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

Section 14.5, 14.6

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Friday, Sept 22nd, 2017

Exam I Monday Evening,No class on Wednesday

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Math 241: Problem of the day

Use the chain rule to calculate

∂w

∂t(1, 0),

where w = x2 + yz2

,

• x = x(s, t) = s cos(t),

• y = s + t,

• z = es2+t2 .

go.illinois.edu/math241fa17.

Math 241: Problem of the day

Use the chain rule to calculate

∂w

∂t(1, 0),

where w = x2 + yz2

,

• x = x(s, t) = s cos(t),

• y = s + t,

• z = es2+t2 .

go.illinois.edu/math241fa17.

Math 241: Problem of the day

Use the chain rule to calculate

∂w

∂t(1, 0),

where w = x2 + yz2

,

• x = x(s, t) = s cos(t),

• y = s + t,

• z = es2+t2 .

go.illinois.edu/math241fa17.

Chain rule

For a function f (x1, . . . , xn) and functionsx1 = x1(t1, . . . , tk), . . . , xn = xn(t1, . . . , tk) the chain rule says:

∂f

∂tj=

∂

∂tjf (x1(t1, . . . , tk), . . . , xn(t1, . . . , tk)) =

n∑i=1

∂f

∂xi

∂xi∂tj

.

go.illinois.edu/math241fa17.

Chain rule

For a function f (x1, . . . , xn) and functionsx1 = x1(t1, . . . , tk), . . . , xn = xn(t1, . . . , tk) the chain rule says:

∂f

∂tj=

∂

∂tjf (x1(t1, . . . , tk), . . . , xn(t1, . . . , tk)) =

n∑i=1

∂f

∂xi

∂xi∂tj

.

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

∂x

dx

dx+

∂F

∂y

dy

dx= 0 =⇒ dy

dx= −

∂F∂x∂F∂y

.

Example 0.1

Suppose that x , y are related by

x3 − y3 − 6xy = 0.

Assume y = y(x) and find dydx .

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

∂x

dx

dx+

∂F

∂y

dy

dx= 0 =⇒ dy

dx= −

∂F∂x∂F∂y

.

Example 0.1

Suppose that x , y are related by

x3 − y3 − 6xy = 0.

Assume y = y(x) and find dydx .

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

∂x

dx

dx+

∂F

∂y

dy

dx= 0 =⇒ dy

dx= −

∂F∂x∂F∂y

.

Example 0.1

Suppose that x , y are related by

x3 − y3 − 6xy = 0.

Assume y = y(x) and find dydx .

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

∂x

dx

dx+

∂F

∂y

dy

dx= 0 =⇒ dy

dx= −

∂F∂x∂F∂y

.

Example 0.1

Suppose that x , y are related by

x3 − y3 − 6xy = 0.

Assume y = y(x) and find dydx .

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 functionz = z(x , y). So z has partial derivatives with respect to x , y . Takepartial derivative first with respect to x and use the chain rule:

∂F

∂x

∂x

∂x+

∂F

∂y

∂y

∂x+

∂F

∂z

∂z

∂x= 0

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

∂z

∂x= −

∂F∂x∂F∂z

, same way:∂z

∂y= −

∂F∂y

∂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 functionz = z(x , y).

So z has partial derivatives with respect to x , y . Takepartial derivative first with respect to x and use the chain rule:

∂F

∂x

∂x

∂x+

∂F

∂y

∂y

∂x+

∂F

∂z

∂z

∂x= 0

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

∂z

∂x= −

∂F∂x∂F∂z

, same way:∂z

∂y= −

∂F∂y

∂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 functionz = z(x , y). So z has partial derivatives with respect to x , y .

Takepartial derivative first with respect to x and use the chain rule:

∂F

∂x

∂x

∂x+

∂F

∂y

∂y

∂x+

∂F

∂z

∂z

∂x= 0

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

∂z

∂x= −

∂F∂x∂F∂z

, same way:∂z

∂y= −

∂F∂y

∂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 functionz = z(x , y). So z has partial derivatives with respect to x , y . Takepartial derivative first with respect to x and use the chain rule:

∂F

∂x

∂x

∂x+

∂F

∂y

∂y

∂x+

∂F

∂z

∂z

∂x= 0

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

∂z

∂x= −

∂F∂x∂F∂z

, same way:∂z

∂y= −

∂F∂y

∂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 functionz = z(x , y). So z has partial derivatives with respect to x , y . Takepartial derivative first with respect to x and use the chain rule:

∂F

∂x

∂x

∂x+

∂F

∂y

∂y

∂x+

∂F

∂z

∂z

∂x= 0

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

∂z

∂x= −

∂F∂x∂F∂z

, same way:∂z

∂y= −

∂F∂y

∂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 functionz = z(x , y). So z has partial derivatives with respect to x , y . Takepartial derivative first with respect to x and use the chain rule:

∂F

∂x

∂x

∂x+

∂F

∂y

∂y

∂x+

∂F

∂z

∂z

∂x= 0

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

∂z

∂x= −

∂F∂x∂F∂z

,

same way:∂z

∂y= −

∂F∂y

∂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 functionz = z(x , y). So z has partial derivatives with respect to x , y . Takepartial derivative first with respect to x and use the chain rule:

∂F

∂x

∂x

∂x+

∂F

∂y

∂y

∂x+

∂F

∂z

∂z

∂x= 0

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

∂z

∂x= −

∂F∂x∂F∂z

, same way:∂z

∂y= −

∂F∂y

∂F∂z

go.illinois.edu/math241fa17.

