Related Rates - Triangles
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A right triangle has legs of 36 inches and 48 inches whose sides are changing. The short leg is increasing by 5 in/sec and the long leg is growing at $3 \mathrm{in} / \mathrm{sec}$. What is the rate of change of the area?
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Step 1 ：The area of a right triangle is given by the formula \( A = \frac{1}{2} \times \text{base} \times \text{height} \)

Step 2 ：Given the legs of the triangle as the base and height, we have \( A = \frac{1}{2} \times 36 \times 48 \)

Step 3 ：The rate of change of the short leg is \( \frac{d(\text{short leg})}{dt} = 5 \frac{\text{in}}{\text{sec}} \) and the long leg is \( \frac{d(\text{long leg})}{dt} = 3 \frac{\text{in}}{\text{sec}} \)

Step 4 ：To find the rate of change of the area, we take the derivative of the area with respect to time: \( \frac{dA}{dt} = \frac{1}{2} \times \left( \frac{d(\text{short leg})}{dt} \times \text{long leg} + \text{short leg} \times \frac{d(\text{long leg})}{dt} \right) \)

Step 5 ：Substituting the given values, we get \( \frac{dA}{dt} = \frac{1}{2} \times (5 \times 48 + 36 \times 3) \)

Step 6 ：Simplifying, we find \( \frac{dA}{dt} = \frac{1}{2} \times (240 + 108) \)

Step 7 ：Thus, \( \frac{dA}{dt} = \frac{1}{2} \times 348 \)

Step 8 ：Finally, \( \frac{dA}{dt} = 174 \frac{\text{in}^2}{\text{sec}} \)

Step 9 ：The rate of change of the area of the triangle is \(\boxed{174} \frac{\text{in}^2}{\text{sec}}\)