Parametric Functions and Rates of Change - AP Precalculus

1. Identify the type of curve: $x = 3 \text{cos}(t), y = 3 \text{sin}(t)$.

   A circle with radius 3. Standard parametric form for a circle centered at origin.

2. Identify the parameter for $x(t) = \text{tan}(t), y(t) = \text{sec}(t)$.

   The parameter is $t$. The independent variable in these parametric equations.

3. What is the rate of change of arc length in parametric form?

   $\frac{ds}{dt} = \sqrt{((\frac{dx}{dt})^2 + (\frac{dy}{dt})^2)}$. Magnitude of velocity vector gives speed of arc traversal.

4. What is the second derivative $\frac{d^2y}{dx^2}$ for parametric equations?

   $\frac{d}{dt}(\frac{dy}{dx}) / \frac{dx}{dt}$. Uses chain rule twice for parametric second derivatives.

5. Find the point at $t = \text{π}$ for $x(t) = \text{cos}(t), y(t) = \text{sin}(t)$.

   Point is $(-1, 0)$.. At $t = \pi$: $\cos(\pi) = -1$, $\sin(\pi) = 0$.

6. Convert $x = 5\text{cos}(t), y = 3\text{sin}(t)$ to Cartesian form.

   $(\frac{x}{5})^2 + (\frac{y}{3})^2 = 1$. Ellipse with semi-major axis 5 and semi-minor axis 3.

7. Find the slope of the tangent line at $t = 1$ for $x(t) = t^2, y(t) = t^3$.

   Slope is $\frac{3}{2}$. Evaluate $\frac{dy}{dx} = \frac{3t^2}{2t}$ at $t = 1$.

8. State the parametric equations for a line with slope $m$ passing through $(x_0, y_0)$.

   $x = x_0 + t, y = y_0 + mt$. Direction vector $(1, m)$ parameterizes the line.

9. Identify the parameter for $x(t) = \text{sin}(t), y(t) = \text{cos}(t)$.

   The parameter is $t$. The independent variable in the parametric representation.

10. Identify the parameter in: $x(t) = 3t + 2, y(t) = 4t - 1$.

    The parameter is $t$. The independent variable in parametric equations.

11. What is the geometric interpretation of $\frac{dy}{dt}$ and $\frac{dx}{dt}$?

    Rates of change of $y$ and $x$ with respect to $t$. Components of velocity vector in parametric motion.

12. What is the parametric form for a line segment from $(x_0, y_0)$ to $(x_1, y_1)$?

    $x = x_0 + t(x_1-x_0), y = y_0 + t(y_1-y_0)$. Linear interpolation between two points using parameter $t$.

13. Determine the point at $t = 0$ for $x(t) = 2t + 1, y(t) = 3t - 2$.

    Point is $(1, -2)$. Substitute $t = 0$ into both equations.

14. Identify the curve: $x = 4\text{cos}(t), y = 4\text{sin}(t)$.

    A circle with radius 4. Parametric circle with radius 4 centered at origin.

15. Find the point at $t = \frac{\text{π}}{2}$ for $x(t) = \text{cos}(t), y(t) = \text{sin}(t)$.

    Point is $(0, 1)$.. At $t = \frac{\pi}{2}$: $\cos(\frac{\pi}{2}) = 0$, $\sin(\frac{\pi}{2}) = 1$.

16. Convert $x = 2t + 3, y = 4t - 5$ to a Cartesian equation.

    $y = 2x - 11$. From $x = 2t + 3$, get $t = \frac{x-3}{2}$, substitute.

17. What is the parametric form for a hyperbola?

    $x = a\text{sec}(t), y = b\text{tan}(t)$. Standard parametrization using secant and tangent functions.

18. Convert $x = 2\text{cos}(t), y = 2\text{sin}(t)$ to Cartesian form.

    $x^2 + y^2 = 4$. Use identity $\cos^2(t) + \sin^2(t) = 1$.

19. Convert $x(t) = t, y(t) = t^2$ to a Cartesian equation.

    $y = x^2$. Direct substitution since $x = t$.

20. Find $\frac{dy}{dx}$ for $x(t) = 2t^2, y(t) = 3t^2$.

    $\frac{dy}{dx} = \frac{3}{2}$. $\frac{dy}{dt} = 6t$, $\frac{dx}{dt} = 4t$, so ratio is constant.

21. Find $\frac{dy}{dx}$ for $x(t) = \text{cos}(t), y(t) = \text{sin}(t)$.

    $\frac{dy}{dx} = -\text{cot}(t)$. $\frac{dy}{dt} = \cos(t)$, $\frac{dx}{dt} = -\sin(t)$, so ratio is $-\cot(t)$.

22. What is the parametric representation for an ellipse?

    $x = a\text{cos}(t), y = b\text{sin}(t)$. General form with semi-axes $a$ and $b$.

23. Find the point at $t = \frac{\pi}{2}$ for $x(t) = \cos(t), y(t) = \sin(t)$.

    Point is $(0, 1)$. At $t = \frac{\pi}{2}$: $\cos(\frac{\pi}{2}) = 0$, $\sin(\frac{\pi}{2}) = 1$.

24. What is the derivative $\frac{dy}{dx}$ for parametric equations $x(t)$ and $y(t)$?

    $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. Chain rule applied to parametric equations.

25. State the parametric equations for a line through $(1, 2)$ with slope 3.

    $x = 1 + t, y = 2 + 3t$. Point $(1,2)$ with direction vector $(1,3)$.

26. Convert $x = 3t + 1, y = 2t - 4$ to Cartesian form.

    $y = \frac{2}{3}(x - 1) - 4$. Solve for $t$ from first equation, substitute into second.

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