A point is moving along the $x$ axis at a constant rate of 5 units per second. At which point is its distance from $(0,1)$ increasing at a rate of 2 units per second? At 4 units per second? At 5 units per second? At 10 units per second? Explain.

Step 1 ：Let the position of the point on the x-axis be x(t), where t is the time in seconds. Since the point is moving at a constant rate of 5 units per second, we can write x(t) = 5t.

Step 2 ：Find the distance between the point (x(t), 0) and (0, 1) using the distance formula: \(distance = \sqrt{(x(t) - 0)^2 + (0 - 1)^2}\)

Step 3 ：Simplify the distance equation: \(distance = \sqrt{(5t)^2 + 1^2}\)

Step 4 ：Find the derivative of the distance with respect to time (ddt) using the chain rule: \(ddt = \frac{1}{2} \cdot \frac{2 \cdot 5t \cdot 5}{\sqrt{(5t)^2 + 1^2}}\)

Step 5 ：Solve for t when ddt is equal to 2, 4, 5, and 10 units per second: \(t \approx 0.087, 0.267, \text{no real solution}, \text{no real solution}\)

Step 6 ：\(\boxed{\text{The point's distance from (0,1) is increasing at a rate of 2 units per second when it is at t ≈ 0.087 seconds, at a rate of 4 units per second when it is at t ≈ 0.267 seconds. The distance cannot increase at a rate of 5 or 10 units per second, as there are no real solutions for t in those cases.}}\)