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This problem involves calculating the rate at which the distance between the boy and the girl is increasing as they move in perpendicular directions. The boy is moving north at 5 ft/sec, and the girl is moving east at 1 ft/sec. We are asked to find how fast they are separating from each other 5 seconds after they started moving.

### Step-by-Step Solution:

Let:
- \( x(t) \) be the distance the girl has walked in the east direction after \( t \) seconds,
- \( y(t) \) be the distance the boy has moved in the north direction after \( t \) seconds,
- \( z(t) \) be the distance between the boy and the girl at time \( t \).

From the Pythagorean theorem, we know that:
\[
z^2 = x^2 + y^2
\]
Since \( x(t) = 1 \cdot t \) and \( y(t) = 5 \cdot t \), we can substitute:
\[
z^2 = (1t)^2 + (5t)^2 = t^2 + 25t^2 = 26t^2
\]
Thus, the distance between them is:
\[
z = \sqrt{26t^2} = t \sqrt{26}
\]

Now, we need to find how fast the distance is changing after 5 seconds, i.e., find \( \frac{dz}{dt} \) at \( t = 5 \).

### Differentiate both sides with respect to time \( t \):
\[
2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]
Simplifying:
\[
z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}
\]
Substitute the values:
- \( x(t) = t \),
- \( y(t) = 5t \),
- \( \frac{dx}{dt} = 1 \) ft/sec (girl's speed),
- \( \frac{dy}{dt} = 5 \) ft/sec (boy's speed).

At \( t = 5 \), we have:
- \( x(5) = 5 \) ft,
- \( y(5) = 25 \) ft,
- \( z(5) = 5 \sqrt{26} \) ft.

Now, substitute into the equation:
\[
(5 \sqrt{26}) \frac{dz}{dt} = (5)(1) + (25)(5)
\]
\[
(5 \sqrt{26}) \frac{dz}{dt} = 5 + 125 = 130
\]
\[
\frac{dz}{dt} = \frac{130}{5 \sqrt{26}} = \frac{130}{5} \times \frac{1}{\sqrt{26}} = \frac{26}{\sqrt{26}} = \sqrt{26}
\]

### Approximate the result:
\[
\sqrt{26} \approx 5.10 \text{ ft/sec}.
\]

### Final Answer:
After 5 seconds, the boy and girl are separating at approximately **5.10 ft/sec**.

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Would you like further details or have any questions?

Here are five related questions you might find useful:
1. What is the significance of applying the Pythagorean theorem to this type of problem?
2. How would the result change if the girl increased her walking speed?
3. What would happen if both individuals were running in the same direction instead of perpendicularly?
4. How can related rates be applied to other real-life problems like this?
5. What are the steps to differentiate implicit functions effectively?

**Tip:** When solving related rates problems, always begin by identifying the geometric relationship between the variables involved, and differentiate that equation with respect to time.

On a rainy day, a girl broke up with her boyfriend after being together for eight long years. They decided to separate at the place where everything about them began, at the same time. The boy is due north crying and running at a rate of 5 ft/sec and the girl is walking due east at a rate of 1 ft/sec thinking if she made the right decision. How fast are they separating from each other 5 seconds after they started moving to a new life without each other?

In this calculus problem, learn how to calculate the rate of separation between two people moving in perpendicular directions using differential calculus. The boy runs north at 5 ft/sec and the girl walks east at 1 ft/sec. Discover how fast they are separating 5 seconds after starting their journey.

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