Example 0.2

Suppose that z is given implicitly as a function z = z(x , y) by theequation:

x2 + 2y2 + 3z2 = 1.

Knowing that z(0, 0) = 1/√

3, find:

∂z

∂x,

∂z

∂y.

go.illinois.edu/math241fa17.

Directional derivatives

f : R2 → R and a UNIT vector u, how does f change in thedirection u at the point (a, b)?

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

Duf (a, b) =d

dt

∣∣∣∣t=0

f (a + tu1, b + tu2)

here u = 〈u1, u2〉. (Evaluate the derivative at t = 0.)

go.illinois.edu/math241fa17.

Directional derivatives

f : R2 → R and a UNIT vector u, how does f change in thedirection u at the point (a, b)?

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

Duf (a, b) =d

dt

∣∣∣∣t=0

f (a + tu1, b + tu2)

here u = 〈u1, u2〉. (Evaluate the derivative at t = 0.)

go.illinois.edu/math241fa17.

Directional derivatives

f : R2 → R and a UNIT vector u, how does f change in thedirection u at the point (a, b)?

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

Duf (a, b) =d

dt

∣∣∣∣t=0

f (a + tu1, b + tu2)

here u = 〈u1, u2〉. (Evaluate the derivative at t = 0.)

go.illinois.edu/math241fa17.

Computation, using gradient and dot product.

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

In general, for u = 〈u1, u2〉, 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 ascomponents:

∇f (a, b) = 〈fx(a, b), fy (a, b)〉.Then the directional derivative is a dot product:Duf (a, b) = ∇f (a, b) · u.

Example: Calculate Duf (2,−1) for f (x , y) = x3y + y2x and

u = 〈 1√2, −1√

2〉.

go.illinois.edu/math241fa17.

Computation, using gradient and dot product.

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

In general, for u = 〈u1, u2〉, 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 ascomponents:

∇f (a, b) = 〈fx(a, b), fy (a, b)〉.Then the directional derivative is a dot product:Duf (a, b) = ∇f (a, b) · u.

Example: Calculate Duf (2,−1) for f (x , y) = x3y + y2x and

u = 〈 1√2, −1√

2〉.

go.illinois.edu/math241fa17.

Computation, using gradient and dot product.

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

In general, for u = 〈u1, u2〉, 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 ascomponents:

∇f (a, b) = 〈fx(a, b), fy (a, b)〉.Then the directional derivative is a dot product:Duf (a, b) = ∇f (a, b) · u.

Example: Calculate Duf (2,−1) for f (x , y) = x3y + y2x and

u = 〈 1√2, −1√

2〉.

go.illinois.edu/math241fa17.

Computation, using gradient and dot product.

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

In general, for u = 〈u1, u2〉, 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 ascomponents:

∇f (a, b) = 〈fx(a, b), fy (a, b)〉.Then the directional derivative is a dot product:Duf (a, b) = ∇f (a, b) · u.

Example: Calculate Duf (2,−1) for f (x , y) = x3y + y2x and

u = 〈 1√2, −1√

2〉.

go.illinois.edu/math241fa17.

Computation, using gradient and dot product.

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

In general, for u = 〈u1, u2〉, 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 ascomponents:

∇f (a, b) = 〈fx(a, b), fy (a, b)〉.Then the directional derivative is a dot product:Duf (a, b) = ∇f (a, b) · u.

Example: Calculate Duf (2,−1) for f (x , y) = x3y + y2x and

u = 〈 1√2, −1√

2〉.

go.illinois.edu/math241fa17.

Computation, using gradient and dot product.

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

In general, for u = 〈u1, u2〉, 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 ascomponents:

∇f (a, b) = 〈fx(a, b), fy (a, b)〉.Then the directional derivative is a dot product:Duf (a, b) = ∇f (a, b) · u.

Example: Calculate Duf (2,−1) for f (x , y) = x3y + y2x and

u = 〈 1√2, −1√

2〉.

go.illinois.edu/math241fa17.

Arbitrary dimension

For f : Rn → R and a unit vector u = 〈u1, . . . , un〉 we can similarlydefine

Duf (a1, . . . , an) =d

dt

∣∣∣∣t=0

f (a1 + tu1, . . . , an + tun).

This is calculated as a dot product with the gradient∇f = 〈fx1 , . . . , fxn〉:

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

go.illinois.edu/math241fa17.

Arbitrary dimension

For f : Rn → R and a unit vector u = 〈u1, . . . , un〉 we can similarlydefine

Duf (a1, . . . , an) =d

dt

∣∣∣∣t=0

f (a1 + tu1, . . . , an + tun).

This is calculated as a dot product with the gradient∇f = 〈fx1 , . . . , fxn〉:

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

go.illinois.edu/math241fa17.

Arbitrary dimension

For f : Rn → R and a unit vector u = 〈u1, . . . , un〉 we can similarlydefine

Duf (a1, . . . , an) =d

dt

∣∣∣∣t=0

f (a1 + tu1, . . . , an + tun).

This is calculated as a dot product with the gradient∇f = 〈fx1 , . . . , fxn〉:

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

go.illinois.edu/math241fa17.