A line through the origin rotates around the origin in such a way that the angle, θ, between the line and the positive x-axis changes at a rate of dθ/dt for time t≥0. Give the expression for the rate at which the slope of the line is changing.

To find the expression for the rate at which the slope of the line is changing, we need to consider the relationship between the slope and the angle θ.

Let's define the slope of the line as m, and the x and y coordinates of any point on the line as (x, y). Since the line passes through the origin, we have x = 0 and y = 0.

Now, to relate the slope m to the angle θ, we can use trigonometry. We know that the tangent of θ is equal to the slope of the line. So, we can write:

tan(θ) = m

To find the relationship between the rate at which the slope is changing and the angle change rate, we need to differentiate both sides of this equation with respect to time t.

Differentiating the left side, we get:

d/dt(tan(θ)) = dθ/dt

On the right side, we need to apply the chain rule, because θ itself depends on time. The chain rule states that if a variable y = f(u) and u = g(t), then dy/dt = (df/du) * (du/dt). In this case, f(u) is tan(u) and u is θ, so we have:

d/dt(tan(θ)) = d(tan(θ))/dθ * dθ/dt

Now, we know that the derivative of the tangent function is the secant squared function:

d(tan(θ))/dθ = sec^2(θ)

Substituting this back into the equation, we have:

sec^2(θ) * dθ/dt = d/dt(tan(θ))

Finally, since we want to find the rate at which the slope of the line is changing, we are interested in the derivative of the slope itself, which is dm/dt. Since the slope is equal to tan(θ), we have:

dm/dt = d(tan(θ))/dt

Substituting the derived equation, we get:

dm/dt = sec^2(θ) * dθ/dt

Therefore, the expression for the rate at which the slope of the line is changing is dm/dt = sec^2(θ) * dθ/dt.