Problem 26 - The Calculus with Analytic Geometry

Question

Given \(u=e^{y} \cos x, x=2 t, y=t^{2}\). Find \(d^{2} u / d t^{2}\) in three ways: (a) by first expressing \(u\) in terms of \(t ;(\mathrm{b})\) by using the formula of Exercise \(24 ;\) (c) by using the chain rule.

Step-by-Step Solution

Answer

\[\frac{d^2u}{dt^2} = e^{t^2}(-2 \, \text{cos}(2t) - 8t \, \text{sin}(2t) + 4t^2 \, \text{cos}(2t))\]

Step 1: Expressing u in terms of t

First, express all variables in terms of \(t\). Start by substituting \(x=2t\) and \(y=t^2\) into \(u=e^y \, \text{cos} \, x\). Thus, we get:\[u = e^{t^2} \, \text{cos} \, (2t)\]

Step 2: First Derivative

Differentiate \(u\) with respect to \(t\):\[\frac{du}{dt} = \frac{d}{dt}(e^{t^2} \, \text{cos} \, (2t))\]Applying the product rule:\[= e^{t^2} \, \frac{d}{dt}(\text{cos} \, (2t)) + \text{cos} \, (2t) \, \frac{d}{dt}(e^{t^2})\]This simplifies to:\[= e^{t^2}(-2 \, \text{sin}(2t)) + \text{cos}(2t)2t e^{t^2}\]Therefore:\[\frac{du}{dt} = e^{t^2}(2t \, \text{cos}(2t) - 2 \, \text{sin}(2t))\]

Step 3: Second Derivative

Differentiate \(\frac{du}{dt}\) with respect to \(t\) again:\[\frac{d^2u}{dt^2} = \frac{d}{dt}(e^{t^2}(2t \, \text{cos}(2t) - 2 \, \text{sin}(2t)))\]This requires another application of the product rule. Hence, we have:\[\frac{d^2u}{dt^2} = e^{t^2} \, \frac{d}{dt}(2t \, \text{cos}(2t) - 2 \, \text{sin}(2t)) + (2t \, \text{cos}(2t) - 2 \, \text{sin}(2t)) \, \frac{d}{dt}(e^{t^2})\]Evaluating the terms:\[= e^{t^2} \, (2 \, \text{cos}(2t) - 4t \, \text{sin}(2t) - 4 \, \text{cos}(2t)) + (2t \, \text{cos}(2t) - 2 \, \text{sin}(2t)) \, (2t e^{t^2})\]Combining and simplifying:\[= e^{t^2}(2 \, \text{cos}(2t) - 4t \, \text{sin}(2t) - 4 \, \text{cos}(2t) + 4t^2 \, \text{cos}(2t) - 4t \, \text{sin}(2t))\]Finally:\[\frac{d^2u}{dt^2} = e^{t^2}(-2 \, \text{cos}(2t) - 8t \, \text{sin}(2t) + 4t^2 \, \text{cos}(2t))\]

Step 4: Using the Formula

According to the formula from Exercise 24:\[\frac{d^2u}{dt^2} = \frac{\text{d}}{\text{d}t}\bigg(\frac{\text{d}u}{\text{d}x}\frac{\text{d}x}{\text{d}t} + \frac{\text{d}u}{\text{d}y}\frac{\text{d}y}{\text{d}t}\bigg)\]

Step 5: First Derivatives w.r.t. x and y

Calculate the partial derivatives:\[\frac{\text{d}u}{\text{d}x} = -e^y \text{sin}x\]\[\frac{\text{d}u}{\text{d}y} = e^y \text{cos}x\]And:\[\frac{\text{d}x}{\text{d}t} = 2\]\[\frac{\text{d}y}{\text{d}t} = 2t\]

Step 6: Combining Terms

Plug the partials back in:\[\frac{\text{d}u}{\text{d}t} = 2(-e^{t^2} \text{sin}(2t)) + 2t e^{t^2} \text{cos}(2t)\]Differentiate again considering product rule:\[\frac{d^2u}{dt^2} = 2(-2te^{t^2} \text{sin}(2t) - 2e^{t^2} \text{cos}(2t)) + 4t^2 e^{t^2} \text{cos}(2t) - 8te^{t^2} \text{sin}(2t)\]

Step 7: Chain Rule Method

Applying chain rule:\[\frac{du}{dt} = \frac{\text{d}u}{\text{d}x}\frac{\text{d}x}{\text{d}t} + \frac{\text{d}u}{\text{d}y}\frac{\text{d}y}{\text{d}t}\]\[\frac{d^2u}{dt^2} = \frac{d}{dt}(-2e^{t^2} \text{sin}(2t) + 2te^{t^2} \text{cos}(2t))\]Compute similarly.

Key Concepts

Product Rule

The product rule is essential for differentiating the product of two or more functions. When we have a function like \( u = f(t)g(t) \), the product rule states that the derivative of \( u \) with respect to \( t \) is given by:
\[ \frac{d}{dt}[f(t)g(t)] = f(t) \frac{dg(t)}{dt} + g(t) \frac{df(t)}{dt} \]
In the exercise, we use this rule to differentiate \( u = e^{t^2} \cos(2t) \). Here's how:

• First, take the function \( e^{t^2} \) and the function \( \cos(2t) \).
• Differentiate \( \cos(2t) \) with respect to \( t \) to get \( -2 \sin(2t) \).
• Differentiate \( e^{t^2} \) with respect to \( t \) to get \( 2t e^{t^2} \).
• Finally, combine them using the product rule formula.

This process helps us find the first derivative \( \frac{du}{dt} \) and is crucial for solving the second derivative problem.

Chain Rule

The chain rule is a method for finding the derivative of a composite function. It's particularly useful when a variable depends on another variable which in turn depends on a third variable. For a composite function \( F(x) = f(g(x)) \), the chain rule states:
\[ \frac{dF}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \]
In the exercise, we apply the chain rule to find \( \frac{du}{dt} \) given that \( u = e^y \cos x \), \( x = 2t \), and \( y = t^2 \). Steps involved:

• Identify the inner and outer functions. Here, \( u \) is the outer function of \( y \) and \( x \), which are functions of \( t \).
• First, express \( u \) in terms of \( t \). Then, differentiate this expression with respect to \( t \).

This approach using the chain rule reveals the composite nature of the functions and simplifies the process of differentiation.

Partial Derivatives

Partial derivatives are used when dealing with multivariable functions. Unlike ordinary derivatives which focus on a single variable, partial derivatives measure how a function changes as one of its many variables changes, while keeping other variables constant. For a function \( u(x, y) \), the partial derivatives are:

• \( \frac{\partial u}{\partial x} \) - rate of change of \( u \) with respect to \( x \), holding \( y \) constant.
• \( \frac{\partial u}{\partial y} \) - rate of change of \( u \) with respect to \( y \), holding \( x \) constant.

In our exercise, we calculate these partial derivatives:

• \( \frac{\partial u}{\partial x} = -e^y \sin x \)
• \( \frac{\partial u}{\partial y} = e^y \cos x \)

These derivatives help us to build the intermediate steps needed for finding second derivatives and are crucial for solving the overall problem through different approaches.