1. **State the problem:** We have a model car moving on a circular track centered at (20, 0) with radius 5 feet. The fixed point is at the origin (0, 0). We want to find the rate of change of the distance $Z$ between the car at $(x, y)$ and the origin when $x=23$, $y=4$, and $\frac{dx}{dt}=2$. 2. **Relevant formulas:** - The circle equation: $$(x - 20)^2 + y^2 = 5^2 = 25$$ - Distance from origin to car: $$Z = \sqrt{x^2 + y^2}$$ 3. **Find $\frac{dy}{dt}$ using the circle equation:** Differentiate both sides with respect to $t$: $$2(x - 20) \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$$ Simplify: $$ (x - 20) \frac{dx}{dt} + y \frac{dy}{dt} = 0$$ Solve for $\frac{dy}{dt}$: $$\frac{dy}{dt} = -\frac{(x - 20)}{y} \frac{dx}{dt}$$ 4. **Substitute known values:** $$x = 23, y = 4, \frac{dx}{dt} = 2$$ $$\frac{dy}{dt} = -\frac{(23 - 20)}{4} \times 2 = -\frac{3}{4} \times 2 = -\frac{3}{2} = -1.5$$ 5. **Find $\frac{dZ}{dt}$:** Differentiate $Z = \sqrt{x^2 + y^2}$: $$Z = (x^2 + y^2)^{1/2}$$ $$\frac{dZ}{dt} = \frac{1}{2}(x^2 + y^2)^{-1/2} \times 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{\sqrt{x^2 + y^2}}$$ 6. **Substitute values:** $$x = 23, y = 4, \frac{dx}{dt} = 2, \frac{dy}{dt} = -1.5$$ Calculate numerator: $$23 \times 2 + 4 \times (-1.5) = 46 - 6 = 40$$ Calculate denominator: $$\sqrt{23^2 + 4^2} = \sqrt{529 + 16} = \sqrt{545}$$ 7. **Final rate of change:** $$\frac{dZ}{dt} = \frac{40}{\sqrt{545}}$$ 8. **Interpretation:** Since $\frac{dZ}{dt} > 0$, the model car is moving away from the fixed point at approximately $\frac{40}{\sqrt{545}} \approx 1.7$ feet per second. **Answer:** B) $\frac{dZ}{dt} = \frac{40}{\sqrt{545}}$, the car is moving away from the fixed point at about 1.7 ft/s.