Question

Find \(t\) -values where the curve defined by the given parametric equations has a horizontal tangent line. \(x=e^{t / 10} \cos t, y=e^{t / 10} \sin t\)

Short answer

The curve has horizontal tangents at \( t = \arctan(-10) + k\pi \), where \( k \) is any integer.

Step-by-step solution

Differentiate Parametric Equations

Differentiate the parametric equations with respect to the parameter \( t \). For the x-component, we have \( \frac{dx}{dt} = \frac{1}{10}e^{t/10}\cos t - e^{t/10}\sin t \). For the y-component, we have \( \frac{dy}{dt} = \frac{1}{10}e^{t/10}\sin t + e^{t/10}\cos t \).

Set Derivative of y to Zero

A horizontal tangent line occurs where the derivative of \( y \) with respect to \( x \) is zero. This means setting \( \frac{dy}{dt} = 0 \). Thus, \( \frac{1}{10}e^{t/10}\sin t + e^{t/10}\cos t = 0 \).

Simplify the Equation

Factor out \( e^{t/10} \) from \( \frac{dy}{dt} = \frac{1}{10}e^{t/10}\sin t + e^{t/10}\cos t = 0 \) to obtain \( e^{t/10}(\frac{1}{10} \sin t + \cos t) = 0 \). Since \( e^{t/10} \) is never zero, the equation simplifies to \( \frac{1}{10} \sin t + \cos t = 0 \).

Solve Trigonometric Equation

Solve the trigonometric equation \( \frac{1}{10} \sin t + \cos t = 0 \). This can be rearranged to \( \cos t = -\frac{1}{10} \sin t \) or \( \tan t = -10 \). Use the inverse tangent function to find \( t = \arctan(-10) + k\pi \), where \( k \) is any integer, since tangent has a period of \( \pi \).

Key concepts

Horizontal Tangent

In the world of parametric equations, a tangent line to a curve is a line that just touches the curve at a certain point. A horizontal tangent is a specific type of tangent line that runs parallel to the x-axis. This is important because it indicates where the slope of the curve is zero, essentially showing flat spots on the curve. Finding these points involves calculating where the derivative of the y-component, with respect to the parameter (often time, denoted as \(t\)), equals zero. In simpler terms, you differentiate the y-value with respect to \(t\) and set the resulting equation to zero. This is a critical step to identify where these horizontal tangents occur on the curve defined by the parametric equations.

Trigonometric Equation

In the context of this exercise, after differentiating the parametric equations and setting the derivative of the y-component equal to zero, we encounter a trigonometric equation. The trigonometric equation here is \( \frac{1}{10} \sin t + \cos t = 0 \), which simplifies from \( e^{t/10}(\frac{1}{10} \sin t + \cos t) = 0 \). Solving it requires rearranging it into a form we can work with more easily: \( \cos t = -\frac{1}{10}\sin t \). This transforms into \( \tan t = -10 \) after dividing both sides by \( \cos t \). Trigonometric equations like these often involve periodic functions, so to find solutions nearly every angle must be considered, specifically \( t = \arctan(-10) + k\pi \), where \( k \) is an integer. This formula considers the periodicity of the tangent function, which repeats every \( \pi \), enabling us to find all potential \( t \)-values fulfilling the condition.

Tangent Line

A tangent line to a curve is a line that just touches the curve at a point along the curve, without crossing it. In the specific realm of parametric equations, understanding tangent lines is vital. They give important information about the behavior and shape of the curve at different points. For a general parametric curve defined by equations for \( x(t) \) and \( y(t) \), the slope of the tangent line at any point \( t \) is given by the ratio \( \frac{dy/dt}{dx/dt} \). When the derivative concerning the x-component also equals zero, one gets into undefined slopes, which result in vertical tangents. Finding a horizontal tangent line specifically implies solving for where the y-derivative equals zero, showcasing where the curve flattens horizontally. This concept extends beyond pure calculus and finds applications in physics, engineering, and computer graphics where understanding curvature is crucial.