SME
Calculus III (Multivariable) · Problem type
Directional derivative
Use this for “How fast does f(x, y) = x² + xy + 2y² change at (1, 2) in the direction of ⟨3, 4⟩?”
Diagram
f(x, y) = x² + xy + 2y².
At (1, 2): f = 11, ∂f/∂x = 4 and ∂f/∂y = 9.
|∇f|
=
√
4
2
+
9
2
=
9.849.
u = ⟨3, 4⟩ ÷ 5 = ⟨0.6, 0.8⟩, so Dᵤf = ∇f·u = 4 × 0.6 + 9 × 0.8 = 9.6, at most |∇f| = 9.849.
This is an example. Type your own values to start a new problem.
f(x,
y)
=
a
x²
+
b
xy
+
c
y²
+
d
x
+
e
y
u
along
⟨
p
,
q
⟩
Dᵤf(
x₀
,
y₀
)
=
Dᵤf
∂f/∂x
Slope in x at the point
worked out (not typed)
∂f/∂y
Slope in y at the point
worked out (not typed)
|∇f|
Length of the gradient
worked out (not typed)
|⟨p, q⟩|
Length of ⟨p, q⟩
worked out (not typed)
u₁
Unit direction, x part
worked out (not typed)
u₂
Unit direction, y part
worked out (not typed)
Formulas
∂f/∂x
=
2
×
a
×
x₀
+
b
×
y₀
+
d
4 = 2 × 1 × 1 + 1 × 2 + 0
∂f/∂y
=
b
×
x₀
+
2
×
c
×
y₀
+
e
9 = 1 × 1 + 2 × 2 × 2 + 0
|∇f|
=
√(
(∂f/∂x)
2
+
(∂f/∂y)
2
)
9.8489
=
√(
(4)
2
+
(9)
2
)
|⟨
p
,
q
⟩|
=
√
p
2
+
q
2
5
=
√
3
2
+
4
2
u₁
=
p
|⟨
p
,
q
⟩|
0.6
=
3
5
u₂
=
q
|⟨
p
,
q
⟩|
0.8
=
4
5
Dᵤf
=
(∂f/∂x)
×
u₁
+
(∂f/∂y)
×
u₂
9.6 = 4 × 0.6 + 9 × 0.8
Assumptions
•
f(x, y) = ax² + bxy + cy² + dx + ey, with ∇f = ⟨∂f/∂x, ∂f/∂y⟩ at the point.
•
The direction ⟨p, q⟩ is made a unit vector u = ⟨p, q⟩ ÷ |⟨p, q⟩| first, so only its direction counts.
•
Dᵤf = ∇f · u. It is largest, |∇f|, along ∇f itself and 0 along the level curve.
Step-by-step
WE KNOW
a = 1, b = 1, c = 2, d = 0, e = 0, x₀ = 1, y₀ = 2, p = 3, q = 4
FIND
slope in x at the point (∂f/∂x), slope in y at the point (∂f/∂y), length of the gradient (|∇f|), length of ⟨p, q⟩ (|⟨p, q⟩|), unit direction, x part (u₁), unit direction, y part (u₂), rate of change toward u (Dᵤf)
Step 1 · Find slope in x at the point (∂f/∂x)
Use ∂f/∂x = 2 × a × x₀ + b × y₀ + d
Hold
y
fixed:
ax²
gives
2ax,
bxy
gives
by,
dx
gives
d
,
and
the
terms
in
y
alone
give
0.
∂f/∂x = 2 × 1 × 1 + 1 × 2 + 0
f
(x,
y)
=
x
2
+
xy
+
2y²
∂f/∂x = 2x + y
∂f/∂x = 4
Step 2 · Find slope in y at the point (∂f/∂y)
Use ∂f/∂y = b × x₀ + 2 × c × y₀ + e
Hold
x
fixed:
bxy
gives
bx,
cy²
gives
2cy,
ey
gives
e
,
and
the
terms
in
x
alone
give
0.
∂f/∂y = 1 × 1 + 2 × 2 × 2 + 0
∂f/∂y = x + 4y
∂f/∂y = 9
Step 3 · Find length of the gradient (|∇f|)
Use |∇f| = √((∂f/∂x)² + (∂f/∂y)²)
∇f = ⟨4, 9⟩ at the point. Its length is the steepest rate of climb there.
|∇f|
=
√(
(4)
2
+
(9)
2
)
|∇f|
=
√
16 + 81
|∇f|
=
√
97
|∇f| = 9.8489
Step 4 · Find length of ⟨p, q⟩ (|⟨p, q⟩|)
Use |⟨p, q⟩| = √(p² + q²)
The length of the direction vector, by Pythagoras.
|⟨
p
,
q
⟩|
=
√
3
2
+
4
2
|⟨
p
,
q
⟩|
=
√
9 + 16
|⟨
p
,
q
⟩|
=
√
25
|⟨
p
,
q
⟩|
=
5
Step 5 · Find unit direction, x part (u₁)
Use u₁ = p ÷ (|⟨p, q⟩|)
Divide
⟨
p
,
q
⟩
by
its
length
to
make
a
unit
vector
u.
u₁
=
3
5
u₁
=
0.6
Step 6 · Find unit direction, y part (u₂)
Use u₂ = q ÷ (|⟨p, q⟩|)
The other part of u, divided by the same length.
u₂
=
4
5
u₂
=
0.8
Step 7 · Find rate of change toward u (Dᵤf)
Use Dᵤf = (∂f/∂x) × u₁ + (∂f/∂y) × u₂
The dot product of ∇f and u: multiply the matching parts and add. It can never pass |∇f| = 9.8489.
Dᵤf = 4 × 0.6 + 9 × 0.8
Dᵤf = 2.4 + 7.2
Dᵤf = 9.6
Check
Put the numbers back in. Both sides should match.
4 = 2 × 1 × 1 + 1 × 2 + 0 ✓
9 = 1 × 1 + 2 × 2 × 2 + 0 ✓
9.8489
=
√(
(4)
2
+
(9)
2
)
✓
5
=
√
3
2
+
4
2
✓
0.6
=
3
5
✓
0.8
=
4
5
✓
9.6 = 4 × 0.6 + 9 × 0.8 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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