Two people start from the same point. One walks east at 4 mi/h and the other walks northeast at 3 mi/h. How fast (in mi/h) is the distance between the people changing after 15 minutes? (Round your answer to three decimal places.)

To solve the problem of determining how fast the distance between the two people is changing after 15 minutes, we'll use related rates in calculus.

### Step 1: Set up the problem
Let's denote:
- The person walking east at \( 4 \) mi/h as A.
- The person walking northeast at \( 3 \) mi/h as B.
- Let \( x(t) \) be the distance of A from the starting point at time \( t \).
- Let \( y(t) \) be the distance of B from the starting point at time \( t \).
- Let \( z(t) \) be the distance between the two people at time \( t \).

### Step 2: Express \( x(t) \) and \( y(t) \)
Since A is walking east, and B is walking northeast (which is at a \( 45^\circ \) angle to the east direction), we have:

\[
x(t) = 4t
\]
\[
y_x(t) = 3t \cos(45^\circ) = \frac{3t}{\sqrt{2}}
\]
\[
y_y(t) = 3t \sin(45^\circ) = \frac{3t}{\sqrt{2}}
\]

The total distance \( y(t) \) can be expressed using the Pythagorean theorem:

\[
y(t) = \sqrt{y_x(t)^2 + y_y(t)^2} = \sqrt{\left(\frac{3t}{\sqrt{2}}\right)^2 + \left(\frac{3t}{\sqrt{2}}\right)^2} = \sqrt{2 \times \frac{9t^2}{2}} = 3t
\]

### Step 3: Use the distance formula for \( z(t) \)
The distance \( z(t) \) between the two people at any time \( t \) is:

\[
z(t) = \sqrt{(x(t))^2 + (y_x(t) - y_y(t))^2}
\]

But since B's movement doesn't change direction along the east-west axis, we use:

\[
z(t) = \sqrt{(x(t) - y_x(t))^2 + (y_y(t))^2}
\]

Simplifying for B:

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

Since \( y(t) = 3t \), \( x(t) = 4t \), we get:

\[
z(t) = \sqrt{(4t)^2 + (3t)^2} = \sqrt{16t^2 + 9t^2} = \sqrt{25t^2} = 5t
\]

### Step 4: Differentiate \( z(t) \) with respect to \( t \)
To find how fast the distance is changing, we need to differentiate \( z(t) \) with respect to \( t \):

\[
\frac{dz}{dt} = \frac{d}{dt}[5t] = 5 \text{ mi/h}
\]

### Step 5: Consider the effect of time after 15 minutes
Since \( t = \frac{15}{60} = \frac{1}{4} \) hours, after 15 minutes, the distance between the two people is:

\[
\frac{dz}{dt} = 5 \text{ mi/h}
\]

### Conclusion:
The distance between the two people is changing at a rate of \( 5 \) mi/h after 15 minutes.

If you'd like to ask more questions or need further details, feel free to ask!

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1. How do you use related rates to solve problems involving distances?
2. What is the significance of differentiating with respect to time?
3. How does the angle of movement impact the rate of distance change?
4. What would happen if both people walked at the same speed?
5. Why do we express time in hours when working with rates in mi/h?

**Tip:** Always convert time to the same units used in the problem (e.g., hours for speed in mi/h) to avoid inconsistencies in calculations.

Learn how to solve a related rates problem involving two people walking at different speeds and directions. Calculate the rate of change of distance between them using calculus and related rates principles. Suitable for advanced high school students.

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