Answer:
None of these
General Formulas and Concepts:
Pre-Calculus
Calculus
Derivatives
Derivative Notation
Derivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Trig Derivative: [tex]\displaystyle \frac{d}{dx}[cos(u)] = -u'sin(u)[/tex]
Polar Derivative: [tex]\displaystyle \frac{dy}{dx} = \frac{rcos(\theta) + r'sin(\theta)}{r'cos(\theta) - rsin(\theta)}[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle r = 6cos(2 \theta) - 5[/tex]
[tex]\displaystyle \theta = \frac{\pi}{4}[/tex]
Step 2: Differentiate
- Trig Derivative [Chain Rule]: [tex]\displaystyle r' = \frac{d}{d\theta}[6cos(2 \theta) - 5] \cdot \frac{d}{d\theta}[2\theta][/tex]
- Rewrite [Derivative Rule - Subtraction]: [tex]\displaystyle r' = \bigg[ \frac{d}{d\theta}[6cos(2 \theta)] - \frac{d}{d\theta}[5] \bigg] \cdot \frac{d}{d\theta}[2\theta][/tex]
- Rewrite [Derivative Rule - Multiplied Constant]: [tex]\displaystyle r' = \bigg[ 6\frac{d}{d\theta}[cos(2 \theta)] - \frac{d}{d\theta}[5] \bigg] \cdot \frac{d}{d\theta}[2\theta][/tex]
- Trig Derivative: [tex]\displaystyle r' = [-6sin(2\theta)] \cdot \frac{d}{d\theta}[2\theta][/tex]
- Basic Power Rule: [tex]\displaystyle r' = -6sin(2\theta) \cdot 2\theta^{1 - 1}[/tex]
- Simplify: [tex]\displaystyle r' = -12sin(2\theta)[/tex]
- Substitute in variables [Polar Derivative]: [tex]\displaystyle \frac{dy}{dx} = \frac{[6cos(2\theta) - 5]cos(\theta) + [-12sin(2\theta)]sin(\theta)}{[-12sin(2\theta)]cos(\theta) - [6cos(2\theta) - 5]sin(\theta)}[/tex]
- [Polar Derivative] Simplify: [tex]\displaystyle \frac{dy}{dx} = \frac{-[6cos(\theta)cos(2\theta) - 12sin(\theta)sin(2\theta) - 5cos(\theta)]}{5sin(\theta)cos(2\theta) + 12cos(\theta)sin(2\theta) - 5sin(\theta)}[/tex]
Step 3: Find Slope
- Substitute in θ [Polar Derivative]: [tex]\displaystyle \frac{dy}{dx} \bigg| \limits_\bigg{\theta = \frac{\pi}{4}} = \frac{-[6cos\frac{\pi}{4})cos(2 \cdot \frac{\pi}{4}) - 12sin(\frac{\pi}{4})sin(2 \cdot \frac{\pi}{4}) - 5cos(\frac{\pi}{4})]}{5sin(\frac{\pi}{4})cos(2 \cdot \frac{\pi}{4}) + 12cos(\frac{\pi}{4})sin(2 \cdot \frac{\pi}{4}) - 5sin(\frac{\pi}{4})}[/tex]
- Evaluate [Unit Circle]: [tex]\displaystyle \frac{dy}{dx} \bigg| \limits_\bigg{\theta = \frac{\pi}{4}} = \frac{17}{7}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Polar Derivatives and Area (BC Only)
Book: College Calculus 10e
