/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Find the points at which the fol... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 16

Find the points at which the following polar curves have a horizontal or a vertical tangent line. $$r=2+2 \sin \theta$$

Short Answer

Expert verified
Question: Determine the points at which the polar curve r = 2+2sin(胃) has horizontal or vertical tangent lines. Answer: To find these points, follow the steps below: Step 1: Convert the polar equation to Cartesian form: x = (2+2sin(胃))cos(胃) y = (2+2sin(胃))sin(胃) Step 2: Compute the derivatives of x and y with respect to 胃: dx/d胃 = -(2+2sin(胃))sin(胃) + 2cos虏(胃) dy/d胃 = (2+2sin(胃))cos(胃) + 2sin(胃)cos(胃) Step 3: Compute the slope of the tangent and its reciprocal: dy/dx = (dy/d胃) / (dx/d胃) dx/dy = (dx/d胃) / (dy/d胃) Step 4: Determine the values of 胃 where the slope is zero or undefined: For horizontal tangents, set dy/dx = 0 and solve for 胃. For vertical tangents, set dx/dy = 0 and solve for 胃. Step 5: Compute the points in polar coordinates: After solving for 胃, replace 胃 in the polar equation r = 2 + 2sin(胃) to find the corresponding r values. Then, combine the values (r, 胃) to find the points in polar coordinates where horizontal or vertical tangents occur.

Step by step solution

01

Convert the polar equation to Cartesian form

Recall the polar to Cartesian conversion formulas: $$x = r \cos \theta$$ $$y = r \sin \theta$$ Given the polar equation: $$r=2+2 \sin \theta$$ Replace r in both formulas: $$x = (2+2 \sin \theta) \cos \theta$$ $$y = (2+2 \sin \theta) \sin \theta$$ Now we have the Cartesian equations.
02

Compute the derivatives of x and y with respect to 胃

Compute dx/d胃 and dy/d胃: $$\frac{dx}{d\theta} = - (2+2 \sin \theta) \sin \theta + 2 \cos^2 \theta$$ $$\frac{dy}{d\theta} = (2+2 \sin \theta) \cos \theta + 2 \sin \theta \cos \theta$$
03

Compute the slope of the tangent and its reciprocal

To find the slope of the tangent, we compute dy/dx by dividing dy/d胃 by dx/d胃: $$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$ Compute the reciprocal of the slope, which is dx/dy: $$\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{\frac{dx}{d\theta}}{\frac{dy}{d\theta}}$$
04

Determine the values of 胃 where the slope is zero or undefined

A tangent is horizontal when the slope is 0, so set dy/dx equal to 0. A tangent is vertical when the slope is undefined, so set dx/dy equal to 0. Horizontal Tangents: $$\frac{dy}{dx} = 0$$ $$\frac{(2+2 \sin \theta) \cos \theta + 2 \sin \theta \cos \theta}{ - (2+2 \sin \theta) \sin \theta + 2 \cos^2 \theta} = 0$$ The numerator must be 0: $$(2+2 \sin \theta) \cos \theta + 2 \sin \theta \cos \theta = 0$$ Now solve for 胃 to find the points at which horizontal tangents occur. Vertical Tangents: $$\frac{dx}{dy} = 0$$ $$\frac{- (2+2 \sin \theta) \sin \theta + 2 \cos^2 \theta}{(2+2 \sin \theta) \cos \theta + 2 \sin \theta \cos \theta} = 0$$ The numerator must be 0: $$- (2+2 \sin \theta) \sin \theta + 2 \cos^2 \theta = 0$$ Now solve for 胃 to find the points at which vertical tangents occur.
05

Compute the points in polar coordinates

After solving for 胃, replace 胃 in the polar equation: $$r = 2 + 2 \sin \theta$$ to find the corresponding r values, and combine those values (r, 胃) to find the points in polar coordinates where horizontal or vertical tangents occur.

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