Math Review Week 5

Highlights of what I (re)learned in week #5 of my math studies. Shorter week due to traveling.

Position, Velocity, and Acceleration

Suppose a car's position (in meters) at time \(t\) (in seconds) is

\[ s(t) = t^3 - 3t^2 + 2t \]

Derivatives describe how these quantities change. Velocity is the derivative of position, and acceleration is the derivative of velocity:

\[ v(t) = s'(t) = 3t^2 - 6t + 2 \]

\[ a(t) = v'(t) = 6t - 6 \]

Antiderivatives let us work backward. Integrating acceleration gives velocity, while integrating velocity gives position:

\[ v(t) = \int (6t - 6)\,dt = 3t^2 - 6t + C \]

\[ s(t) = \int (3t^2 - 6t + 2)\,dt = t^3 - 3t^2 + 2t + C \]

The constant of integration is determined by the initial conditions, such as the car's initial velocity and position.

In short, differentiation takes us from position to velocity to acceleration; integration takes us in the reverse direction.

Finding Volume Using the Disc Method

Suppose we rotate the parabola \(y=x^2\) around the \(y\)-axis on the interval \(y\in [0,4]\) . The resulting solid is a paraboloid.

To use the disc method, we express \(x\) in terms of \(y\):

\[ x=\sqrt{y} \]

At each height \(y\), a cross-section perpendicular to the \(y\)-axis is a circle with radius \(r(y)=\sqrt{y}\). Its area is

\[ A(y)=\pi r(y)^2=\pi y \]

Since \(y\) ranges from \(0\) to \(4\), the volume is

\[ V=\int_0^4 \pi y\,dy \]

Evaluating the integral gives

\[ \begin{aligned} V&=\pi\left[\frac{y^2}{2}\right]_0^4\\[6pt] &=\pi\left(\frac{16}{2}-0\right)\\[6pt] &=\boxed{8\pi} \end{aligned} \]

The solid has a volume of (\(8\pi\)) cubic units.

The key idea behind the disc method is to integrate the areas of infinitely many thin circular cross-sections to obtain the volume of the entire solid. In general, when rotating \(x=f(y)\) around the \(y\)-axis,

\[ V=\pi\int_c^d [f(y)]^2\,dy \]

Finding Volume Using the Washer Method

The washer method is useful when rotating the region between two curves around an axis, creating a solid with a hole through its center.

Suppose we rotate the region between \(y=x\) and \(y=x^2\) on the interval \([0,1]\) around the \(x\)-axis. Each cross-section is a washer, with outer radius \(R(x)=x\) and inner radius \(r(x)=x^2\).

The area of a washer is the area of the outer circle minus the area of the inner circle:

\[ A(x)=\pi R(x)^2-\pi r(x)^2 \]

Substituting the functions, we obtain the volume integral:

\[ V=\pi\int_0^1 (x^2-x^4)\,dx \]

Evaluating the integral gives

\[ \begin{aligned} V&=\pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1\\[6pt] &=\pi\left(\frac13-\frac15\right)\\[6pt] &=\boxed{\frac{2\pi}{15}} \end{aligned} \]

The solid has a volume of (\(\frac{2\pi}{15}\)) cubic units.

The key idea is to subtract the inner circle's area from the outer circle's area before integrating. In general, the washer method uses

\[ V=\pi\int_a^b \left(R(x)^2-r(x)^2\right)\,dx \]