Answer: \(-\frac{7}{4} \quad \text{ft}^{2}/\sec\)
Step-by-step explanation:
Since ABC is a right triangle, at any moment it holds that
\(5^2=(AC)^2+(BC)^2\)
Moreover, the area A of the triangle is given by
\(A= \frac{1}{2}(AC)(BC)\)
and we know that the rate of change of the length (AC) is
constant decreasing 2, which may be written using the Leibniz
notation as
\(\dfrac{d(AC)}{dt}=-2\).
Using the chain rule and the product rule for derivation, the two
first equations tell us
that
\(0 = 2 \dfrac{d(AC)}{dt}(AC) + 2 \dfrac{d(BC)}{dt}(BC)\)
and
\(\dfrac{dA}{dt} = \dfrac{1}{2} \left( \dfrac{d(AC)}{dt} \cdot (BC) + \dfrac{d(BC)}{dt} \cdot (AC)\right)\)
Moreover, using the first of the last two equations we get
\((AC)\dfrac{d(AC)}{dt} = -(BC)\dfrac{d(BC)}{dt} \Rightarrow\\\\\\(AC)(-2) = -(BC) \dfrac{d(BC)}{dt} \quad \Rightarrow \quad \dfrac{d(BC)}{dt}=2 \dfrac{(AC)}{(BC)}\)
Now, when (AC)=3, we have that
\(25=(AC)^2 + (BC)^2 \quad \Rightarrow 25 = 9 + (BC)^2\\ \\\Rightarrow \quad 16=(BC)^2 \quad \Rightarrow (BC)=4\)
and
\(\dfrac{d(BC)}{dt} = 2 \dfrac{(AC)}{(BC)}=2 \dfrac{3}{4}=\dfrac{3}{2}\).
Hence, at this moment the rate of change of the area of the triangle is
\(\dfrac{dA}{dt} = \dfrac{1}{2} \left( \dfrac{d(AC)}{dt} \cdot (BC) + \dfrac{d(BC)}{dt} \cdot(AC) \right)=\dfrac{1}{2}\left( -2 \cdot 4 + \dfrac{3}{2}\cdot 3\right ) = -\dfrac{7}{4}\)
