Two people are at an elevator. At the same time one person starts to walk away from the elevator at a rate of 2 ft/sec and the other person starts going up in the elevator at a rate of 7 ft/sec. What rate is the distance between the two people changing 15 seconds later?

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Find Rate of Change: We need to find the rate at which the distance between the two people is changing after 15 seconds. Since one person is moving horizontally away from the elevator at 2 ft/sec and the other is moving vertically in the elevator at 7 ft/sec, we can use the Pythagorean theorem to model the situation. Let's denote the horizontal distance from the elevator as $ x $ and the vertical distance traveled by the elevator as $ y $. The rate of change of $ x $ is $ dx/dt = 2 $ ft/sec and the rate of change of $ y $ is $ dy/dt = 7 $ ft/sec. The distance between the two people at any time $ t $ is given by $ d = \sqrt{x^2 + y^2} $. We need to find $ dd/dt $ when $ t = 15 $ seconds.Calculate Distances: First, let's calculate the horizontal and vertical distances after 15 seconds. For the horizontal distance, $ x = 2 \cdot 15 $ ft, and for the vertical distance, $ y = 7 \cdot 15 $ ft.Differentiate Equation: Now we calculate the actual distances: $ x = 2 \cdot 15 = 30 $ ft and $ y = 7 \cdot 15 = 105 $ ft.Substitute Values: Next, we differentiate the equation $ d = \sqrt{x^2 + y^2} $ with respect to time $ t $ to find $ dd/dt $. Using the chain rule, we get $ dd/dt = (1/2)(2x dx/dt + 2y dy/dt) / \sqrt{x^2 + y^2} $.Perform Calculations: Substitute the known values into the differentiated equation: $ dd/dt = (1/2)(2 \cdot 30 \cdot 2 + 2 \cdot 105 \cdot 7) / \sqrt{30^2 + 105^2} $ ft/sec.Simplify Numerator: Now we perform the calculations: $ dd/dt = (1/2)(60 \cdot 2 + 210 \cdot 7) / \sqrt{900 + 11025} $ ft/sec.Calculate Square Root: Simplify the numerator: $ dd/dt = (1/2)(120 + 1470) / \sqrt{11925} $ ft/sec.Divide Numerator: Add the terms in the numerator: $ dd/dt = (1/2)(1590) / \sqrt{11925} $ ft/sec.Perform Division: Calculate the square root of the denominator: $ \sqrt{11925} = 109.204 $ ft (rounded to three decimal places).Calculate Final Value: Now divide the numerator by the denominator: $ dd/dt = (1/2)(1590) / 109.204 $ ft/sec.Finally, perform the division to find the rate of change of distance: $ dd/dt = 795 / 109.204 $ ft/sec.Calculate the final value: $ dd/dt ≈ 7.28 $ ft/sec (rounded to two decimal places).

Two people are at an elevator. At the same time one person starts to walk away from the elevator at a rate of $2\,\text{ft/sec}$ and the other person starts going up in the elevator at a rate of $7\,\text{ft/sec}$ . What rate is the distance between the two people changing $15$ seconds later?

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