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The problem in the image describes two ships leaving a port at the same time: one travels north at 90 knots, and the other travels west at 90 knots. The question asks for the rate at which the distance between the two ships is increasing.

### Solution:

We can model the situation using related rates in a right triangle. The ships are moving perpendicular to each other (north and west directions), so their paths form a right triangle. Let:
- \( x \) be the distance the ship traveling west has covered,
- \( y \) be the distance the ship traveling north has covered,
- \( z \) be the distance between the two ships.

We know the following:
- The speed of the ship traveling west is \( \frac{dx}{dt} = 90 \) knots.
- The speed of the ship traveling north is \( \frac{dy}{dt} = 90 \) knots.

The Pythagorean theorem relates \( x \), \( y \), and \( z \) as:
\[
z^2 = x^2 + y^2
\]
Differentiate both sides with respect to time \( t \):
\[
2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]
Simplify:
\[
z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}
\]
Since both ships travel at the same speed (90 knots), at any given time, \( x = y \). Let \( x = y = 90t \), where \( t \) is the time after the ships leave port. Then, \( z \) becomes:
\[
z = \sqrt{x^2 + y^2} = \sqrt{(90t)^2 + (90t)^2} = \sqrt{2(90t)^2} = 90t\sqrt{2}
\]
Substitute these values into the differentiated equation:
\[
90t\sqrt{2} \frac{dz}{dt} = 90t \cdot 90 + 90t \cdot 90
\]
\[
90t\sqrt{2} \frac{dz}{dt} = 16200t
\]
Solve for \( \frac{dz}{dt} \):
\[
\frac{dz}{dt} = \frac{16200t}{90t\sqrt{2}} = \frac{16200}{90\sqrt{2}} = \frac{180}{\sqrt{2}} = 90\sqrt{2} \approx 127.28 \text{ knots}
\]

Thus, the distance between the two ships is increasing at approximately **127.28 knots**.

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Would you like further details or explanations on any step?

Here are 5 related questions:
1. How would the solution change if the ships were traveling at different speeds?
2. What is the rate of change in distance if one ship travels east and the other west?
3. How would you solve this problem if the two ships didn't leave at the same time?
4. What happens to the rate of change in distance if the ships change their courses after a while?
5. Can this related rates problem be applied to airplanes flying in different directions?

**Tip:** Related rates problems often require a clear understanding of how different quantities are changing with respect to time, so always ensure the relationships between variables are correctly identified before differentiating.

Two ships leave port at the same time. One travels north at 90 knots (that is, 90 nautical miles per hour), and the other west at 90 knots. At what rate is the distance between the two ships increasing?

This problem explores the rate at which the distance between two ships is increasing. Both ships travel at 90 knots, one north and one west, forming a right triangle. The solution uses related rates and the Pythagorean theorem to find the rate of change in distance. Ideal for students in Grades 11-12.

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