math 241: multivariable calculus, lecture 11nirobles/files241/... · 2017. 9. 22. · math 241:...
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Math 241: Multivariable calculus, Lecture 11Implicit 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.
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.