Study Notes: Sine and Cosine Functions, Periodic Modeling, and Linear Approximation

Sine and Cosine Functions

Domain and Range

The sine and cosine functions are fundamental periodic functions in calculus, commonly used to model oscillatory phenomena.

• Domain: Both $y = \sin x$ and $y = \cos x$ are defined for all real numbers: $x \in (-\infty, \infty)$.

• Range: Both functions take values between -1 and 1: $y \in [-1, 1]$.

Periodicity

Sine and cosine functions repeat their values in regular intervals, known as their period.

• Period: $T = 2\pi$ for both $\sin x$ and $\cos x$.

• General Form:

  • $y = \sin x = \sin(x + 2\pi)$

  • $y = \cos x = \cos(x + 2\pi)$

Phase Shift and Relationship

Sine and cosine are shifted versions of each other. The graph of $y = \cos x$ is the same as $y = \sin(x + \frac{\pi}{2})$.

• Phase Shift: $\cos x = \sin(x + \frac{\pi}{2})$

Intercepts

• Sine x-intercepts: $y = \sin x$ crosses the x-axis at $x = k\pi$, where $k$ is any integer.

• Cosine x-intercepts: $y = \cos x$ crosses the x-axis at $x = \pm (2k+1)\frac{\pi}{2}$, where $k$ is any integer (odd multiples of $\frac{\pi}{2}$).

Modeling Periodic Phenomena with Sine and Cosine

General Steps for Modeling

Periodic functions are used to model phenomena that repeat at regular intervals, such as sound waves, tides, or hormone levels.

1. Choose a function: Select sine or cosine to model the phenomenon.

2. Find the period (T): Determine how long it takes for one full cycle.

3. Find the amplitude (A): Amplitude is half the difference between the maximum and minimum values:

   • $A = \frac{\text{max} - \text{min}}{2}$

4. Find the vertical shift (C): The vertical shift moves the graph up or down:

   • $C = \frac{\text{max} + \text{min}}{2}$

5. Determine the horizontal shift (d): The horizontal shift is based on how many units the graph has moved to the right or left.

General Equation

• $y(t) = A \sin(\omega (t - d)) + C$

• $y(t) = A \cos(\omega (t - d)) + C$

• Where $\omega = \frac{2\pi}{T}$ is the angular frequency.

Example: Hormone Level Modeling

Problem: The level of a certain hormone in the bloodstream fluctuates between undetectable and 100 ng/ml, with a period of 24 hours. If the cycle starts at midnight, what is the hormone level at midnight?

• Let $H(t)$ be the hormone level as a function of time $t$ (in hours).

• Maximum = 100, Minimum = 0

• Amplitude: $A = \frac{100 - 0}{2} = 50$

• Vertical shift: $C = \frac{100 + 0}{2} = 50$

• Period: $T = 24$ hours, so $\omega = \frac{2\pi}{24} = \frac{\pi}{12}$

• General equation: $H(t) = 50 \sin\left(\frac{\pi}{12}(t - d)\right) + 50$

• To find $d$, use the time when the hormone is at its minimum or maximum.

Linear Approximation

Definition and Purpose

Linear approximation uses the tangent line at a point to estimate the value of a function near that point. This is useful for approximating complicated functions with simpler linear functions.

• Formula: For a differentiable function $f(x)$ at $x = a$:

  • $L(x) = f(a) + f'(a)(x - a)$

• Tangent Line: The tangent line at $x = a$ is given by the same formula.

• Application: For values of $x$ near $a$, $L(x)$ gives a good approximation to $f(x)$.

Example: Linear Approximation of $\sqrt{x}$ Near $x = 1$

• $f(x) = \sqrt{x}$

• $f(1) = 1$

• $f'(x) = \frac{1}{2\sqrt{x}}$

• $f'(1) = \frac{1}{2}$

• Linear approximation near $x = 1$:

  • $L(x) = 1 + \frac{1}{2}(x - 1)$

• For $x = 2$, $L(2) = 1 + \frac{1}{2}(2 - 1) = 1.5$

• Actual $\sqrt{2} \approx 1.414$

Example: Linear Approximation of $f(x) = \sqrt{x + 4}$ Near $x = 5$

• $f(x) = \sqrt{x + 4}$

• $f(5) = \sqrt{9} = 3$

• $f'(x) = \frac{1}{2\sqrt{x + 4}}$

• $f'(5) = \frac{1}{2 \times 3} = \frac{1}{6}$

• Linear approximation near $x = 5$:

  • $L(x) = 3 + \frac{1}{6}(x - 5)$

• For $x = 6$, $L(6) = 3 + \frac{1}{6}(6 - 5) = 3 + \frac{1}{6} \approx 3.166$

• Actual $\sqrt{10} \approx 3.162$

Summary Table: Properties of Sine and Cosine Functions

Property	Sine ($\sin x$)	Cosine ($\cos x$)
Domain	$(-\infty, \infty)$	$(-\infty, \infty)$
Range	$[-1, 1]$	$[-1, 1]$
Period	$2\pi$	$2\pi$
x-intercepts	$x = k\pi$	$x = \pm (2k+1)\frac{\pi}{2}$
Phase Shift	None	$\sin(x + \frac{\pi}{2})$

Application Problem

Surface Area Rate of Change: (from the last page) "Is the surface area of the sphere increasing when the radius is 8 cm?"

• Let $S = 4\pi r^2$ be the surface area of a sphere.

• The rate of change of surface area with respect to radius is $\frac{dS}{dr} = 8\pi r$.

• At $r = 8$ cm, $\frac{dS}{dr} = 8\pi \times 8 = 64\pi$ cm2/cm.

• Since $\frac{dS}{dr} > 0$, the surface area is increasing as the radius increases.

Additional info: Linear approximation and modeling with sine/cosine are key topics in Calculus, especially in applications involving rates of change and periodic phenomena.