Question

A boy is standing on a dock watching a boat moving north away from him at a speed of 5000ft/min. A girl is standing 1000ft east of the boy and is watching the same boat. How fast is the boat moving away from the girl when it is 12500ft away from the boy?

Answer 1

When the boat is 12500 ft away from the boy, it is moving away from the girl at a rate of 62500 ft/min. This is determined using the Pythagorean theorem and related rates.

To solve this problem, we can use the concept of related rates and apply the Pythagorean theorem.

Let:

- x be the distance between the boat and the boy (along the north direction),

- y be the distance between the boat and the girl (along the east direction).

According to the Pythagorean theorem:
`\( x^2 + y^2 = d^2 \)`, where d is the distance between the boy and the girl.

Differentiating both sides with respect to time t:

`\[ 2x (dx)/(dt) + 2y (dy)/(dt) = 0 \]`

Now, we are given that
`\( (dx)/(dt) = 5000 \, \text{ft/min} \)`,
`\( x = 12500 \, \text{ft} \)`, and
`\( y = 1000 \, \text{ft} \)`.

Substitute these values into the equation and solve for
`\( (dy)/(dt) \)`:

`\[ 2(12500) \cdot 5000 + 2(1000) \cdot (dy)/(dt) = 0 \]\[ 25000 \cdot 5000 + 2000 \cdot (dy)/(dt) = 0 \]\[ (dy)/(dt) = -(25000 \cdot 5000)/(2000) \]\[ (dy)/(dt) = -62500 \, \text{ft/min} \]`

The negative sign indicates that y is decreasing, which makes sense as the boat moves away from the girl.

Therefore, the boat is moving away from the girl at a rate of 62500
`\, \text{ft/min} \)` when it is 12500 ft away from the boy.