Atriangle has two constant sides of length 3 feet and 5 feet. the angle between these two sides is increasing at a rate of 0.1 radians per second. find the rate at which the area of the triangle is changing when the angle between the two sides is π/6.

The rate of change of the area is 0.65 feet²/second

Step-by-step explanation:

* Lets explain how to solve the problem

- A triangle has two constant sides of length 3 feet and 5 feet

∵ The angle between them is Ф

- The angle between these two sides is increasing at a rate of

 0.1 radians per second

∴ dФ/dt = 0.1 radians/second

- The rule of the area of a triangle whose sides are a , b and the angle

  between them is Ф is ⇒ A = 1/2 × a × b × sin Ф

∵ The two sides of the triangle are 3 feet and 5 feet

∵ The area of the triangle = 1/2 × 3 × 5 × sin Ф

∴ A = 7.5 sin Ф

- The rate of change of the area is dA/dt

- Lets find the differentiation of the area with respect to angle Ф

∵ A = 7.5 sin Ф

- Remember the differentiation of sin Ф is cos Ф

∴ dA/dФ = 7.5 cos Ф

- The angle between the two sides is π/6

∴ Ф = π/6

∴ dA/dФ = 7.5 cos(π/6)

- The rate of change of the area is dA/dt

∵ dA/dФ = 7.5 cos(π/6) ⇒ (1)

∵ dФ/dt = 0.1 ⇒ (2)

- Multiply (1) and (2) to find dA/dt

∴ dA/dФ × dФ/dt = 7.5 cos(π/6) × 0.1 ⇒ (dФ up cancel dФ down)

∴ dA/dt = 0.75 cos(π/6) = (3√3)/8 = 0.649519

∴ dA/dt ≅ 0.65 feet²/second

* The rate of change of the area is 0.65 feet²/second

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8

b.

step-by-step explanation: