the legs of an isosceles right triangle increase in length at a rate of 3 m/s.\na. at what rate is the area…

the legs of an isosceles right triangle increase in length at a rate of 3 m/s.\na. at what rate is the area of the triangle changing when the legs are 3 m long?\nb. at what rate is the area of the triangle changing when the hypotenuse is 1 m long?\nc. at what rate is the length of the hypotenuse changing?\na. write an equation relating the area of an isosceles right triangle, a, and the length of the legs of the triangle, x.\n$a = \\frac{1}{2}x^{2}$\ndifferentiate both sides of the equation with respect to t.\n$\\frac{da}{dt}=x(\\frac{dx}{dt})$\nwhen the legs are 3 m long, the area of the triangle is changing at a rate of 9 $m^{2}/s$.\n(type an exact answer, using radicals as needed.)\nb. when the hypotenuse is 1 m long, the area of the triangle is changing at a rate of \n(type an exact answer, using radicals as needed.)

the legs of an isosceles right triangle increase in length at a rate of 3 m/s.\na. at what rate is the area of the triangle changing when the legs are 3 m long?\nb. at what rate is the area of the triangle changing when the hypotenuse is 1 m long?\nc. at what rate is the length of the hypotenuse changing?\na. write an equation relating the area of an isosceles right triangle, a, and the length of the legs of the triangle, x.\n$a = \\frac{1}{2}x^{2}$\ndifferentiate both sides of the equation with respect to t.\n$\\frac{da}{dt}=x(\\frac{dx}{dt})$\nwhen the legs are 3 m long, the area of the triangle is changing at a rate of 9 $m^{2}/s$.\n(type an exact answer, using radicals as needed.)\nb. when the hypotenuse is 1 m long, the area of the triangle is changing at a rate of \n(type an exact answer, using radicals as needed.)

Answer

Explanation:

Step1: Find the length of the leg when hypotenuse is 1m

For an isosceles right - triangle, if the hypotenuse is (h) and the leg is (x), by Pythagorean theorem (h^{2}=x^{2}+x^{2}=2x^{2}). Given (h = 1), then (1^{2}=2x^{2}), so (x=\frac{1}{\sqrt{2}})

Step2: Use the formula (\frac{dA}{dt}=x\frac{dx}{dt})

We know that (\frac{dx}{dt}=3) m/s. Substitute (x = \frac{1}{\sqrt{2}}) and (\frac{dx}{dt}=3) into (\frac{dA}{dt}=x\frac{dx}{dt})

[ \begin{align*} \frac{dA}{dt}&=\frac{1}{\sqrt{2}}\times3\ &=\frac{3\sqrt{2}}{2} \end{align*} ]

Answer:

(\frac{3\sqrt{2}}{2}\text{ m}^2/\text{s})