Calculus 2

Given a parametric function with a single input $T$ and a vector output defined as: $x = T \cdot \cos(T)$ and $y = T \cdot \sin(T)$, evaluate the function at different points and trace the curve it forms.

Show that a parametric function tracing a curve can be re-parameterized to alter the rate at which the curve is traced without changing its shape.

Find the slope of the tangent line to a parameterized curve given functions $x(t)$ and $y(t)$.

Find the second derivative of a parameterized curve given functions $x(t)$ and $y(t)$.

Find the area under the curve of a parametric function where x = t^2, y = 4t^2 - t^4, and t is bounded between 0 and 2.

Parametrize the parabola $Y = X^2$ that lies in the plane $Z = 1$ from the starting point where $X = -1$ to the ending point where $X = 2$ using $R(T) = (T, T^2, 1)$ for $T \in [-1, 2]$.

Parametrize the curve of intersection of surfaces $Z=X^3$ and $Y=X^2$ from the origin to the point $(2,4,8)$ using $X=T$, $Y=T^2$, $Z=T^3$ for $T\in [0,2]$.

Parametrize the intersection of the cylinder $X^2 + Y^2 = 4$ and the plane $2X + 3Y + 5Z = 10$ using polar coordinates for X and Y, and solving for Z. Use $R(T) = (2\cos(T), 2\sin(T), 2 - \frac{4}{5}\cos(T) - \frac{6}{5}\sin(T))$ for T in $[0, 2\pi]$.

Parametrize the curve $y = 2x^2 + 1$ by letting $x = t$, and then find the parameterization when $x = 2t$ and $x = t + 3$, and discuss the practical differences between these parameterizations.

Graph the polar equation $r = 2 \cos(\theta)$, and verify by converting to rectangular form.

Find the area of a circle using polar coordinates.

Graph the equation where $r = a \cos(2\theta)$.

Find the equation of the tangent line for the polar equation $r = 2 + 3\cos(\theta)$ when $\theta = \frac{\pi}{2}$.

Write the equation of the tangent line for the polar equation $r = 3 - 2\sin(\theta)$ when $\theta = \pi$.

Evaluate the double integral $\displaystyle \int_{0}^{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} e^{-r^2} r \, dr \, d\theta$ by converting it to polar coordinates in the region bounded by $x = \sqrt{4 - y^2}$ and the y-axis.

Solve for y given $\frac{dy}{dx} = \frac{x^2}{y^2}$.

Solve for $y$ given $\frac{dy}{dx} = \frac{6x^2}{2y + \cos(y)}$.

dy/dx = xe$y$. Given the initial condition $y(0) = 0$, solve for $y$.

Solve the separable differential equation $\frac{dY}{dX} = -\frac{X}{Y}e^{X^2}$ given the initial condition that the solution must pass through the point (0,1).

Solve the differential equation: $\frac{dy}{dx} = x + x y^2$ with the initial condition $y(0) = -1$.