SME
Calculus III (Multivariable) · Problem type
Curvature of y = ax² at a point
Use this for “Find the curvature and the radius of curvature of y = x² at x = 1.”
Chart
f(1) = 1
Tangent at x = 1: slope f′(1) = 2, y = 2x − 1
This is an example. Type your own values to start a new problem.
y
=
a
x²,
x
=
x
κ
=
|
y″
|
÷
(1
+
y′
2
)
3
2
=
κ
y
Height y
worked out (not typed)
R
Radius of curvature
worked out (not typed)
Formulas
y
=
a
×
x
2
1
=
1
×
1
2
y
′
=
2
×
a
×
x
2 = 2 × 1 × 1
y
″
=
2
×
a
2 = 2 × 1
κ
=
|
y
″|
÷
(1 +
y
′²)
3
2
0.1789
=
|2|
÷
(1 +
2
2
)
3
2
R
=
1
κ
5.5902
=
1
0.1789
Assumptions
•
κ = |y″| ÷ (1 + y′²)3/2 is the curvature of a graph y = f(x): it is |r′ × r″| ÷ |r′|³ for r(x) = ⟨x, f(x)⟩.
•
The parabola bends most at its vertex x = 0, where κ = 2|a|, and flattens as |x| grows.
•
R = 1 ÷ κ is the radius of the circle that hugs the curve at the point.
Step-by-step
WE KNOW
a = 1, x = 1
FIND
height y (y), slope y′ (y′), second derivative y″ (y″), curvature (κ), radius of curvature (R)
Step 1 · Find height y (y)
Use y = a × x²
Put
x
into
y
=
ax²
to
find
the
point
on
the
curve.
y
=
1
×
1
2
y
=
1
Step 2 · Find slope y′ (y′)
Use y′ = 2 × a × x
The power rule: the slope of ax² is 2ax.
y
′
=
2
×
1
×
1
y
′
=
2
Step 3 · Find second derivative y″ (y″)
Use y″ = 2 × a
Differentiate
2ax
once
more:
the
slope
of
the
slope
is
2
a
.
y
″
=
2
×
1
y
″
=
2
Step 4 · Find curvature (κ)
Use κ = |y″| ÷ (1 + y′²)3/2
The
bending
y
″
is
shared
over
the
length:
(1 +
y
′²)
3
2
grows
as
the
curve
gets
steep.
κ
=
|2|
÷
(1 +
2
2
)
3
2
κ
=
|2|
÷
(1 + 4)
3
2
κ
=
|2|
÷
5
3
2
κ
=
2
11.18
κ
=
0.178885
Step 5 · Find radius of curvature (R)
Use R = 1 ÷ κ
The
circle
that
hugs
the
curve
bends
as
much
as
it
does,
so
its
radius
is
1
κ
.
R
=
1
0.178885
R
=
5.5902
Check
Put the numbers back in. Both sides should match.
1
=
1
×
1
2
✓
2 = 2 × 1 × 1 ✓
2 = 2 × 1 ✓
0.178885
=
|2|
÷
(1 +
2
2
)
3
2
✓
5.5902
=
1
0.178885
✓
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