The legs of an isosceles right triangle increase in length at a rate of $$ 6 \mathrm{~m} / \mathrm{s} $$. a. At what rate is the area of the triangle changing when the legs are 5 m long? b. At what rate is the area of the triangle changing when the hypotenuse is 4 m long? c. At what rate is the length of the hypotenuse changing? Uiterentuate both sides of the equation with respect to t . $$ \frac{\mathrm{dA}}{\mathrm{dt}}=\mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}} $$ When the legs are 5 m long, the area of the triangle is changing at a rate of $$ 30 \mathrm{~m}^{2} / \mathrm{s} $$. (Type an exact answer, using radicals as needed.) b. When the hypotenuse is 4 m long, the area of the triangle is changing at a rate of $$ 12 \sqrt{2} \mathrm{~m} / \mathrm{s} $$. (Type an exact answer, using radicals as needed.) c. Write an equation relating the length of the legs of an isosceles triangle, x, to the length of the hypotenuse of the triangle, h . $$ \square $$ ?

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**a. Calculation of $$\frac{dA}{dt}$$ when $$x = 5\, \text{m}$$:**  
The area of an isosceles right triangle with leg length $$x$$ is given by  
$$
A = \frac{1}{2}x^2.
$$  
Differentiating both sides with respect to time $$t$$ yields  
$$
\frac{dA}{dt} = x \frac{dx}{dt}.
$$  
Given that $$\frac{dx}{dt} = 6\, \text{m/s}$$ and $$x = 5\, \text{m}$$, we have  
$$
\frac{dA}{dt} = 5 \times 6 = 30\, \text{m}^2/\text{s}.
$$

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**b. Calculation of $$\frac{dA}{dt}$$ when $$h = 4\, \text{m}$$:**  
For an isosceles right triangle, the relationship between the leg $$x$$ and the hypotenuse $$h$$ is  
$$
h = x\sqrt{2}.
$$  
When $$h = 4\, \text{m}$$,
$$
x = \frac{h}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\, \text{m}.
$$  
Using the differentiated area formula  
$$
\frac{dA}{dt} = x \frac{dx}{dt},
$$  
with $$\frac{dx}{dt} = 6\, \text{m/s}$$ and $$x = 2\sqrt{2}\, \text{m}$$, we obtain  
$$
\frac{dA}{dt} = 2\sqrt{2} \times 6 = 12\sqrt{2}\, \text{m}^2/\text{s}.
$$

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**c. Equation Relating $$x$$ and $$h$$:**  
In an isosceles right triangle, applying the Pythagorean theorem gives  
$$
h^2 = x^2 + x^2 = 2x^2.
$$  
Taking the square root of both sides (and considering only the positive root as lengths are positive) yields  
$$
h = x\sqrt{2}.
$$
**a.** The area is changing at a rate of $$30\, \text{m}^2/\text{s}$$ when the legs are 5 m long. **b.** The area is changing at a rate of $$12\sqrt{2}\, \text{m}^2/\text{s}$$ when the hypotenuse is 4 m long. **c.** The relationship between the leg length $$x$$ and the hypotenuse $$h$$ is $$ h = x\sqrt{2}. $$

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