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 $3\mathrm{in}/\sec$. 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: $\frac{dA}{dt}=\frac{1}{2}\times (\frac{d(\text{short leg})}{dt}\times \text{long leg}+\text{short leg}\times \frac{d(\text{long leg})}{dt})$

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 $\boxed{{\displaystyle 174}}\frac{{\text{in}}^2}{\text{sec}}$