Ch.12 - Parametric and Polar Curves

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch.12 - Parametric and Polar Curves

Problem 12.1.114

Chapter 12, Problem 12.1.114

Second derivative Assume a curve is given by the parametric equations x=f(t) and y=g(t), where f and g are twice differentiable. Use the Chain Rule to show that y″x=(fʹ(t)g″(t)−gʹ(t)f″(t))/(fʹ(t))³.

Verified step by step guidance

Recall that for parametric equations \(x = f(t)\) and \(y = g(t)\), the first derivative of \(y\) with respect to \(x\) is given by the chain rule as \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{g'(t)}{f'(t)}\).

To find the second derivative \(\frac{d^2y}{dx^2}\), we differentiate \(\frac{dy}{dx}\) with respect to \(x\). Using the chain rule, this is \(\frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( \frac{g'(t)}{f'(t)} \right) \cdot \frac{dt}{dx}\).

Calculate \(\frac{d}{dt} \left( \frac{g'(t)}{f'(t)} \right)\) using the quotient rule: \(\frac{d}{dt} \left( \frac{g'(t)}{f'(t)} \right) = \frac{g''(t) f'(t) - g'(t) f''(t)}{(f'(t))^2}\).

Since \(\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{f'(t)}\), substitute this into the expression for the second derivative to get \(\frac{d^2y}{dx^2} = \frac{g''(t) f'(t) - g'(t) f''(t)}{(f'(t))^2} \cdot \frac{1}{f'(t)}\).

Simplify the expression to obtain \(\frac{d^2y}{dx^2} = \frac{f'(t) g''(t) - g'(t) f''(t)}{(f'(t))^3}\), which is the desired formula.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

Parametric Equations and Derivatives

Parametric equations express coordinates as functions of a parameter, typically t. To analyze curves defined parametrically, derivatives of x and y with respect to t are used. Understanding how to compute first and second derivatives of these functions is essential for studying the curve's behavior.

Chain Rule in Differentiation

The Chain Rule allows differentiation of composite functions by relating the derivative of the outer function to the derivative of the inner function. In parametric contexts, it helps find derivatives of y with respect to x by linking dy/dt and dx/dt, crucial for expressing higher-order derivatives like y''.

Second Derivative of Parametric Curves

The second derivative y'' with respect to x measures the curvature of a parametric curve. It is found by differentiating dy/dx with respect to x, often using the chain rule to convert derivatives with respect to t into derivatives with respect to x. The formula involves first and second derivatives of both x and y with respect to t.

Recommended video:

06:49

Differentiation of Parametric Curves

Related Practice

Textbook Question

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

r² = 4 sin θ

Textbook Question

Circles in general Show that the polar equation

r² - 2r r₀ cos(θ - θ₀) = R² - r₀²

describes a circle of radius R whose center has polar coordinates (r₀, θ₀)

Textbook Question

63–66. Tracing hyperbolas and parabolas Graph the following equations. Then use arrows and labeled points to indicate how the curve is generated as θ increases from 0 to 2π.

r = 3/(1 - cos θ)

views

Textbook Question

9–13. Graph the points with the following polar coordinates. Give two alternative representations of the points in polar coordinates.

(-1, -π/3)

views

Textbook Question

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

The region inside the rose r = 4 sin 2θ and inside the circle r = 2

Textbook Question

37–52. Curves to parametric equations Find parametric equations for the following curves. Include an interval for the parameter values. Answers are not unique.

The left half of the parabola y=x ² +1, originating at (0, 1)

views