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? a. Write an equation relating the area of an isosceles right triangle, A, and the length of the legs of the triangle, x . A \( =\frac{1}{2} \mathrm{x}^{2} \) Differentiate both sides of the equation with respect to t . ?

Given the area of the triangle as \[ A = \frac{1}{2} x^2, \] differentiate both sides with respect to \( t \): \[ \frac{dA}{dt} = \frac{1}{2} \cdot 2x \frac{dx}{dt}. \] Simplify: \[ \frac{dA}{dt} = x \frac{dx}{dt}. \] --- **(a) Rate of change of the area when the legs are \(5\) m long** We are given that \[ \frac{dx}{dt} = 6\ \text{m/s}. \] When \( x = 5 \) m, substitute into the derivative: \[ \frac{dA}{dt} = 5 \times 6 = 30\ \text{m}^2/\text{s}. \] --- **(b) Rate of change of the area when the hypotenuse is \(4\) m long** For an isosceles right triangle, the hypotenuse \( c \) is related to the leg \( x \) by \[ c = x\sqrt{2}. \] Solve for \( x \) when \( c = 4 \) m: \[ x = \frac{4}{\sqrt{2}} = 2\sqrt{2}\ \text{m}. \] Now, substitute \( x = 2\sqrt{2} \) m into the derivative: \[ \frac{dA}{dt} = x \frac{dx}{dt} = 2\sqrt{2} \times 6 = 12\sqrt{2}\ \text{m}^2/\text{s}. \] --- **(c) Rate at which the length of the hypotenuse is changing** Recall the relationship: \[ c = x\sqrt{2}. \] Differentiate with respect to \( t \): \[ \frac{dc}{dt} = \sqrt{2}\frac{dx}{dt}. \] Substitute \(\frac{dx}{dt} = 6\ \text{m/s}\): \[ \frac{dc}{dt} = \sqrt{2} \times 6 = 6\sqrt{2}\ \text{m/s}. \]