SME
Calculus I · Problem type
Two cars moving apart
Use this for “Two cars leave a crossroads, one east at 15 km/h and one north at 20 km/h. How fast is the gap between them growing when they are 12 km and 5 km out?”
Diagram
dy/dt = 80 km/h, dx/dt = 60 km/h, dD/dt = 100 km/h
y² + x² = D², so y·dy/dt + x·dx/dt = D·dD/dt: 40 × 80 + 30 × 60 = 5,000 = 50 × 100
This is an example. Type your own values to start a new problem.
(
D (km)
)
2
=
(
x (km)
)
2
+
(
y (km)
)
2
D
·
dD
dt
=
x
·
dx
dt
+
y
·
dy
dt
D (km)
(
dD/dt (km/h)
)
=
x (km)
(
dx/dt (km/h)
)
+
y (km)
(
dy/dt (km/h)
)
Formulas
D
2
=
x
2
+
y
2
50
2
=
30
2
+
40
2
D
×
(dD/dt)
=
x
×
(dx/dt)
+
y
×
(dy/dt)
50 × 100 = 30 × 60 + 40 × 80
Assumptions
•
The roads cross at a right angle, so the cars and the crossing make a right triangle: D² = x² + y² at every instant.
•
Differentiating both sides in time t and dividing by 2 gives D·(dD/dt) = x·(dx/dt) + y·(dy/dt).
•
A positive rate means that distance is growing; a car driving back toward the crossing has a negative rate.
•
Distances are in km and rates in km/h.
Step-by-step
WE KNOW
x = 30 km, y = 40 km, dx/dt = 60 km/h, dy/dt = 80 km/h
FIND
distance between the cars (D), rate the distance between them grows (dD/dt)
Step 1 · Find distance between the cars (D)
Use D² = x² + y²
The distance between the cars is the hypotenuse: add the squares of the two road distances, then the square root.
D
=
√
x
2
+
y
2
D
=
√
30
2
+
40
2
D
=
√
900 + 1,600
D
=
√
2
,500
D
=
50
km
Step 2 · Find rate the distance between them grows (dD/dt)
Use D × (dD/dt) = x × (dx/dt) + y × (dy/dt)
Differentiate
D
2
=
x
2
+
y
2
in
t,
divide
by
2,
then
divide
by
D
.
dD/dt
=
x
× (dx/dt) +
y
× (dy/dt)
D
dD/dt
=
30 × 60 + 40 × 80
50
dD/dt
=
1,800 + 3,200
50
dD/dt
=
5,000
50
dD/dt = 100 km/h
Check
Put the numbers back in. Both sides should match.
50
2
=
30
2
+
40
2
✓
50 × 100 = 30 × 60 + 40 × 80 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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