Question

For the following exercises, find \(d^{2} y / d x^{2}$$x=e^{-t}, y=t e^{2 t}\)

Short answer

The second derivative is \(d^{2} y / d x^{2} = -e^{4t}(5 + 6t)\).

Step-by-step solution

Find Expression for the First Derivative

First, express both variables in terms of the parameter \(t\). We have \(x = e^{-t}\) and \(y = t e^{2t}\). Compute \(dx/dt\) and \(dy/dt\). We have:\[ \frac{dx}{dt} = -e^{-t} \]\[ \frac{dy}{dt} = e^{2t} + 2te^{2t} \]

Apply the Chain Rule to Find \(dy/dx\)

Use the chain rule to find \(dy/dx\). The chain rule states that:\[ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} \]We know that \( \frac{dt}{dx} = \frac{1}{dx/dt} = -e^{t} \). Therefore,\[ \frac{dy}{dx} = (e^{2t} + 2te^{2t})(-e^{t}) = -(e^{3t} + 2te^{3t}) \]

Differentiate \(dy/dx\) to Find \(d^2y/dx^2\)

Differentiate the expression for \(dy/dx\) with respect to \(x\) again. We first differentiate with respect to \(t\) using the product rule and then multiply by \(dt/dx\):First, differentiate:\[ \frac{d}{dt}[-(e^{3t} + 2te^{3t})] = -[3e^{3t} + (6te^{3t} + 2e^{3t})] = -[5e^{3t} + 6te^{3t}] \]Multiply by \(dt/dx\):\[ \frac{d^2y}{dx^2} = -[5e^{3t} + 6te^{3t}](e^t) = -5e^{4t} - 6te^{4t} \]

Simplify the Expression for \(d^2y/dx^2\)

Combine terms to simplify the expression:\[ \frac{d^2y}{dx^2} = -e^{4t}(5 + 6t) \]

Key concepts

Parametric Equations

In mathematics, sometimes quantities are determined by an external parameter rather than directly related to each other. This leads us to parametric equations.
Parametric equations express the coordinates of the points that make up a geometric object in terms of one or more variables called parameters.For example, we consider the parametric equations given in this exercise:

• \(x = e^{-t}\)
• \(y = t e^{2t}\)

Here, both \(x\) and \(y\) depend on the parameter \(t\). This allows us to explore the relationship between \(x\) and \(y\) in a more flexible and powerful way.
Instead of describing a curve by a function \(y=f(x)\), one can use a parameter to define both \(x\) and \(y\).
This approach is particularly useful in dealing with complex curves where one-to-one functions may not suffice.

Chain Rule

The chain rule is a fundamental concept in calculus. It helps us differentiate composite functions.
When dealing with parametric equations, the chain rule becomes indispensable.To compute \(\frac{dy}{dx}\) when both \(x\) and \(y\) are given in terms of \(t\), we use:

• \(\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}\)

In the exercise provided, this method allows us to find the rate of change of \(y\) with respect to \(x\) even though \(x\) and \(y\) were originally expressed in terms of \(t\).
The chain rule is efficient in linking the changes across multiple related variables, providing a chain-like connection of rates.

Product Rule

Another important rule for differentiation is the product rule. It is used when you need to differentiate a product of functions.
The product rule states:

• If \(u(t)\) and \(v(t)\) are functions of \(t\), then \(\frac{d}{dt}[u(t) \cdot v(t)] = u'(t)\cdot v(t) + u(t) \cdot v'(t)\).

In our exercise, when finding the second derivative \(\frac{d^2y}{dx^2}\), the product rule helps differentiate each term within \(\frac{d}{dt}[-(e^{3t} + 2te^{3t})]\).
This highlights the power of the product rule in breaking down complex derivatives into manageable parts.

Differentiation

Differentiation is a core operation in calculus that involves computing the derivative of a function.
The derivative represents the rate of change of a function at any given point and is crucial for understanding how functions behave.In this exercise, differentiation is used in several parts:

• First, to find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\).
• Then, to determine \(\frac{dy}{dx}\) using the chain rule.
• Lastly, to compute the second derivative \(\frac{d^2 y}{dx^2}\) using the first derivative.

Differentiation allows us to solve complex problems by computing how one variable changes with respect to another.
Its applications are vast, ranging from mathematics to physics and engineering.