13. expanding isosceles triangle the legs of an isosceles right tri - angle increase in length at a rate of…

13. expanding isosceles triangle the legs of an isosceles right tri - angle increase in length at a rate of 2 m/s.\na. at what rate is the area of the triangle changing when the legs are 2 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?

13. expanding isosceles triangle the legs of an isosceles right tri - angle increase in length at a rate of 2 m/s.\na. at what rate is the area of the triangle changing when the legs are 2 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?

Answer

Explanation:

Step1: Recall the formula for the area of a triangle

The area of a right - angled triangle (A=\frac{1}{2}xy). For an isosceles right - triangle (x = y), so (A=\frac{1}{2}x^{2}).

Step2: Differentiate the area formula with respect to time (t)

Using the chain rule, (\frac{dA}{dt}=x\frac{dx}{dt}). Given (\frac{dx}{dt} = 2\ m/s).

Part (a)

When (x = 2\ m) Substitute (x = 2) and (\frac{dx}{dt}=2) into (\frac{dA}{dt}=x\frac{dx}{dt}) (\frac{dA}{dt}=2\times2)

Part (b)

First, find (x) when the hypotenuse (h) is given. For an isosceles right - triangle (h=\sqrt{x^{2}+x^{2}}=\sqrt{2}x). When (h = 1), then (x=\frac{1}{\sqrt{2}}) Substitute (x=\frac{1}{\sqrt{2}}) and (\frac{dx}{dt}=2) into (\frac{dA}{dt}=x\frac{dx}{dt}) (\frac{dA}{dt}=\frac{1}{\sqrt{2}}\times2=\sqrt{2})

Part (c)

Recall (h=\sqrt{2}x). Differentiate with respect to (t) using the chain rule (\frac{dh}{dt}=\sqrt{2}\frac{dx}{dt}) Since (\frac{dx}{dt}=2\ m/s), then (\frac{dh}{dt}=2\sqrt{2}\ m/s)

Answer:

a. (4\ m^{2}/s) b. (\sqrt{2}\ m^{2}/s) c. (2\sqrt{2}\ m/s)