SME
Calculus III (Multivariable)
Partial derivatives and gradients
Use this for “f(x, y) = x² + xy + 2y². Find the gradient and the tangent plane at (1, 2).”
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.
The tangent plane is z = 11 + 4(x − 1) + 9(y − 2).
Heights are squeezed to fit: one unit of z is drawn 0.075 as long as one of x or y.
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
at
(
x₀
,
y₀
):
∇f
=
⟨
∂f/∂x
,
∂f/∂y
⟩,
|∇f|
=
z
=
f(x₀, y₀)
+
∂f/∂x
(x
−
x₀
)
+
∂f/∂y
(y
−
y₀
)
x
Nearby point, x
example
y
Nearby point, y
example
L(x, y)
Tangent plane height there
worked out (not typed)
f(x, y)
True height there
worked out (not typed)
Formulas
f
(
x₀
,
y₀
)
=
a
×
x₀
2
+
b
×
x₀
×
y₀
+
c
×
y₀
2
+
d
×
x₀
+
e
×
y₀
11
=
1
×
1
2
+
1
×
1
×
2
+
2
×
2
2
+
0
×
1
+
0
×
2
∂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
)
L(
x
,
y
)
=
f
(
x₀
,
y₀
)
+
(∂f/∂
x
)
×
(
x
−
x₀
)
+
(∂f/∂
y
)
×
(
y
−
y₀
)
10.5 = 11 + 4 × (1.1 − 1) + 9 × (1.9 − 2)
f
(
x
,
y
)
=
a
×
x
2
+
b
×
x
×
y
+
c
×
y
2
+
d
×
x
+
e
×
y
10.52
=
1
×
1.1
2
+
1
×
1.1
×
1.9
+
2
×
1.9
2
+
0
×
1.1
+
0
×
1.9
Assumptions
•
f(x, y) = ax² + bxy + cy² + dx + ey. ∂f/∂x holds y fixed and takes the slope in x; ∂f/∂y holds x fixed.
•
∇f = ⟨∂f/∂x, ∂f/∂y⟩ points uphill fastest, at the rate |∇f|, and is square to the level curve through the point.
•
The tangent plane z = f(x₀, y₀) + (∂f/∂x)(x − x₀) + (∂f/∂y)(y − y₀) is the linear approximation L: near the point, f ≈ L.
Step-by-step
WE KNOW
a = 1, b = 1, c = 2, d = 0, e = 0, x₀ = 1, y₀ = 2, x = 1.1, y = 1.9
FIND
height at the point (f(x₀, y₀)), slope in x at the point (∂f/∂x), slope in y at the point (∂f/∂y), length of the gradient (|∇f|), tangent plane height there (L(x, y)), true height there (f(x, y))
Step 1 · Find height at the point (f(x₀, y₀))
Use f(x₀, y₀) = a × x₀² + b × x₀ × y₀ + c × y₀² + d × x₀ + e × y₀
Put
x₀
and
y₀
into
f.
f
(
x₀
,
y₀
)
=
1
×
1
2
+
1
×
1
×
2
+
2
×
2
2
+
0
×
1
+
0
×
2
f
(
x₀
,
y₀
)
=
1
+
2
+
8
f
(
x₀
,
y₀
)
=
11
Step 2 · 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
+
2
y
²
∂f/∂
x
=
2
x
+
y
∂f/∂
x
=
4
Step 3 · 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
+
4
y
∂f/∂
y
=
9
Step 4 · 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 5 · Find tangent plane height there (L(x, y))
Use L(x, y) = f(x₀, y₀) + (∂f/∂x) × (x − x₀) + (∂f/∂y) × (y − y₀)
Start
at
the
height
of
the
point
and
follow
each
slope
for
its
step
in
x
and
in
y
.
L(
x
,
y
)
=
11
+
4
×
(1.1
−
1)
+
9
×
(1.9
−
2)
L(
x
,
y
)
=
11
+
4
×
0.1
+
9
×
(−0.1)
L(
x
,
y
)
=
11
+
0.4
+
(−0.9)
L(
x
,
y
)
=
10.5
→
The
tangent
plane:
z
=
11
+
4(
x
−
1)
+
9(
y
−
2)
Step 6 · Find true height there (f(x, y))
Use f(x, y) = a × x² + b × x × y + c × y² + d × x + e × y
f itself at the nearby point, to compare: the closer the point, the closer L is to f.
f
(
x
,
y
)
=
1
×
1.1
2
+
1
×
1.1
×
1.9
+
2
×
1.9
2
+
0
×
1.1
+
0
×
1.9
f
(
x
,
y
)
=
1.21
+
2.09
+
7.22
f
(
x
,
y
)
=
10.52
Check
Put the numbers back in. Both sides should match.
11
=
1
×
1
2
+
1
×
1
×
2
+
2
×
2
2
+
0
×
1
+
0
×
2
✓
4 = 2 × 1 × 1 + 1 × 2 + 0 ✓
9 = 1 × 1 + 2 × 2 × 2 + 0 ✓
9.8489
=
√(
(4)
2
+
(9)
2
)
✓
10.5 = 11 + 4 × (1.1 − 1) + 9 × (1.9 − 2) ✓
10.52
=
1
×
1.1
2
+
1
×
1.1
×
1.9
+
2
×
1.9
2
+
0
×
1.1
+
0
×
1.9
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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