Skip to main content
Ch.12 - Parametric and Polar Curves
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh.12 - Parametric and Polar CurvesProblem 12.3.20
Chapter 12, Problem 12.3.20

11–20. Slopes of tangent lines Find the slope of the line tangent to the following polar curves at the given points.


r = 2θ; (π/2, π/4)

Verified step by step guidance
1
Recall that for a polar curve given by \(r = f(\theta)\), the slope of the tangent line \(\frac{dy}{dx}\) can be found using the formulas \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), and then applying the chain rule to find \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\).
Write down the expressions for \(x\) and \(y\) in terms of \(\theta\): \(x = r \cos(\theta) = 2\theta \cos(\theta)\) \(y = r \sin(\theta) = 2\theta \sin(\theta)\).
Differentiate both \(x\) and \(y\) with respect to \(\theta\) using the product rule: \(\frac{dx}{d\theta} = \frac{d}{d\theta} (2\theta \cos(\theta))\) \(\frac{dy}{d\theta} = \frac{d}{d\theta} (2\theta \sin(\theta))\).
Calculate the slope of the tangent line using the formula \(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\).
Evaluate \(\frac{dy}{dx}\) at the given point \(\theta = \frac{\pi}{4}\) to find the slope of the tangent line at that point.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3m

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Polar Coordinates and Curves

Polar coordinates represent points using a radius and an angle (r, θ) instead of Cartesian (x, y). Understanding how curves are defined in polar form, such as r = 2θ, is essential for analyzing their geometric properties and converting between coordinate systems.
Recommended video:
05:32
Intro to Polar Coordinates

Slope of Tangent Lines in Polar Coordinates

The slope of a tangent line to a polar curve at a point is found by converting the polar equation to Cartesian coordinates or using the formula dy/dx = (dr/dθ sinθ + r cosθ) / (dr/dθ cosθ - r sinθ). This allows determination of the tangent's slope without explicit Cartesian conversion.
Recommended video:
Guided course
05:13
Slopes of Tangent Lines

Differentiation with Respect to θ

Calculating the slope requires differentiating r with respect to θ (dr/dθ). This derivative captures how the radius changes as the angle varies, which is crucial for applying the tangent slope formula in polar coordinates.
Recommended video:
Guided course
05:53
Finding Differentials
Related Practice
Textbook Question

73–76. Tangent lines Find an equation of the line tangent to the curve at the point corresponding to the given value of t.


x=t ²−1, y=t ³ +t; t=2

1
views
Textbook Question

77–80. Slopes of tangent lines Find all points at which the following curves have the given slope.


x = 2 cos t, y = 8 sin t; slope = -1

1
views
Textbook Question

41–44. Intersection points and area  Find all the intersection points of the following curves. Find the area of the entire region that lies within both curves


r = 1 + sin θ and r = 1 + cos θ

1
views
Textbook Question

45–60. Areas of regions Find the area of the following regions.


The region inside the limaçon r = 4 - 2 cos θ

Textbook Question

31–36. Eliminating the parameter Eliminate the parameter to express the following parametric equations as a single equation in x and y.


x=t,y= √(4−t²) a

1
views
Textbook Question

57–64. Graphing polar curves Graph the following equations. Use a graphing utility to check your work and produce a final graph.


r = 2 - 2 sin θ  b