Solve the problem.
-A piece of land is shaped like a right triangle. Two people start at the right angle of the triangle at the same time, and walk at the same speed along different legs of the triangle. If the area formed by the positions of the two people and their starting point (the right angle) is changing at $5 \mathrm {~m} 2 / \mathrm { s }$, then how fast are the people moving when they are $4 \mathrm {~m}$ from the right angle? (Round your answer to two decimal places.)
A) $0.63 \mathrm {~m} / \mathrm { s }$
B) $3.20 \mathrm {~m} / \mathrm { s }$
C) $1.25 \mathrm {~m} / \mathrm { s }$
D) $2.50 \mathrm {~m} / \mathrm { s }$

Question

Quizplus · Accepted Answer

The area of the triangle formed by the positions of the two people and the starting point is given by $ \frac{1}{2}xy $, where $x$ and $y$ are the distances of the two people from the right angle. Since both people are moving at the same speed and are $4 \mathrm{~m}$ from the right angle, we have $x = y = 4 \mathrm{~m}$. The rate of change of the area is given by the derivative of the area with respect to time, $ \frac{dA}{dt} = \frac{1}{2}(x\frac{dy}{dt} + y\frac{dx}{dt}) $. Given that $ \frac{dA}{dt} = 5 \mathrm{~m}^2/\mathrm{s} $ and $x = y = 4 \mathrm{~m}$, and knowing that $ \frac{dx}{dt} = \frac{dy}{dt} $ because the speeds are the same, we can solve for $ \frac{dx}{dt} $ (or $ \frac{dy}{dt} $). Substituting the given values, we get $ 5 = \frac{1}{2}(4\frac{dy}{dt} + 4\frac{dx}{dt}) = 4\frac{dx}{dt} $, leading to $ \frac{dx}{dt} = 1.25 \mathrm{~m}/\mathrm{s} $.

Solve the Problem 5 m2/s5 \mathrm {~m} 2 / \mathrm { s }5 m2/s

Solve the Problem $5 \mathrm {~m} 2 / \mathrm { s }$