Parametric Equations, Polar Coordinates, and Vector–Valued Functions (BC Only)

Correct

Period

Assess through the p-test, noting that the type of denominators doesn't meet the criteria of being greater than 1.

Apply the Comparison Test with the harmonic series. If the terms of the given series are smaller than the terms of the harmonic series, and the harmonic series converges, then the given series must also converge.

Use the Integral Test, since the integration of the function $y=\frac{1}{x}$ shows that the area under the curve is infinite, corresponding to a divergent series.

Determine using the Root Test, because squaring each term provides insight into the behavior for large values enforced.

$A = \frac{1}{2} \int_{0}^{2\pi} r(\theta)^2 d\theta$

$A = \frac{1}{2} \int_{0}^{\theta} r(\theta)^3 d\theta$

$A = \frac{1}{2} \int_{0}^{2\pi} r(\theta)^3 d\theta$

$A = \frac{1}{2} \int_{0}^{\theta} r(\theta)^2 d\theta$

Finding the derivative of the curves

Subtracting the area under one curve from the area under the other

Adding the areas of the two curves

Dividing the area by the radius

$\frac{1}{6}\int_{0}^{\pi}(9\cos^6(\theta))d\theta$

$\frac{1}{6}\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}(3\cos(3\theta))^2d\theta$

$\frac{1}{6}\int_{0}^{\pi}(3\cos(3\theta))^2d\theta$

$\frac{1}{6}\int_{0}^{\frac{\pi}{3}}(3\cos(3\theta))^2d\theta$

Take two large steps directly from $(1,2)$ assuming linearity throughout each interval spanning multiple units along x-axis.

Directly calculate from $(5,y(5))$ backward two steps since midpoints provide better estimates than endpoints or beginnings.

Use one step from $(1,2)$ giving an estimate at $(3,y(3))$ based on a linear approximation guided by $\frac{dy}{dx}$ at $(1,2).

Apply four small steps forward but average all intermediate values for $y(x)$ before estimating actual value at $x=3$.

Plot the initial point on the slope field and follow the pattern of slopes to approximate the solution curve.

Integrate the slopes at each point starting from $(x_0, y_0)$ without considering the overall pattern.

Use Euler's method starting from $(x_0, y_0)$ without plotting any points on the slope field.

Solve for $y$ algebraically using only the differential equation without referring to the slope field.

Exactly $0.25$

Approximately $0.312$

Approximately $0.75$

Approximately $0.5$

The length of one loop in the function

The total area under the curve

The rate of change of the function

The interval for integration

Integral from $0$ to $\pi/2$ and multiply the result by two since the curve has two petals

Interval from $0$ to $\pi/2$, the result will be the same as it is an even function and the whole loop is covered

Integral from $0$ to $\pi/2$ and add the area of the first half petal to get the total area (which will be incorrect)

Integrate from $0$ to $\pi/2$ and double the result since it is an odd function