Question Statement

At 8:00 am ship $S_1$ is 20 km due north of $S_2$. Ship $S_1$ sails south at a rate of $9\mathrm{km}/\mathrm{hr}$ and $S_2$ sails west at a rate of $12\mathrm{km}/\mathrm{hr}$. At what rate is the distance between the two ships changing at 9:20 am?

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Background and Explanation

This is a related rates problem in calculus, where we determine how quickly the distance between two moving objects changes over time. The solution requires setting up a geometric relationship (typically using the Pythagorean theorem) between the positions of the ships, then differentiating with respect to time to find the rate of change.

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Solution

Setting Up the Geometry

At 8:00 am, ship $S_1$ is 20 km due north of ship $S_2$. Let's denote the initial position of $S_2$ as point $Q$ and the initial position of $S_1$ as point $P$, so $|PQ|=20$ km.

After time $t$ hours (measured from 8:00 am):

• Ship $S_1$ sails south, so it moves toward point $Q$
• Ship $S_2$ sails west, perpendicular to the original north-south line

Determining the Time Variable

At 9:20 am, the time elapsed from 8:00 am is: $t=1\text{ hour }20\text{ minutes}=1+\frac{20}{60}=\frac{4}{3}\text{ hours}$

Expressing Positions as Functions of Time

Using the formula $\text{Distance}=\text{Velocity}\times \text{Time}$:

For ship $S_1$ (moving south):

• Distance traveled south from $P$: $|PA|=9t$
• Remaining distance from $Q$: $|AQ|=|PQ|-|PA|=20-9t$

For ship $S_2$ (moving west):

• Distance traveled west from $Q$: $|BQ|=12t$

Finding the Distance Between Ships

Let $D$ be the distance between the ships at time $t$. By the Pythagorean theorem applied to right triangle $AQB$:

$D=\sqrt{|AQ{|}^2+|BQ{|}^2}$

Substituting the expressions: $D=\sqrt{(20-9t{)}^2+(12t{)}^2}$

Expanding the squared terms: $D=\sqrt{400-360t+81t^2+144t^2}$

$D=\sqrt{225t^2-360t+400}$

Differentiating to Find the Rate of Change

To find how fast the distance is changing, we differentiate $D$ with respect to $t$ using the chain rule:

$\frac{dD}{dt}=\frac{1}{2}(225t^2-360t+400{)}^{-1/2}\cdot \frac{d}{dt}(225t^2-360t+400)$

$\frac{dD}{dt}=\frac{1}{2\sqrt{225t^2-360t+400}}\cdot (450t-360)$

Simplifying by factoring out 2 from the numerator: $\frac{dD}{dt}=\frac{225t-180}{\sqrt{225t^2-360t+400}}$

Evaluating at 9:20 am

At 9:20 am, we have $t=\frac{4}{3}$. Substituting this value:

Numerator: $225\left(\frac{4}{3}\right)-180=300-180=120$

Denominator: $\sqrt{225{\left(\frac{4}{3}\right)}^2-360\left(\frac{4}{3}\right)+400}$

$=\sqrt{225\cdot \frac{16}{9}-480+400}$

$=\sqrt{400-480+400}$

$=\sqrt{320}$

Therefore: $\frac{dD}{dt}=\frac{120}{\sqrt{320}}=\frac{120}{8\sqrt{5}}=\frac{15}{\sqrt{5}}=3\sqrt{5}\approx 6.708\text{ km/hr}$

The distance between the two ships is increasing at a rate of approximately $6.708\mathrm{km}/\mathrm{hr}$ (or exactly $3\sqrt{5}$ km/hr) at 9:20 am.

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Key Formulas or Methods Used

• Distance formula: $d=v\times t$ (where $v$ is velocity and $t$ is time)
• Pythagorean theorem: $c=\sqrt{a^2+b^2}$ for right triangles
• Chain rule for differentiation: $\frac{d}{dt}[f(g(t))]=f^{\prime }(g(t))\cdot g^{\prime }(t)$
• Related rates method: Differentiating a geometric constraint equation with respect to time to relate the rates of change of different variables

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Summary of Steps

1. Set up coordinates: Establish that at $t=0$ (8:00 am), the vertical separation is 20 km with $S_1$ north of $S_2$.
2. Define time variable: Calculate $t=\frac{4}{3}$ hours for 9:20 am.
3. Express positions: Write $|AQ|=20-9t$ (remaining north-south separation) and $|BQ|=12t$ (westward displacement).
4. Apply Pythagorean theorem: $D=\sqrt{(20-9t{)}^2+(12t{)}^2}=\sqrt{225t^2-360t+400}$.
5. Differentiate: Use chain rule to get $\frac{dD}{dt}=\frac{225t-180}{\sqrt{225t^2-360t+400}}$.
6. Substitute $t=\frac{4}{3}$: Calculate numerator ($120$) and denominator ($\sqrt{320}$) to get the final rate of $3\sqrt{5}\approx 6.708$ km/hr.