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. \(x=t^{2}-t, \quad y=t^{2}+t\)

Short answer

Concave down for \( t < \frac{1}{2} \), concave up for \( t > \frac{1}{2} \).

Step-by-step solution

Differentiate with respect to parameter t

To find the first derivative \( \frac{dy}{dx} \), we begin by finding \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).For \( x = t^2 - t \), we differentiate with respect to \( t \) to get:\[ \frac{dx}{dt} = 2t - 1 \]For \( y = t^2 + t \), we differentiate with respect to \( t \) to get:\[ \frac{dy}{dt} = 2t + 1 \].

Find the first derivative \( \frac{dy}{dx} \)

Now, find \( \frac{dy}{dx} \) using:\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t + 1}{2t - 1} \].

Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)

To find \( \frac{d^2y}{dx^2} \), differentiate \( \frac{dy}{dx} \) with respect to \( t \) and use the parametric form:First, differentiate \( \frac{dy}{dx} = \frac{2t + 1}{2t - 1} \):Let \( u = 2t + 1 \) and \( v = 2t - 1 \), find the derivatives \( u' = 2 \) and \( v' = 2 \):\[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2} = \frac{(2t - 1)(2) - (2t + 1)(2)}{(2t - 1)^2} = \frac{4t - 2 - 4t - 2}{(2t - 1)^2} \]This simplifies to: \[ \frac{-4}{(2t - 1)^2}. \]Then, use the chain rule to get \( \frac{d^2 y}{dx^2} \):\[ \frac{d^2 y}{dx^2} = \frac{\frac{d}{dt} \left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\frac{-4}{(2t - 1)^2}}{2t - 1} = \frac{-4}{(2t - 1)^3} \].

Determine concavity using \( \frac{d^2y}{dx^2} \)

The graph is concave up when \( \frac{d^2y}{dx^2} > 0 \) and concave down when \( \frac{d^2y}{dx^2} < 0 \).Since \( \frac{d^2y}{dx^2} = \frac{-4}{(2t - 1)^3} \), we observe that it depends on the sign of \( (2t - 1)^3 \).1. \( (2t - 1)^3 > 0 \) when \( t > \frac{1}{2} \).2. \( (2t - 1)^3 < 0 \) when \( t < \frac{1}{2} \).Hence:- The curve is concave down on \( t < \frac{1}{2} \).- The curve is concave up on \( t > \frac{1}{2} \).

Key concepts

Concavity

Concavity helps us understand the 'bending' nature of a curve. We talk about a curve being "concave up" when it bends upwards like a smile, and "concave down" when it bends down like a frown. This idea is key because it shows you how the slope of the curve is changing.

To determine this, we use the second derivative of a function. If it's positive, the graph is concave up in that interval. On the other hand, a negative second derivative means the graph is concave down. In the context of parametric equations, the process is the same - just a bit more algebraically involved.

Once we find the second derivative, we assess its sign in different intervals. This tells us precisely where the curve changes its concavity. In the exercise, since the second derivative, \( \frac{d^2 y}{dx^2} = \frac{-4}{(2t - 1)^3} \), is negative for \( t < \frac{1}{2} \) and positive for \( t > \frac{1}{2} \), we quickly conclude on the curve’s concavity behavior.

Second Derivative

The second derivative is the derivative of the derivative. While the first derivative gives us the slope of the tangent to the curve, the second derivative tells us how this slope is changing. Essentially, it measures the curvature or bending of the curve. If the second derivative is positive, the slope is getting steeper, indicating a concave-up shape. If it's negative, the slope is decreasing, showing a concave-down shape.

In parametric equations, we find the second derivative \( \frac{d^2 y}{dx^2} \) by differentiating \( \frac{dy}{dx} \) again with respect to the parameter, typically \( t \). Then, we use the chain rule to adjust for how \( x \) changes with \( t \). This involves some more algebra but follows the same principles as for single-variable calculus.

Differentiation

Differentiation is the process of finding the derivative of a function. A derivative is a fundamental concept in calculus as it gives the rate at which one quantity changes with respect to another. When dealing with parametric equations, each variable \( x \) and \( y \) is expressed with respect to a third variable, usually \( t \) (the parameter). This requires us to differentiate \( x \) and \( y \) independently with respect to \( t \).

For the given parametric equations \( x = t^2 - t \) and \( y = t^2 + t \), we find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) separately. These derivatives then help us find \( \frac{dy}{dx} \), the rate of change of \( y \) with respect to \( x \). This involves a simple quotient: \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \).

This process is integral in revealing how the curve is moving. Especially in problems involving concavity and curve sketching, understanding differentiation becomes crucial.

Chain Rule

The chain rule is a powerful tool in calculus that lets us differentiate composite functions. This rule becomes especially handy when dealing with parametric equations, as both \( x \) and \( y \) are expressed in terms of another variable, \( t \).

The chain rule formula in the context of these equations is: if we have functions \( y = f(t) \) and \( x = g(t) \), then \( \frac{d^2 y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) / \frac{dx}{dt} \). Essentially, we first differentiate the expression \( \frac{dy}{dx} \) with respect to \( t \), and then divide by \( \frac{dx}{dt} \).

The exercise shows this clearly with \( \frac{d^2 y}{dx^2} \) being calculated as \( \frac{\frac{-4}{(2t - 1)^2}}{2t - 1} = \frac{-4}{(2t - 1)^3} \). This lets us analyze the concavity of the curve effectively. The application of the chain rule is crucial, providing a streamlined method to handle derivatives in multi-variable situations.