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