SME
Calculus I · Problem type
A sliding ladder
Use this for “A 10 m ladder leans on a wall and its foot slides away at 1 m/s. How fast is the top sliding down when the foot is 6 m from the wall?”
Diagram
dy/dt = −0.375 m/s, dx/dt = 0.5 m/s
y² + x² = L², so y·dy/dt + x·dx/dt = L·L′: 4 × (−0.375) + 3 × 0.5 = 0 = 5 × 0 (L is fixed)
L is fixed (L′ = 0)
This is an example. Type your own values to start a new problem.
(
x (m)
)
2
+
(
y (m)
)
2
=
(
L (m)
)
2
x (m)
(
dx/dt (m/s)
)
+
y (m)
(
dy/dt (m/s)
)
=
0
Formulas
x
2
+
y
2
=
L
2
3
2
+
4
2
=
5
2
x
×
(dx/dt)
+
y
×
(dy/dt)
=
0
3 × 0.5 + 4 × (−0.375) = 0
Assumptions
•
The wall is vertical and the ground level, so the ladder, the wall and the ground make a right triangle: x² + y² = L² at every instant.
•
The ladder’s length L never changes. Differentiating both sides in time t gives 2x·(dx/dt) + 2y·(dy/dt) = 0.
•
A positive dx/dt means the foot moves away from the wall; a negative dy/dt means the top slides down. Lengths are in m and rates in m/s.
•
Near the ground (y → 0) the model gives a top speed without bound: the top leaves the wall before that.
Step-by-step
WE KNOW
L = 5 m, x = 3 m, dx/dt = 0.5 m/s
FIND
top’s height up the wall (y), top’s rate up the wall (dy/dt)
Step 1 · Find top’s height up the wall (y)
Use x² + y² = L²
The
ladder
is
the
hypotenuse:
take
x
2
from
L
2
,
then
the
square
root.
y
=
√
L
2
−
x
2
y
=
√
5
2
−
3
2
y
=
√
25 − 9
y
=
√
16
y
=
4
m
Step 2 · Find top’s rate up the wall (dy/dt)
Use x × (dx/dt) + y × (dy/dt) = 0
Differentiate
x
2
+
y
2
=
L
2
in
t
with
L
fixed,
divide
by
2,
then
solve
for
dy/dt.
dy/dt
=
−
x
× (dx/dt)
y
dy/dt
=
−
3 × 0.5
4
dy/dt
=
−
1.5
4
dy/dt = −0.375 m/s
Check
Put the numbers back in. Both sides should match.
3
2
+
4
2
=
5
2
✓
3 × 0.5 + 4 × (−0.375) = 0 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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