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