Problem 11 - Calculus Early Transcendentals

Question

Find \(d y / d x\) and \(d^{2} y / d x^{2} .\) For which values of \(t\) is the curve concave upward? $$x=4+t^{2}, \quad y=t^{2}+t^{3}$$

Step-by-Step Solution

Answer

The curve is concave upward for \(t > 0\).

Step 1: Compute the first derivatives dx/dt and dy/dt

Start by differentiating the given parametric equations with respect to t. We have: \(x = 4 + t^2\) and \(y = t^2 + t^3\). The derivatives are \(\frac{dx}{dt} = 2t\) and \(\frac{dy}{dt} = 2t + 3t^2\).

Step 2: Compute dy/dx using parametric derivatives

To find \(\frac{dy}{dx}\), use the formula \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\). Substitute the expressions from Step 1: \(\frac{dy}{dx} = \frac{2t + 3t^2}{2t}\). Simplify the expression: \(\frac{dy}{dx} = 1 + \frac{3}{2}t\).

Step 3: Compute the second derivative d^2y/dx^2

The second derivative \(\frac{d^2y}{dx^2}\) can be found using \(\frac{d}{dt}(\frac{dy}{dx}) / \frac{dx}{dt}\). Differentiate \(\frac{dy}{dx} = 1 + \frac{3}{2}t\) with respect to \(t\): \(\frac{d}{dt}(\frac{dy}{dx}) = \frac{3}{2}\). Therefore, \(\frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}\).

Step 4: Determine concavity of the curve

The curve is concave upward when \(\frac{d^2y}{dx^2} > 0\). Using \(\frac{d^2y}{dx^2} = \frac{3}{4t}\), this inequality holds when \(4t > 0\). Solving this, we find \(t > 0\).

Key Concepts

First Derivative

The first derivative is a fundamental concept in calculus that helps us understand how a function changes. When working with parametric equations, like in the case of this exercise, we need to find the first derivative \(\frac{dy}{dx}\) using the parametric derivatives \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\). This is done because the functions x and y are both expressed in terms of another parameter, in this case, t. By applying the chain rule, we find \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\). - For the given equations, \(x = 4 + t^2\) and \(y = t^2 + t^3\): - Differentiate the x-equation with respect to t to get \(\frac{dx}{dt} = 2t\). - Differentiate the y-equation with respect to t to get \(\frac{dy}{dt} = 2t + 3t^2\).Substitute these derivatives into the formula for \(\frac{dy}{dx}\) to get \(\frac{dy}{dx} = \frac{2t + 3t^2}{2t} = 1 + \frac{3}{2}t\).This first derivative \(\frac{dy}{dx}\) indicates the slope of the tangent to the curve at any point \(t\). It's especially useful when analyzing the behavior of the curve.

Second Derivative

The second derivative \(\frac{d^2y}{dx^2}\) helps us understand the curvature of a function. It tells us how the slope of the first derivative is changing. By looking at the second derivative, we can gain insights into the function's concavity.To compute the second derivative when dealing with parametric equations, use the expression \(\frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy}{dx}) / \frac{dx}{dt}\). This involves differentiating the first derivative \(\frac{dy}{dx}\) with respect to t and dividing by \(\frac{dx}{dt}\).- For the given first derivative \(\frac{dy}{dx} = 1 + \frac{3}{2}t\): - Differentiate with respect to parameter t to obtain \(\frac{d}{dt}(\frac{dy}{dx}) = \frac{3}{2}\). - Divide by \(\frac{dx}{dt} = 2t\) to find \(\frac{d^2y}{dx^2} = \frac{\frac{3}{2}}{2t} = \frac{3}{4t}\).The second derivative provides valuable information regarding the curve’s behavior such as concavity and potential inflection points.

Concavity

Concavity is a concept which indicates the direction in which a curve bends. Understanding concavity involves analyzing the second derivative \(\frac{d^2y}{dx^2}\). A curve is concave upward if \(\frac{d^2y}{dx^2} > 0\), meaning it bends upwards like a cup. Conversely, a curve is concave downward if \(\frac{d^2y}{dx^2} < 0\).- For the curve described by the equations \(x = 4 + t^2\) and \(y = t^2 + t^3\): - The second derivative is \(\frac{3}{4t}\). - To determine concavity, set the second derivative greater than zero: \(\frac{3}{4t} > 0\).Solving the inequality \(4t > 0\), we find \(t > 0\). This means the curve is concave upward when the parameter \(t\) is positive.

Differentiation

Differentiation is a key tool in calculus used to find the rate at which one quantity changes with respect to another. For parametric equations, differentiation helps in identifying these rates with respect to a parameter, which in this case is t.- Start by differentiating each parametric equation: - For \(x = 4 + t^2\), differentiate with respect to t to get \(\frac{dx}{dt} = 2t\). - For \(y = t^2 + t^3\), differentiate with respect to t to get \(\frac{dy}{dt} = 2t + 3t^2\).This process of differentiation using parametric derivatives is important in converting rates of change in terms of t into a more conventional form, such as \(\frac{dy}{dx}\), which is essential for understanding vector paths, curves, and physics applications.Through differentiating these parametric equations, we derive the first and second derivatives, which then provide deeper insights into the behavior of the function such as slopes and concavity.