example: Calculate Duf (2, −1) for f (x, y) = x y + y x and u = h √1 , √−1 i. 2 2

Computation, using gradient and dot product.

go.illinois.edu/math241fa17. In general, for u = hu1, u2i, Duf (a, b) = ?

Duf (a, b) = fx (a, b)u1 + fy (a, b)u2

Define the gradient of f at (a, b) as the vector with partials as components: ∇f (a, b) = hfx (a, b), fy (a, b)i. Then the directional derivative is a dot product: Duf (a, b) = ∇f (a, b) · u.

3 2 Example: Calculate Duf (2, −1) for f (x, y) = x y + y x and u = h √1 , √−1 i. 2 2

Computation, using gradient and dot product.

Example: What are Dif (a, b) and Djf (a, b)?

go.illinois.edu/math241fa17. Duf (a, b) = fx (a, b)u1 + fy (a, b)u2

Define the gradient of f at (a, b) as the vector with partials as components: ∇f (a, b) = hfx (a, b), fy (a, b)i. Then the directional derivative is a dot product: Duf (a, b) = ∇f (a, b) · u.

3 2 Example: Calculate Duf (2, −1) for f (x, y) = x y + y x and u = h √1 , √−1 i. 2 2

Computation, using gradient and dot product.

Example: What are Dif (a, b) and Djf (a, b)?

In general, for u = hu1, u2i, Duf (a, b) = ?

go.illinois.edu/math241fa17. Define the gradient of f at (a, b) as the vector with partials as components: ∇f (a, b) = hfx (a, b), fy (a, b)i. Then the directional derivative is a dot product: Duf (a, b) = ∇f (a, b) · u.

3 2 Example: Calculate Duf (2, −1) for f (x, y) = x y + y x and u = h √1 , √−1 i. 2 2

Computation, using gradient and dot product.

Example: What are Dif (a, b) and Djf (a, b)?

In general, for u = hu1, u2i, Duf (a, b) = ?

Duf (a, b) = fx (a, b)u1 + fy (a, b)u2

go.illinois.edu/math241fa17. 3 2 Example: Calculate Duf (2, −1) for f (x, y) = x y + y x and u = h √1 , √−1 i. 2 2

Computation, using gradient and dot product.

Example: What are Dif (a, b) and Djf (a, b)?

In general, for u = hu1, u2i, Duf (a, b) = ?

Duf (a, b) = fx (a, b)u1 + fy (a, b)u2

Define the gradient of f at (a, b) as the vector with partials as components: ∇f (a, b) = hfx (a, b), fy (a, b)i. Then the directional derivative is a dot product: Duf (a, b) = ∇f (a, b) · u.

go.illinois.edu/math241fa17. Computation, using gradient and dot product.

Example: What are Dif (a, b) and Djf (a, b)?

In general, for u = hu1, u2i, Duf (a, b) = ?

Duf (a, b) = fx (a, b)u1 + fy (a, b)u2

Define the gradient of f at (a, b) as the vector with partials as components: ∇f (a, b) = hfx (a, b), fy (a, b)i. Then the directional derivative is a dot product: Duf (a, b) = ∇f (a, b) · u.

3 2 Example: Calculate Duf (2, −1) for f (x, y) = x y + y x and u = h √1 , √−1 i. 2 2 go.illinois.edu/math241fa17. n For f : R → R and a unit vector u = hu1,..., uni we can similarly define

d Duf (a1,..., an) = f (a1 + tu1,..., an + tun). dt t=0

This is calculated as a dot product with the gradient

∇f = hfx1 ,..., fxn i:

Duf (a1,..., an) = ∇f (a1,..., an) · u

Arbitrary dimension

go.illinois.edu/math241fa17. This is calculated as a dot product with the gradient

∇f = hfx1 ,..., fxn i:

Duf (a1,..., an) = ∇f (a1,..., an) · u

Arbitrary dimension

n For f : R → R and a unit vector u = hu1,..., uni we can similarly define

d Duf (a1,..., an) = f (a1 + tu1,..., an + tun). dt t=0

go.illinois.edu/math241fa17. Arbitrary dimension

n For f : R → R and a unit vector u = hu1,..., uni we can similarly define

d Duf (a1,..., an) = f (a1 + tu1,..., an + tun). dt t=0

This is calculated as a dot product with the gradient

∇f = hfx1 ,..., fxn i:

Duf (a1,..., an) = ∇f (a1,..., an) · u

go.illinois.edu/math241fa17.