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One leg of a right triangle is decreasing at a rate of 5 kilometers per hour and the other leg of the triangle is increasing at a rate of 14 kilometers per hour.
At a certain instant, the decreasing leg is 3 kilometers and the increasing leg is 9 kilometers.
What is the rate of change of the area of the right triangle at that instant (in square kilometers per hour)?
Choose 1 answer:
(A) 1.5
(B) 111
(C) -1.5
(D) -111

One leg of a right triangle is decreasing at a rate of $5$ kilometers per hour and the other leg of the triangle is increasing at a rate of $14$ kilometers per hour. $\newline$ At a certain instant, the decreasing leg is $3$ kilometers and the increasing leg is $9$ kilometers. $\newline$ What is the rate of change of the area of the right triangle at that instant (in square kilometers per hour)? $\newline$ Choose $1$ answer: $\newline$ (A) $1$ . $5$ $\newline$ (B) $111$ $\newline$ (C) $-1$ . $5$ $\newline$ (D) $-111$

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Find instantaneous rates of change

Full solution

Q. One leg of a right triangle is decreasing at a rate of

5

kilometers per hour and the other leg of the triangle is increasing at a rate of

14

kilometers per hour.

\newline

At a certain instant, the decreasing leg is

3

kilometers and the increasing leg is

9

kilometers.

\newline

What is the rate of change of the area of the right triangle at that instant (in square kilometers per hour)?

\newline

Choose

1

answer:

\newline

(A)

1

.

5

\newline

(B)

111

\newline

(C)

-1

.

5

\newline

(D)

-111

Area Formula Explanation: The area of a right triangle is given by the formula $A = (\frac{1}{2}) \times \text{base} \times \text{height}$ . Here, one leg is the base and the other is the height.
Legs Notation: Let's denote the decreasing leg as ' $b$ ' and the increasing leg as ' $h$ '. The rate of change of ' $b$ ' is $-5$ km/h and the rate of ' $h$ ' is $14$ km/h.
Given Rates: At the instant in question, $b = 3 \, \text{km}$ and $h = 9 \, \text{km}$ . The rate of change of the area $A$ with respect to time $t$ can be found by differentiating the area formula with respect to $t$ .
Instant Values: $\frac{dA}{dt} = \frac{1}{2} \cdot \left(\frac{db}{dt} \cdot h + b \cdot \frac{dh}{dt}\right)$ . Now we plug in the values: $\frac{db}{dt} = -5 \, \text{km/h}$ , $\frac{dh}{dt} = 14 \, \text{km/h}$ , $b = 3 \, \text{km}$ , and $h = 9 \, \text{km}$ .
Differentiation: $\frac{dA}{dt} = \frac{1}{2} \times (-5 \times 9 + 3 \times 14)$ .
Substitute Values: $\frac{dA}{dt} = \frac{1}{2} \times (-45 + 42)$ .
Calculate: $\frac{dA}{dt} = \frac{1}{2} \times (-3)$ .
Final Result: $\frac{dA}{dt} = -1.5$ square kilometers per hour. So, the correct answer is (C) $-1.5$ .

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