The problem involves finding the speed of a particle moving along a curve in the xy-plane at time $t = 1$. The speed is the magnitude of the velocity vector, which is determined using the derivatives of the position functions $x(t)$ and $y(t)$. Here's the step-by-step solution:

Problem Setup

We are given: $x'(t) = (t - 10)^2, \quad y'(t) = 9t^2 - 7t$

At $t = 1$, the speed is calculated as: $\text{Speed} = \sqrt{x'(t)^2 + y'(t)^2}$

Step 1: Calculate $x'(1)$

Substitute $t = 1$ into $x'(t)$: $x'(1) = (1 - 10)^2 = (-9)^2 = 81$

Step 2: Calculate $y'(1)$

Substitute $t = 1$ into $y'(t)$: $y'(1) = 9(1)^2 - 7(1) = 9 - 7 = 2$

Step 3: Compute Speed

The speed is given by: $\text{Speed} = \sqrt{x'(1)^2 + y'(1)^2}$ Substitute $x'(1) = 81$ and $y'(1) = 2$: $\text{Speed} = \sqrt{81^2 + 2^2} = \sqrt{81 + 4} = \sqrt{85}$ Using a calculator: $\text{Speed} \approx 9.220$

Final Answer:

The speed of the particle at $t = 1$ is approximately: $\boxed{9.220}$

Would you like a detailed explanation of any of these steps? Here are five related questions:

1. What is the physical significance of the magnitude of the velocity vector?
2. How does the derivative of position relate to the velocity vector in general?
3. Can you find the particle's speed at a different time $t$?
4. How does the speed change over time in this example?
5. What is the relationship between acceleration and speed in this problem?

Tip: Always double-check your derivative evaluations to ensure accuracy in physics-based problems!