Question

A goose is flying horizontally at an altitude of 300 ft and a speed of 50 ft/sec passes directly over a pond. Find the rate at which the distance from the goose to the pond is increasing when it is 500 ft away from the pond.

Answer 1

The rate at which the distance from the goose to the pond is increasing is approximately 42.875 feet per second.

Explanation:

We assume that goose is flying at constant velocity. After reading carefully the statement of the problem, we create the following geometrical diagram, which is included at the end of this explanation and represents the following Pythagorean identity:

`x^(2)+y^(2) = r^(2)` (Eq. 1)

Where:

`x` - Horizontal distance of the goose from pond, measured in feet.

`y` - Vertical distance of the goose from pond, measured in feet.

`r` - Resultant distance of the goose from pond, measured in feet.

We find the rate of change in time at which resultant distance is increasing by differentiating on (Eq. 1):

`2\cdot x \cdot \dot x + 2\cdot y \cdot \dot y = 2\cdot r \cdot \dot r`

`x\cdot \dot x + y\cdot \dot y = r\cdot \dot r`

`\dot r = \frac{x\cdot \dot x + y\cdot \dot y}{\sqrt{x^(2)+y^(2)}}` (Eq. 2)

Where:

`\dot x` - Rate of change of horizontal distance, measured in feet per second.

`\dot y` - Rate of change of vertical distance, measured in feet per second.

`\dot r` - Rate of change of resulting distance, measured in feet per second.

If we know that
`x = 500\,ft`,
`y = 300\,ft`,
`\dot x = 50\,(ft)/(s)` and
`\dot y = 0\,(ft)/(s)`, the rate at which the distance from the goose to the pond is increasing is:

`\dot r = \frac{(500\,ft)\cdot \left(50\,(ft)/(s)\right)+(300\,ft)\cdot \left(0\,(ft)/(s) \right) }{\sqrt{(500\,ft)^(2)+(300\,ft)^(2)}}`

`\dot r \approx 42.875\,(ft)/(s)`

The rate at which the distance from the goose to the pond is increasing is approximately 42.875 feet per second.