Two people start from the same point. One walks east at 9 mi/h and the other walks northeast at 8 mi/h. How fast is the distance between the people changing after 15 minutes? (Round your answer to three decimal places.)


We can solve this problem using the Pythagorean theorem and related rates, as it involves two people walking in different directions and we are asked to find how fast the distance between them is changing.

### Step-by-step solution:

#### 1. Define Variables:
- Let \( t \) be the time in hours.
- The person walking east travels at 9 mi/h, so after time \( t \), they have traveled \( x(t) = 9t \) miles.
- The person walking northeast travels at 8 mi/h. The direction "northeast" forms a 45° angle with the east direction. Therefore, we can break down the movement into its horizontal and vertical components:
  - Horizontal component: \( x_{\text{NE}}(t) = 8t \cdot \cos(45^\circ) = 8t \cdot \frac{\sqrt{2}}{2} = 4\sqrt{2}t \)
  - Vertical component: \( y_{\text{NE}}(t) = 8t \cdot \sin(45^\circ) = 4\sqrt{2}t \)

#### 2. Distance Between the Two People:
The total distance between the two people after time \( t \) can be expressed using the Pythagorean theorem as:
\[
d(t) = \sqrt{x(t)^2 + \left( y_{\text{NE}}(t) \right)^2}
\]
where:
- \( x(t) = 9t \) (east distance)
- \( y_{\text{NE}}(t) = 4\sqrt{2}t \) (northeast vertical distance)

Thus, the distance function becomes:
\[
d(t) = \sqrt{(9t)^2 + (4\sqrt{2}t)^2}
\]

#### 3. Differentiate to Find the Rate of Change of Distance:
We need to differentiate \( d(t) \) with respect to \( t \) to find how fast the distance is changing. Using the chain rule:

\[
\frac{dd}{dt} = \frac{d}{dt} \left( \sqrt{(9t)^2 + (4\sqrt{2}t)^2} \right)
\]

Let's compute this step-by-step. First, simplify the distance expression, then differentiate.

Let me do the calculations.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, I can explain the next steps in solving the problem, so you can compute it manually if you'd like.

### Continuing the process:
The distance function we derived earlier was:
\[
d(t) = \sqrt{(9t)^2 + (4\sqrt{2}t)^2}
\]
which simplifies to:
\[
d(t) = \sqrt{81t^2 + 32t^2} = \sqrt{113t^2} = t\sqrt{113}
\]

Now, to find the rate at which the distance is changing after 15 minutes (which is \( t = \frac{1}{4} \) hours), we differentiate \( d(t) \):
\[
\frac{dd}{dt} = \frac{d}{dt} \left( t\sqrt{113} \right) = \sqrt{113}
\]

Thus, the rate of change of the distance between the two people is:
\[
\frac{dd}{dt} = \sqrt{113} \text{ miles per hour}
\]

Finally, calculate \( \sqrt{113} \approx 10.630 \) mi/h.

### Final Answer:
The distance between the two people is changing at approximately 10.630 mi/h after 15 minutes.

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Do you have any questions or would you like me to elaborate further? Here are 5 related questions to expand your understanding:

1. How would the result change if one person walked at a different speed?
2. Can this method apply to situations where one person walks west instead of east?
3. What happens to the rate of change if both people walk in the same direction?
4. How would you calculate the distance after different amounts of time (e.g., 30 minutes or an hour)?
5. Can you model the problem using vectors instead of right triangles?

**Tip:** Related rates problems often require a strong understanding of the chain rule in calculus!

Two people start from the same point. One walks east at 9 mi/h and the other walks northeast at 8 mi/h. How fast is the distance between the people changing after 15 minutes? (Round your answer to three decimal places.)

Solve a related rates problem involving two people walking in different directions. One person walks east at 9 mi/h, and the other walks northeast at 8 mi/h. Learn how to apply the Pythagorean theorem, trigonometry, and the chain rule to calculate how fast the distance between them is changing after 15 minutes.

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