One leg 
\[x(t)\] of a right triangle is increasing at a rate of 
\[5\,\text{km}/\text{h}\] and the other leg 
\[y(t)\] of the triangle is decreasing at a rate of 
\[14\,\text{km}/\text{h}\]. At a certain instant 
\[t_0\], the increasing leg is 
\[3\,\text{km}\] and the decreasing leg is 
\[9\,\text{km}\]. What is the rate of change of the area 
\[A(t)\] of the right triangle at that instant?
Match each expression with its units.
\[\text{km}\]
\[\text{km}/\text{h}\]
\[\text{km}^2\]
\[\text{km}^2/\text{h}\]
\[x'(t)\]
\[A(t_0)\]
\[y'(t_0)\]


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.

Discover how to calculate the rate of change of the area of a right triangle at a given instant using given rates of change in the lengths of its legs. Step-by-step solution with formulas and examples.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation