Mathematics, 16.10.2019 23:00 madisonsolis05

Calculus question. urgent needed. due today.

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Answer from: HomeWorkForLife

None of these

General Formulas and Concepts:

Pre-Calculus

Unit Circle

Calculus

Derivatives

Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Trig Derivative: $\displaystyle \frac{d}{dx}[cos(u)] = -u'sin(u)$

Polar Derivative: $\displaystyle \frac{dy}{dx} = \frac{rcos(\theta) + r'sin(\theta)}{r'cos(\theta) - rsin(\theta)}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle r = 6cos(2 \theta) - 5$

$\displaystyle \theta = \frac{\pi}{4}$

Step 2: Differentiate

Trig Derivative [Chain Rule]: $\displaystyle r' = \frac{d}{d\theta}[6cos(2 \theta) - 5] \cdot \frac{d}{d\theta}[2\theta]$ Rewrite [Derivative Rule - Subtraction]: $\displaystyle r' = \bigg[ \frac{d}{d\theta}[6cos(2 \theta)] - \frac{d}{d\theta}[5] \bigg] \cdot \frac{d}{d\theta}[2\theta]$ Rewrite [Derivative Rule - Multiplied Constant]: $\displaystyle r' = \bigg[ 6\frac{d}{d\theta}[cos(2 \theta)] - \frac{d}{d\theta}[5] \bigg] \cdot \frac{d}{d\theta}[2\theta]$ Trig Derivative: $\displaystyle r' = [-6sin(2\theta)] \cdot \frac{d}{d\theta}[2\theta]$ Basic Power Rule: $\displaystyle r' = -6sin(2\theta) \cdot 2\theta^{1 - 1}$ Simplify: $\displaystyle r' = -12sin(2\theta)$ Substitute in variables [Polar Derivative]: $\displaystyle \frac{dy}{dx} = \frac{[6cos(2\theta) - 5]cos(\theta) + [-12sin(2\theta)]sin(\theta)}{[-12sin(2\theta)]cos(\theta) - [6cos(2\theta) - 5]sin(\theta)}$ [Polar Derivative] Simplify: $\displaystyle \frac{dy}{dx} = \frac{-[6cos(\theta)cos(2\theta) - 12sin(\theta)sin(2\theta) - 5cos(\theta)]}{5sin(\theta)cos(2\theta) + 12cos(\theta)sin(2\theta) - 5sin(\theta)}$

Step 3: Find Slope

Substitute in θ [Polar Derivative]: $\displaystyle \frac{dy}{dx} \bigg| \limits_\bigg{\theta = \frac{\pi}{4}} = \frac{-[6cos\frac{\pi}{4})cos(2 \cdot \frac{\pi}{4}) - 12sin(\frac{\pi}{4})sin(2 \cdot \frac{\pi}{4}) - 5cos(\frac{\pi}{4})]}{5sin(\frac{\pi}{4})cos(2 \cdot \frac{\pi}{4}) + 12cos(\frac{\pi}{4})sin(2 \cdot \frac{\pi}{4}) - 5sin(\frac{\pi}{4})}$ Evaluate [Unit Circle]: $\displaystyle \frac{dy}{dx} \bigg| \limits_\bigg{\theta = \frac{\pi}{4}} = \frac{17}{7}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Polar Derivatives and Area (BC Only)

Book: College Calculus 10e

Answer from: TelestoisaMoon4437

12 (none of these)

Step-by-step explanation:

To find the slope at a point, we need to find the derivative of the function at the point, then evaluate at that point

derivative with respect to theta ( r = 6 cos 2theta - 5) evaluated at theta = pi/4

dr/dtheta = 6 * dr dtheta (cos 2theta -5)

We need to find the derivative of cos 2 theta

d /dtheta = -2 sin 2 theta

dr/dtheta = 6 * (-2sin (2theta) -0)