Math Review Week 6

Highlights of what I (re)learned in week #6 of my math studies.

Arc Length

The formula for computing the arc length of a curve from a to b is:

\[ \int_a^b \sqrt{1 +(\frac{dy}{dx})^2} \; dx \]

or

\[ \int_a^b \sqrt{1 +(f'(x))^2} \; dx \]

Given

\(x = 2\sin(1+3t)\) and \(y = 2t^3\)

To find \(\frac{dy}{dx}\) when \(t=-\frac{1}{3}\), use the formula for parametric equations:

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]

Step 1: Find the derivatives with respect to \(t\).

Using the chain rule,

\[ \frac{dx}{dt} = 6\cos(1+3t) \]

\[ \frac{dy}{dt} = 6t^2 \]

Step 2: Calculate \(\frac{dy}{dx}\).

Substitute the derivatives into the formula:

\[ \frac{dy}{dx} = \frac{6t^2}{6\cos(1+3t)} = \frac{t^2}{\cos(1+3t)} \]

Step 3: Evaluate at \(t=-\frac{1}{3}\).

\[ \frac{dy}{dx} = \frac{\left(-\frac{1}{3}\right)^2} {\cos\left(1+3\left(-\frac{1}{3}\right)\right)} \]

Simplify:

\[ = \frac{\frac{1}{9}}{\cos(0)} = \frac{\frac{1}{9}}{1} \]

Therefore, the final answer is

\[ \boxed{\frac{dy}{dx} = \frac{1}{9}} \]

Given the vector valued function:

\(\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}\)

The derivative is:

\(\vec{r}^{\,\prime}(t) = x'(t)\hat{i} + y'(t)\hat{j}\)

\(x=r\cdot cos(\theta)\) and \(y=r\cdot sin(\theta)\)

Given

\(r=sin(2\theta)\)

Then

\(x=sin(2\theta)\cdot cos(\theta)\) and \(y=sin(2\theta)\cdot sin(\theta)\)

Polar Curve Sector

The area of a sector of angle \(\theta\) of a circle is \(\pi\cdot r^2\cdot \frac{\theta}{2\pi}\) or \(\frac{r^2 \theta}{2}\)

So the area from angle \(\alpha\) to \(\beta\) of a polar curve is:

\(\int_\alpha^\beta \frac{1}{2}(r(\theta))^2 \; d\theta\)