Question

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

Short answer

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

Step-by-step solution

Define Horizontal Tangent Line Condition

For a parametric curve defined as \(x=f(t)\) and \(y=g(t)\), the curve has a horizontal tangent line where the derivative \(\frac{dy}{dt}\) is zero and \(\frac{dx}{dt}\) is not zero.

Calculate Derivatives

Calculate \(\frac{dx}{dt}\) which is the derivative of \(x=t\), and \(\frac{dy}{dt}\) which is the derivative of \(y=t^2\).\[\frac{dx}{dt} = \frac{d}{dt}(t) = 1\]\[\frac{dy}{dt} = \frac{d}{dt}(t^2) = 2t\]

Set Derivative of \(y\) to Zero

For a horizontal tangent line, \(\frac{dy}{dt} = 0\). Set the equation of \(\frac{dy}{dt}\) to zero and solve for \(t\).\[2t = 0\]Solve for \(t\):\[t = 0\]

Verify \(\frac{dx}{dt} \neq 0\)

Ensure that \(\frac{dx}{dt} = 1\) is not zero, which it never is, hence valid for \(t=0\). The curve truly has a horizontal tangent line at this point.

Key concepts

Horizontal Tangent

A tangent line is a straight line that just touches a curve at a particular point, without crossing it at that point. It represents the slope or direction of the curve at that specific location. A horizontal tangent line is special because it means that at that point, the slope (or incline) of the tangent line is zero.
This occurs when the rate of change of the dependent variable (often y), \(rac{dy}{dt}\), is zero. However, for a point to truly be horizontal, the rate of change of the independent variable, \(rac{dx}{dt}\), must not be zero to ensure that the line doesn’t become undefined or vertical.
To find these specific points on a curve defined by parametric equations, you need to solve \(rac{dy}{dt} = 0\) while ensuring \(rac{dx}{dt}\) does not equal zero.

Derivatives

Derivatives are fundamental tools in calculus, used to measure how a function changes as its input changes.
They help in understanding concepts related to rate of change, such as velocity or speed in motion, or in this case, the slope of a line with respect to the parametric equations.
For parametric equations, you find derivatives by differentiating the functions with respect to the parameter, often denoted as \('t'\).
In our exercise, for \(x = t\) and \(y = t^2\), differentiating gives us \[\frac{dx}{dt} = 1, \quad and \quad \frac{dy}{dt} = 2t\] These represent how x and y change with respect to t.

Parametric Curve

A parametric curve is a curve that is described using one or more equations and a parameter, often \(t\).
Unlike standard functions which use a one-to-one relationship between x and y, parametric equations allow both x and y to be described in terms of another variable.
This is useful in several fields including physics, computer graphics, and anywhere complex systems require a detailed path or curve.

• Parametric equations provide a convenient way to describe motion and changes over time.
• They separate the control of the horizontal and vertical components, often making complex motion much easier to describe.

In the context of our exercise, the curve \((x=t, y=t^2)\) obeys the laws of parametric equations by defining both x and y in terms of t.

Calculus

Calculus is a branch of mathematics that studies change and motion, through methods such as differentiation and integration.
It's essential for understanding the behaviour of curves, especially those that are defined in a more complex manner, like parametric equations.
In calculus, derivatives are used to understand the rate of change at any given point on a function or a curve.
This is key in our exercise; to determine when the curve has a horizontal tangent, we rely on the derivative of the parametric equations.
By calculating and analyzing these, calculus helps us establish that the curve only has a horizontal tangent when \(t = 0\), since the \(rac{dy}{dt}\) becomes zero, while \(rac{dx}{dt}\) remains nonzero.