To solve this problem, we'll find the rate of change of the area $A(t)$ of the right triangle at the given instant $t_0$.

Step 1: Expression for the Area

The area $A(t)$ of a right triangle with legs $x(t)$ and $y(t)$ is given by: $A(t) = \frac{1}{2} x(t) y(t)$

Step 2: Differentiate with Respect to Time

We differentiate $A(t)$ with respect to time $t$ to find $\frac{dA(t)}{dt}$: $\frac{dA(t)}{dt} = \frac{1}{2} \left[ x'(t) y(t) + x(t) y'(t) \right]$

Step 3: Substitute Given Values

Now, we substitute the given values at the instant $t_0$:

• $x(t_0) = 3\,\text{km}$
• $y(t_0) = 9\,\text{km}$
• $x'(t) = 5\,\text{km}/\text{h}$ (rate of increase of $x(t)$)
• $y'(t) = -14\,\text{km}/\text{h}$ (rate of decrease of $y(t)$)

Substitute these values into the differentiated equation: $\frac{dA(t)}{dt}\bigg|_{t_0} = \frac{1}{2} \left[ 5\,\text{km}/\text{h} \times 9\,\text{km} + 3\,\text{km} \times (-14\,\text{km}/\text{h}) \right]$ $\frac{dA(t)}{dt}\bigg|_{t_0} = \frac{1}{2} \left[ 45\,\text{km}^2/\text{h} - 42\,\text{km}^2/\text{h} \right]$ $\frac{dA(t)}{dt}\bigg|_{t_0} = \frac{1}{2} \times 3\,\text{km}^2/\text{h}$ $\frac{dA(t)}{dt}\bigg|_{t_0} = 1.5\,\text{km}^2/\text{h}$

Answer: Rate of Change of the Area

The rate of change of the area $A(t)$ at the instant $t_0$ is $1.5\,\text{km}^2/\text{h}$.

Step 4: Matching Expressions with Units

1. $x'(t)$: $\text{km}/\text{h}$ (rate of change of $x(t)$)
2. $y'(t_0)$: $\text{km}/\text{h}$ (rate of change of $y(t)$)
3. $A(t_0)$: $\text{km}^2$ (area of the triangle at time $t_0$)
4. $\text{km}$: Unit of distance, for $x(t)$ or $y(t)$
5. $\text{km}^2/\text{h}$: Rate of change of area, for $\frac{dA(t)}{dt}$

Would you like further details on any part of the solution or explanation?

Here are five related questions:

1. How would the rate of change of the area $\frac{dA(t)}{dt}$ be affected if both legs were increasing?
2. What would the area $A(t)$ be at $t_0$?
3. Can you express $\frac{dA(t)}{dt}$ as a function of time $t$ without substituting the specific values?
4. How does the sign of $y'(t)$ (negative, since $y(t)$ is decreasing) affect the rate of change of the area?
5. What happens to $\frac{dA(t)}{dt}$ if $y(t)$ becomes constant instead of decreasing?

Tip: When differentiating products involving functions of time, always apply the product rule and carefully substitute the correct values.