Welcome back to class! As we approach the end of the semester, it is time to consolidate everything we have learned in Chapters 10 and 12. This review session is designed to test your understanding of parametric equations, polar coordinates, and vector geometry. Below, we walk through the solutions to the review test, highlighting the essential formulas and techniques you need to succeed.

Part 1: Calculus with Parametric Equations

We started the review by looking at a particle moving defined by the parametric equations $x = t^2$ and $y = t^3$. Parametric calculus requires us to think about how $x$ and $y$ change independently with respect to the parameter $t$.

  • Finding the Tangent Line: Remember that to find the slope in parametric mode, we use the chain rule: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$ In our example, at the point $(\frac{1}{4}, \frac{1}{8})$, we found the slope to be $\frac{3}{4}$, leading to the linear equation $y - \frac{1}{8} = \frac{3}{4}(x - \frac{1}{4})$.
  • Surface Area of Revolution: When rotating around the x-axis, the integral formula becomes somewhat complex. Don't forget the arc length element! The formula we used was: $$S = \int_{\alpha}^{\beta} 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$$

Part 2: Polar Coordinates and Integration

Next, we tackled the polar curve $r = 2\sin\theta$. A critical skill in Chapter 10 is converting between polar and rectangular forms and performing calculus on polar curves.

Key Takeaway: When calculating the Area under the curve in polar coordinates, we use the formula $A = \int \frac{1}{2}r^2 d\theta$. In this specific problem, we had to integrate $\sin^2\theta$. This is a classic calculus trap! Always remember your power-reducing identity:

$$\sin^2\theta = \frac{1}{2}(1 - \cos 2\theta)$$

Using this identity made the integration straightforward, resulting in an area of $\pi$. We also calculated the Arc Length using $L = \int \sqrt{r^2 + (dr/d\theta)^2} d\theta$, which simplified beautifully to $2\pi$.

Part 3: Vectors and Static Equilibrium

Moving into Chapter 12, we applied vectors to a real-world physics problem involving a crane suspending a 400-lb steel beam. This problem required us to find the tension in cables acting at $60^{\circ}$ angles.

To solve this, we broke the tension vectors ($T_1$ and $T_2$) into their horizontal ($i$) and vertical ($j$) components. Since the beam is in static equilibrium, the sum of the forces must equal zero:

  • Sum of horizontal forces = 0
  • Sum of vertical forces - Weight = 0

By setting up this system of equations, we determined the magnitude of the tension was approximately $230.9$ lbs.

Part 4: Vector Operations and Geometry

Finally, we reviewed pure vector geometry operations, which are essential for understanding 3D space.

  1. Angle Between Vectors: To find the angle between vector $\mathbf{u}$ and $\mathbf{v}$, we utilized the dot product formula: $$\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| ||\mathbf{v}||}$$ After calculating the dot product (which was 3) and the magnitudes, we found the angle to be approximately $79.9^{\circ}$.
  2. The Scalar Triple Product: The final challenge asked if four points lay in the same plane (coplanar). We used vectors $\vec{AB}, \vec{AC},$ and $\vec{AD}$ to form a determinant. The geometric interpretation of the scalar triple product is the volume of a parallelepiped defined by those vectors.

    Since our calculated determinant was $3$ (and not $0$), there is volume, which means the points do not lie in the same plane.

Keep practicing these formulas, especially the trigonometric identities for integration and the geometric interpretations of vector products. Good luck studying!