Question

Calculus question. Urgent help needed. Due today.

Answer 1

None of these

General Formulas and Concepts:

Pre-Calculus

• Unit Circle

Calculus

Derivatives

Derivative Notation

Derivative Property [Multiplied Constant]:
`\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)`

Derivative Property [Addition/Subtraction]:
`\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]`

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
`\displaystyle (d)/(dx)[f(g(x))] =f'(g(x)) \cdot g'(x)`

Trig Derivative:
`\displaystyle (d)/(dx)[cos(u)] = -u'sin(u)`

Polar Derivative:
`\displaystyle (dy)/(dx) = (rcos(\theta) + r'sin(\theta))/(r'cos(\theta) - rsin(\theta))`

Explanation:

Step 1: Define

Identify

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

`\displaystyle \theta = (\pi)/(4)`

Step 2: Differentiate

1. Trig Derivative [Chain Rule]:
   `\displaystyle r' = (d)/(d\theta)[6cos(2 \theta) - 5] \cdot (d)/(d\theta)[2\theta]`
2. Rewrite [Derivative Rule - Subtraction]:
   `\displaystyle r' = \bigg[ (d)/(d\theta)[6cos(2 \theta)] - (d)/(d\theta)[5] \bigg] \cdot (d)/(d\theta)[2\theta]`
3. Rewrite [Derivative Rule - Multiplied Constant]:
   `\displaystyle r' = \bigg[ 6(d)/(d\theta)[cos(2 \theta)] - (d)/(d\theta)[5] \bigg] \cdot (d)/(d\theta)[2\theta]`
4. Trig Derivative:
   `\displaystyle r' = [-6sin(2\theta)] \cdot (d)/(d\theta)[2\theta]`
5. Basic Power Rule:
   `\displaystyle r' = -6sin(2\theta) \cdot 2\theta^(1 - 1)`
6. Simplify:
   `\displaystyle r' = -12sin(2\theta)`
7. Substitute in variables [Polar Derivative]:
   `\displaystyle (dy)/(dx) = ([6cos(2\theta) - 5]cos(\theta) + [-12sin(2\theta)]sin(\theta))/([-12sin(2\theta)]cos(\theta) - [6cos(2\theta) - 5]sin(\theta))`
8. [Polar Derivative] Simplify:
   `\displaystyle (dy)/(dx) = (-[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

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

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

Unit: Polar Derivatives and Area (BC Only)

Book: College Calculus 10e

Answer 2

12 (none of these)

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)

dr/dtheta = -12 sin (2theta) evaluated at theta= pi/4

dr/dtheta = -12 sin (2*pi/4)

dr/ dtheta = -12 sin (pi/2)

= -12(-1)

=12