Question

Parametric equations for a curve are given. Find \(\frac{d^{2} y}{d x^{2}},\) then determine the intervals on which the graph of the curve is concave up/down. \(x=t^{2}-1, \quad y=t^{3}-t\)

Short answer

The curve is concave up where \( t > 0 \) and concave down where \( t < 0 \).

Step-by-step solution

Find First Derivative

To find the first derivative \( \frac{dy}{dx} \), we need to find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). First, differentiate \( x = t^{2} - 1 \) with respect to \( t \):\[ \frac{dx}{dt} = 2t \].Next, differentiate \( y = t^{3} - t \) with respect to \( t \): \[ \frac{dy}{dt} = 3t^{2} - 1 \].Thus, the first derivative is given by:\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^{2} - 1}{2t} \].

Simplify First Derivative

Simplify \( \frac{dy}{dx} = \frac{3t^{2} - 1}{2t} \) by dividing each term in the numerator by \( 2t \):\[ \frac{dy}{dx} = \frac{3t^{2}}{2t} - \frac{1}{2t} = \frac{3t}{2} - \frac{1}{2t} \].

Find Second Derivative

Find the second derivative \( \frac{d^{2}y}{dx^{2}} \) by differentiating \( \frac{dy}{dx} \) with respect to \( t \), then multiply by \( \frac{dt}{dx} \):First, differentiate \( \frac{3t}{2} - \frac{1}{2t} \) with respect to \( t \):\[ \frac{d}{dt}\left(\frac{3t}{2} - \frac{1}{2t}\right) = \frac{3}{2} + \frac{1}{2t^{2}} \].Next, find \( \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{2t} \).Thus, \( \frac{d^{2}y}{dx^{2}} = \left(\frac{3}{2} + \frac{1}{2t^{2}}\right) \cdot \frac{1}{2t} \).Simplify:\[ \frac{d^{2}y}{dx^{2}} = \frac{3}{4t} + \frac{1}{4t^{3}} \].

Determine Concavity Intervals

The concavity is determined by the sign of \( \frac{d^{2}y}{dx^{2}} \). A graph is concave up where \( \frac{d^{2}y}{dx^{2}} > 0 \) and concave down where \( \frac{d^{2}y}{dx^{2}} < 0 \).Since \( \frac{d^{2}y}{dx^{2}} = \frac{3}{4t} + \frac{1}{4t^{3}} \), the expression is positive if both terms are positive. Given both coefficients \( \frac{3}{4} \) and \( \frac{1}{4} \) are positive, the graph is concave up when both terms, \( \frac{1}{t} \) and \( \frac{1}{t^3} \), are positive, i.e., \( t > 0 \).The graph is concave down when both terms are negative, i.e., \( t < 0 \).

Key concepts

First Derivative

When dealing with parametric equations, understanding derivatives is key to analyzing curves. To find the first derivative, often noted as \( \frac{dy}{dx} \), we differentiate the given parametric equations with respect to the parameter. In the exercise, we have equations:

• \( x = t^2 - 1 \)
• \( y = t^3 - t \)

First, calculate \( \frac{dx}{dt} = 2t \) and \( \frac{dy}{dt} = 3t^2 - 1 \). The first derivative \( \frac{dy}{dx} \) combines these derivatives as: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2 - 1}{2t} \] Understanding this derivative helps determine the slope of the curve as it is traced by the parameter \( t \), allowing us to analyze the curve's behavior.

Second Derivative

The second derivative, \( \frac{d^2y}{dx^2} \), offers insight into the curve's concavity and points of inflection. Once you've found the first derivative \( \frac{dy}{dx} = \frac{3t}{2} - \frac{1}{2t} \), differentiate it again with respect to \( t \) to find the rate of change of the slope. This gives: \[ \frac{d}{dt}\left(\frac{3t}{2} - \frac{1}{2t}\right) = \frac{3}{2} + \frac{1}{2t^2} \] To convert this to \( \frac{d^2y}{dx^2} \), multiply by \( \frac{dt}{dx} \), which is \( \frac{1}{2t} \): \[ \frac{d^2y}{dx^2} = \left(\frac{3}{2} + \frac{1}{2t^2}\right) \cdot \frac{1}{2t} = \frac{3}{4t} + \frac{1}{4t^3} \] This leads to a formula that helps understand how sharply the curve bends.

Concavity

The concept of concavity is closely linked to the second derivative \( \frac{d^2y}{dx^2} \). It tells us how the curve is oriented, whether it opens upwards or downwards, and it affects the shape and characteristics of the curve.

• If \( \frac{d^2y}{dx^2} > 0 \), the graph is considered to be concave up.
• If \( \frac{d^2y}{dx^2} < 0 \), the graph is considered to be concave down.

Using the expression \( \frac{3}{4t} + \frac{1}{4t^3} \), we determine where the curve is concave up or down by examining the sign of this expression across different intervals of \( t \).

Differentiation

Differentiation is the mathematical process used to calculate derivatives, and it is essential for analyzing parametric equations. It applies to various fields of science and engineering as it handles rates of change and allows for the investigation of function behavior. For parametric equations, differentiation involves deriving \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \), leading to expressions for \( \frac{dy}{dx} \) using the chain rule: \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \] This efficient method provides a direct way to explore properties of curves defined parametricallu and gives us insights on curves' behaviors without needing explicit functions in standard \( x, y \) form.

Concave Up

A curve is considered concave up if it bends upwards like a U shape. Mathematically, this occurs when the second derivative \( \frac{d^2y}{dx^2} \) is positive. In the exercise, the expression \( \frac{3}{4t} + \frac{1}{4t^3} \) determines concavity:

• For \( t > 0 \), both terms are positive, making \( \frac{d^2y}{dx^2} > 0 \).
• This condition signifies the graph is concave up on these intervals, implying that the slope of the tangent line is increasing.

Understanding where this occurs helps in predicting the curve's orientation and possible points of interest such as minimums.

Concave Down

Conversely, a curve is concave down if it bends downwards like an upside-down U. This is observed when the second derivative, \( \frac{d^2y}{dx^2} \), is negative. In our specific exercise, analyzing \( \frac{3}{4t} + \frac{1}{4t^3} \) shows:

• For \( t < 0 \), both terms are negative, leading to \( \frac{d^2y}{dx^2} < 0 \).
• This denotes the graph is concave down in these intervals, suggesting that the slope of the tangent line is decreasing.

Identifying these sections can inform about maximum values or other critical points of the graph.