Question

If the length of the side of the rhombus is 4 in and is increasing at a rate of 1/2 in/s, and the angle is π/4 rad, and the area of the rhombus is decreasing at a rate of 1/4 in²/s, at what rate is the angle changing, and is it getting larger or smaller?

Answer 1

The rate at which the angle is changing is - 0.322 radians per second. As - 0.322 is negative, the angle is decreasing.

How to calculate the rate at which the angle is changing

Given:

Length of side of the rhombus (s) = 4 inches

Rate of increase of the side (ds/dt) = 1/2 inches per second

Angle of the rhombus (θ) = π/4 radians

Rate of decrease of the area of the rhombus (dA/dt) = 1/4 square inches per second

We want to find:

The rate at which the angle is changing (dθ/dt)

The area of a rhombus is given by the formula: A =
`s^2` * sin(θ)

Given that the area is decreasing at a rate of 1/4 square inches per second, differentiate the area formula with respect to time to find the rate of change of the area:

dA/dt = d/dt (
`s^2` * sin(θ))

Now, substitute the values:

- 1/4 = 2s * sin(θ) * ds/dt +
`s^2` * cos(θ) * dθ/dt

Here, ds/dt = 1/2 (the rate of increase of the side) and s = 4.

1/4 = 2 * 4 * sin(π/4) * 1/2 +
`4^2` * cos(π/4) * dθ/dt

Simplify:

1/4 = 4 * √2/2 + 16 * √2/2 * dθ/dt

1/4 = 2√2 + 8√2 * dθ/dt

Now solve for dθ/dt:

8√2 * dθ/dt = 1/4 - 2√2

dθ/dt = (-1/4 - 2√2) / (8√2)

dθ/dt = (1 - 8√2) / 32

dθ/dt = - 0.322

So, the rate at which the angle is changing is - 0.322 radians per second. As - 0.322 is negative, the angle is decreasing.