Question Statement

A woman jogging at a constant rate of $10\mathrm{km}/\mathrm{hr}$ crosses a point $A$ heading north. Ten minutes later a man jogging at a constant rate of $9\mathrm{km}/\mathrm{hr}$ crosses the same point heading east. How fast is the distance between the joggers changing 20 minutes after the man crosses $A$?

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Background and Explanation

This is a related rates problem where two objects move in perpendicular directions, and we need to find how fast the distance between them changes. The key is to establish a coordinate system, express the positions as functions of time, use the Pythagorean theorem to relate the distance, and then differentiate with respect to time.

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Solution

Setting Up the Coordinate System

Let point $A$ be the origin $(0,0)$. We align the axes such that:

• The woman travels along the positive $y$-axis (north) at $10km/hr$
• The man travels along the positive $x$-axis (east) at $9km/hr$

Let $t$ represent the time in hours since the man crossed point $A$.

Expressing Positions as Functions of Time

Since the woman had a 10-minute head start (which equals $\frac{10}{60}=\frac{1}{6}$ hours), at time $t$ she has been jogging for $\left(t+\frac{1}{6}\right)$ hours.

Using the formula $\text{distance}=\text{velocity}\times \text{time}$:

• Woman's position (north of A): $y(t)=10\left(t+\frac{1}{6}\right)$

• Man's position (east of A): $x(t)=9t$

Distance Between the Joggers

Let $z(t)$ be the distance between the two joggers. By the Pythagorean theorem:

$z(t)=\sqrt{x(t{)}^2+y(t{)}^2}$

Substituting our expressions:

$z(t)=\sqrt{(9t{)}^2+{\left(10\left(t+\frac{1}{6}\right)\right)}^2}$

$z(t)=\sqrt{81t^2+100{\left(t+\frac{1}{6}\right)}^2}$

Finding the Rate of Change

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

$\frac{dz}{dt}=\frac{1}{2}{\left[81t^2+100{\left(t+\frac{1}{6}\right)}^2\right]}^{-1/2}\cdot \frac{d}{dt}\left[81t^2+100{\left(t+\frac{1}{6}\right)}^2\right]$

Differentiating the expression inside the brackets:

$\frac{d}{dt}\left[81t^2+100{\left(t+\frac{1}{6}\right)}^2\right]=162t+200\left(t+\frac{1}{6}\right)$

Therefore:

$\frac{dz}{dt}=\frac{162t+200\left(t+\frac{1}{6}\right)}{2\sqrt{81t^2+100{\left(t+\frac{1}{6}\right)}^2}}$

Simplifying the numerator: $162t+200t+\frac{200}{6}=362t+\frac{100}{3}$

So: $\frac{dz}{dt}=\frac{362t+\frac{100}{3}}{2\sqrt{81t^2+100{\left(t+\frac{1}{6}\right)}^2}}=\frac{181t+\frac{50}{3}}{\sqrt{81t^2+100{\left(t+\frac{1}{6}\right)}^2}}$

Evaluating at $t=20$ Minutes

We need to find the rate when $t=20$ minutes $=\frac{20}{60}=\frac{1}{3}$ hour.

Calculate the numerator: $181\left(\frac{1}{3}\right)+\frac{50}{3}=\frac{181}{3}+\frac{50}{3}=\frac{231}{3}=77$

Calculate the denominator: $\sqrt{81{\left(\frac{1}{3}\right)}^2+100{\left(\frac{1}{3}+\frac{1}{6}\right)}^2}$

$=\sqrt{81\left(\frac{1}{9}\right)+100{\left(\frac{2}{6}+\frac{1}{6}\right)}^2}$

$=\sqrt{9+100{\left(\frac{1}{2}\right)}^2}$

$=\sqrt{9+100\left(\frac{1}{4}\right)}$

$=\sqrt{9+25}=\sqrt{34}$

Final calculation: $\frac{dz}{dt}=\frac{77}{\sqrt{34}}\approx 13.2km/hr$

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Key Formulas or Methods Used

• Distance formula (Pythagorean theorem): $z=\sqrt{x^2+y^2}$ for perpendicular motion
• Chain rule for differentiation: $\frac{d}{dt}[\sqrt{u}]=\frac{1}{2\sqrt{u}}\cdot \frac{du}{dt}$
• Related rates: Differentiating the position relationship to find how distance changes over time
• Alternative implicit differentiation method: $2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$, yielding $\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{z}$

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Summary of Steps

1. Set up coordinates: Place $A$ at origin $(0,0)$ with woman on $y$-axis and man on $x$-axis
2. Define time variable: Let $t$ = hours since man crossed $A$ (woman started at $t=-\frac{1}{6}$)
3. Write position functions: $x(t)=9t$ and $y(t)=10(t+\frac{1}{6})$
4. Apply Pythagorean theorem: $z(t)=\sqrt{x^2+y^2}=\sqrt{81t^2+100(t+\frac{1}{6}{)}^2}$
5. Differentiate: Use chain rule to find $\frac{dz}{dt}=\frac{181t+\frac{50}{3}}{\sqrt{81t^2+100(t+\frac{1}{6}{)}^2}}$
6. Substitute $t=\frac{1}{3}$: Calculate numerator $=77$ and denominator $=\sqrt{34}$
7. Compute final answer: $\frac{77}{\sqrt{34}}\approx 13.2km/hr$