At 1:00pm, Ship A is 50 miles west of Ship B. Ship A is sailing north at 25 mph and Ship B is sailing south at 20 mph. How fast is the distance between the ships changing at 3:30pm?
At 1:00pm, Ship A is 50 miles west of Ship B. Ship A is sailing north at 25 mph and Ship B is sailing south at 20 mph. How fast is the distance between the ships changing at 3:30pm?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
At 1:00pm, Ship A is 50 miles west of Ship B. Ship A is sailing north at 25 mph and Ship B is sailing south at 20 mph. How fast is the distance between the ships changing at 3:30pm?
**Explanation:**
To find how fast the distance between the two ships is changing, we can use the Pythagorean theorem. Let's set up a coordinate system where Ship B is at the origin at 1:00 pm, and Ship A is 50 miles to the west on the x-axis.
Let:
- \( x(t) \) be the east-west distance between the ships.
- \( y(t) \) be the north-south distance between the ships.
- \( z(t) \) be the straight-line distance between the ships.
Initially, \( x(0) = 50 \) miles and \( y(0) = 0 \) miles.
**Ships' Movement:**
- Ship A moves north at 25 mph: \( y(t) = 25t \)
- Ship B moves south at 20 mph: \( y(t) = 20t \)
**Calculating Distance:**
The distance between Ship A and Ship B at any time \( t \) is given by the Pythagorean theorem:
\[ z(t) = \sqrt{x(t)^2 + y(t)^2} \]
**At 3:30pm (2.5 hours later):**
- \( x(t) = 50 \)
- \( y(t) = 25t + 20t = 45t = 45 \times 2.5 = 112.5 \)
Substitute these into the Pythagorean formula to find \( z(t) \).
**Finding Rate of Change: \( \frac{dz}{dt} \):**
Differentiate the distance function with respect to time to find the rate at which the distance between the ships is changing at 3:30 pm.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8d9f2dd3-4242-4da1-8488-f4a8f54e7aef%2F0837b275-6978-45ca-9e7e-cf6812f6cd50%2F886nw4i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
At 1:00pm, Ship A is 50 miles west of Ship B. Ship A is sailing north at 25 mph and Ship B is sailing south at 20 mph. How fast is the distance between the ships changing at 3:30pm?
**Explanation:**
To find how fast the distance between the two ships is changing, we can use the Pythagorean theorem. Let's set up a coordinate system where Ship B is at the origin at 1:00 pm, and Ship A is 50 miles to the west on the x-axis.
Let:
- \( x(t) \) be the east-west distance between the ships.
- \( y(t) \) be the north-south distance between the ships.
- \( z(t) \) be the straight-line distance between the ships.
Initially, \( x(0) = 50 \) miles and \( y(0) = 0 \) miles.
**Ships' Movement:**
- Ship A moves north at 25 mph: \( y(t) = 25t \)
- Ship B moves south at 20 mph: \( y(t) = 20t \)
**Calculating Distance:**
The distance between Ship A and Ship B at any time \( t \) is given by the Pythagorean theorem:
\[ z(t) = \sqrt{x(t)^2 + y(t)^2} \]
**At 3:30pm (2.5 hours later):**
- \( x(t) = 50 \)
- \( y(t) = 25t + 20t = 45t = 45 \times 2.5 = 112.5 \)
Substitute these into the Pythagorean formula to find \( z(t) \).
**Finding Rate of Change: \( \frac{dz}{dt} \):**
Differentiate the distance function with respect to time to find the rate at which the distance between the ships is changing at 3:30 pm.
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