Question

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a person is standing 400 feet away from a model rocket that is fired straight up into the air at a rate of 12 ft/sec. at what rate is the distance between the person and the rocket increasing 1 minute after liftoff?

a. 4.3
b. 10.5
c. 68.6
d. 78.5

Explanation:

Set up the geometric model

We model the scenario using a right triangle.
Let \(x\) be the constant horizontal distance from the observer to the launch pad:

$$x = 400\text{ ft}$$

Let \(y(t)\) be the height of the rocket at time \(t\).
Let \(s(t)\) be the distance between the observer and the rocket at time \(t\).
By the Pythagorean theorem:

$$s^2 = x^2 + y^2$$

Determine values at the given instant

The rocket rises at a constant rate:

$$\frac{dy}{dt} = 12\text{ ft/sec}$$

At \(t = 1\text{ minute} = 60\text{ seconds}\), the height \(y\) is:

$$y = 12\text{ ft/sec} \times 60\text{ sec} = 720\text{ ft}$$

Using the Pythagorean theorem, find the distance \(s\) at this instant:

$$s = \sqrt{400^2 + 720^2} = \sqrt{160000 + 518400} = \sqrt{678400} \approx 823.65\text{ ft}$$

Differentiate with respect to time

Differentiate both sides of the relation \(s^2 = x^2 + y^2\) with respect to \(t\):

$$2s\frac{ds}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$$

Since the observer is stationary, \(x\) is constant, so \(\frac{dx}{dt} = 0\):

$$s\frac{ds}{dt} = y\frac{dy}{dt}$$

Solve for the rate of change of distance

Isolate \(\frac{ds}{dt}\):

$$\frac{ds}{dt} = \frac{y}{s}\frac{dy}{dt}$$

Substitute the known values at \(t = 60\text{ seconds}\):

$$\frac{ds}{dt} = \frac{720}{823.65} \times 12 \approx 10.49\text{ ft/sec}$$

Match with the given options

Rounding to one decimal place:

$$\frac{ds}{dt} \approx 10.5\text{ ft/sec}$$

This matches option B.

Answer:

• (A) 4.3
• (B) 10.5 (Correct answer)
• (C) 68.6
• (D) 78.5