The cars are approaching each other at a rate of mi/h.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![y
B
Video Example())
EXAMPLE 4 Car A is traveling west at 60 mi/h and car B is traveling north at 40 mi/h. Both are headed for the intersection of the two roads. At what rate are
the cars approaching each other when car A is 0.4 mi and car B is 0.3 mi from the intersection?
SOLUTION We draw the figure to the left, where C is the intersection of the roads. At a given time t, let x be the distance from car A to C, let y be the distance
from car B to C, and let z be the distance between the cars, where x, y, and z are measured in miles.
We are given that dx/dt = -60 mi/h and dy/dt = -40 mi/h. (The derivatives are negative because x and y are decreasing.) We are asked to find dz/dt. The
equation that relates x, y, and z is given by the Pythagorean Theorem:
z² = x² + y².
Differentiating each side with respect to t, we have
2z
z dz = 2x
dt
dz = ²(x dx + y dx).
dt
When x = 0.4 mi and y = 0.3 mi, the Pythagorean Theorem gives z = 0.5 mi, so
(-60) +
(-40)]
dz
dt
mi/h.
The cars are approaching each other at a rate of
mi/h.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e533fec-51fd-4870-9b3f-7eb1d9840991%2Ffccb86d9-6391-492f-bd6e-794cb4cd7428%2Fpbg494_processed.png&w=3840&q=75)
Transcribed Image Text:y
B
Video Example())
EXAMPLE 4 Car A is traveling west at 60 mi/h and car B is traveling north at 40 mi/h. Both are headed for the intersection of the two roads. At what rate are
the cars approaching each other when car A is 0.4 mi and car B is 0.3 mi from the intersection?
SOLUTION We draw the figure to the left, where C is the intersection of the roads. At a given time t, let x be the distance from car A to C, let y be the distance
from car B to C, and let z be the distance between the cars, where x, y, and z are measured in miles.
We are given that dx/dt = -60 mi/h and dy/dt = -40 mi/h. (The derivatives are negative because x and y are decreasing.) We are asked to find dz/dt. The
equation that relates x, y, and z is given by the Pythagorean Theorem:
z² = x² + y².
Differentiating each side with respect to t, we have
2z
z dz = 2x
dt
dz = ²(x dx + y dx).
dt
When x = 0.4 mi and y = 0.3 mi, the Pythagorean Theorem gives z = 0.5 mi, so
(-60) +
(-40)]
dz
dt
mi/h.
The cars are approaching each other at a rate of
mi/h.
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