Question

Find \(t\) -values where the curve defined by the given parametric equations has a horizontal tangent line. \(x=\sqrt{t}, y=5 t+2\)

Short answer

The curve has a horizontal tangent line when \( t = 0 \).

Step-by-step solution

Understand Horizontal Tangents

A curve has a horizontal tangent line where the slope of the tangent line is zero. Slope in the context of parametric equations is given by \( \frac{dy}{dx} \). For a horizontal tangent, \( \frac{dy}{dx} = 0 \).

Differentiate Y with Respect to T

Find \( \frac{dy}{dt} \) from the equation \( y = 5t + 2 \). The derivative is the rate of change of \( y \) with respect to \( t \). Calculate \( \frac{dy}{dt} = \frac{d}{dt}(5t + 2) = 5 \).

Differentiate X with Respect to T

Find \( \frac{dx}{dt} \) from the equation \( x = \sqrt{t} \). This is \( \frac{dx}{dt} = \frac{d}{dt}(t^{1/2}) = \frac{1}{2} t^{-1/2} \).

Calculate \( \frac{dy}{dx} \)

The derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Substitute the derivatives found: \( \frac{dy}{dx} = \frac{5}{\frac{1}{2} t^{-1/2}} = 10 t^{1/2} \).

Set \( \frac{dy}{dx} = 0 \) for Horizontal Tangents

Since \( \frac{dy}{dx} = 10 \sqrt{t} \) must be zero for a horizontal tangent, set \( 10 \sqrt{t} = 0 \) and solve for \( t \).

Solve for T

The equation \( 10 \sqrt{t} = 0 \) gives \( \sqrt{t} = 0 \). Hence, \( t = 0 \), as it is the only value for which the square root equals zero.

Key concepts

Horizontal Tangent Line

A horizontal tangent line is a key concept in calculus that signifies where a curve flattens out and momentarily stops increasing or decreasing. For a curve defined by parametric equations, this happens when the slope of the tangent is zero. - To find where a curve has a horizontal tangent, you need \(\frac{dy}{dx} = 0\). - This means the curve isn't rising or falling at that exact point.In our example, to locate horizontal tangents in the curve defined by \(x=\sqrt{t}\) and \(y=5t+2\), we set the equation for the slope \(\frac{dy}{dx}\) to zero. Understanding the point where this occurs helps gain insight into the behavior of the curve.

Differentiation

Differentiation is the process of finding the derivative, which describes how one quantity changes in relation to another. In the context of parametric equations, we have to differentiate both equations concerning the parameter, often denoted by \(t\).- First, differentiate \(y = 5t + 2\) with respect to \(t\) to get \(\frac{dy}{dt} = 5\).- Then, differentiate \(x = \sqrt{t}\), resulting in \(\frac{dx}{dt} = \frac{1}{2}t^{-1/2}\).These derivatives help us find \(\frac{dy}{dx}\), which is crucial for further calculations involving the slope of the curve. Differentiation provides the necessary tools to analyze how the position of the point changes as \(t\) changes.

Slope of a Curve

The slope of a curve at any given point indicates how steep the curve is at that point. For parametric equations, the slope is determined by the ratio of the changes in \(y\) to the changes in \(x\), which is \(\frac{dy}{dx}\).- Calculate \(\frac{dy}{dx}\) using the formula \(\frac{dy}{dt} \div \frac{dx}{dt}\). - In our example, it becomes \(10t^{1/2}\).A horizontal tangent appears where this slope equals zero, indicating no vertical change at that position on the curve. Understanding this slope is essential in recognizing various features and behaviors of the curve.

Calculus Problem Solving

Solving calculus problems, like identifying where a curve has a horizontal tangent, often involves a series of logical steps. Mastering these steps helps tackle many varying calculus issues effectively.- First, understand the problem in terms of derivatives and what you're solving for.- Evaluate the derivatives: For our problem, examine \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) to establish \(\frac{dy}{dx}\).- Set the derivative of interest equal to zero for conditions like finding horizontal tangents.- Solve the resulting equation as demonstrated through \(10\sqrt{t} = 0\), which simplifies to \(t = 0\).By following a clear strategy, you build confidence in solving similar problems across calculus, whether involving parametric equations or other concepts.