Problem 242 - Calculus 3

Question

Find \(\frac{d z}{d t}\) by the chain rule where \(z=\cosh ^{2}(x y), x=\frac{1}{2} t\) and \(y=e^{t}\)

Step-by-Step Solution

Answer

\(\frac{dz}{dt} = \sinh(te^{t})\left(\frac{e^{t}}{2} + \frac{1}{2}te^{2t}\right)\)."

Step 1: Differentiate z with respect to x

First, we find the derivative of \(z\) with respect to \(x\). We have \(z = \cosh^2(xy)\). By the chain rule, we differentiate \(\cosh^2(u)\) where \(u = xy\), yielding \(\frac{d z}{d u} = 2\cosh(u)\sinh(u) = \sinh(2u)\), and so \(\frac{d z}{d x} = \sinh(2xy) \cdot y\) since \(\frac{d u}{d x} = y\).

Step 2: Differentiate z with respect to y

Next, differentiate \(z\) with respect to \(y\). Using the derivative found earlier, \(\frac{d z}{d y} = \sinh(2xy) \cdot x\) as \(\frac{d u}{d y} = x\).

Step 3: Differentiate x and y with respect to t

Determine \(\frac{d x}{d t}\) and \(\frac{d y}{d t}\). We have \(x = \frac{1}{2}t\), so \(\frac{d x}{d t} = \frac{1}{2}\). For \(y = e^{t}\), \(\frac{d y}{d t} = e^{t}\).

Step 4: Apply the Chain Rule to Find dz/dt

Use the chain rule to find \(\frac{dz}{dt}\). According to the chain rule, \(\frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}\). Substituting the results from previous steps, this becomes: \(\frac{dz}{dt} = \sinh(2xy) \cdot y \cdot \frac{1}{2} + \sinh(2xy) \cdot x \cdot e^{t}\). Simplifying, we get: \(\frac{dz}{dt} = \sinh(2xy) \left(\frac{y}{2} + xe^{t}\right)\).

Step 5: Substitute x and y

Substitute \(x=\frac{1}{2}t\) and \(y=e^{t}\) into the expression to find \(\frac{d z}{d t}\). This gives: \(\frac{dz}{dt} = \sinh\left(t e^{t}\right)\left(\frac{e^{t}}{2} + \frac{1}{2}te^{2t}\right)\).

Key Concepts

Partial Derivative

Understanding partial derivatives is essential when dealing with multivariable functions. In this type of calculus, a function depends on several variables, and a partial derivative tells us how this function changes with respect to one of these variables, while keeping the others constant.
Partial derivatives are denoted by symbols such as \( \frac{\partial z}{\partial x} \). They represent the rate of change of the function \( z \) in response to changes in \( x \), ignoring other variables like \( y \).

• For our problem, finding \( \frac{\partial z}{\partial x} \) involves differentiating \( z = \cosh^2(xy) \) using the chain rule. We assume all values of \( y \) are constant during this differentiation process.
• Likewise, \( \frac{\partial z}{\partial y} \) requires the same consideration for the \( x \) variable, treating \( x \) as a constant, while differentiating by \( y \).
• These derivatives highlight how \( z \) behaves separately under changes to either \( x \) or \( y \), critical for solving the full problem where changes from both variables are combined.

Hyperbolic Functions

Hyperbolic functions, analogous to trigonometric functions, are frequently encountered in various calculus problems.
Two key hyperbolic functions are \( \cosh(x) \) and \( \sinh(x) \). These functions resemble \( \cos(x) \) and \( \sin(x) \) but with crucial differences suited to different contexts.

• The hyperbolic cosine, \( \cosh(x) \), is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).
• The hyperbolic sine, \( \sinh(x) \), is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \).

These functions are important because their derivatives have neat, predictable patterns. For instance:

• The derivative of \( \cosh(x) \) is \( \sinh(x) \).
• The derivative of \( \sinh(x) \) is \( \cosh(x) \).

In the exercise, recognizing these derivatives allows for the use of the chain rule efficiently when differentiating \( z = \cosh^2(xy) \) with respect to \( x \) and \( y \).

Differentiation with Respect to a Parameter

Differentiating with respect to a parameter means understanding how a function changes when the parameter, instead of a variable, changes.
In our given problem, the parameter is \( t \), around which the expressions for \( x \) and \( y \) are built. Their change with regard to \( t \) affects \( z \) in turn, because \( x = \frac{1}{2}t \) and \( y = e^t \).

• The derivative \( \frac{dx}{dt} = \frac{1}{2} \) indicates how \( x \) changes with each increment in \( t \).
• Similarly, \( \frac{dy}{dt} = e^t \) shows the instantaneous rate of change of \( y \) as \( t \) varies.

Differentiation with respect to a parameter is crucial when performing the final steps using the chain rule, incorporating how each function changes as \( t \) changes, thereby impacting the overall expression for \( \frac{dz}{dt} \).

Multivariable Calculus

Multivariable calculus is an extension of basic calculus to functions of more than one variable. It opens the door to a wide range of applications and fields of study, allowing for analysis of systems where multiple inputs affect the outputs simultaneously.
In multivariable calculus, each variable can independently affect the function's outcome. In our exercise, \( z \) depended on both \( x \) and \( y \), which are further dependent on \( t \).

• Using partial derivatives, we can examine the input changes separately, leading to thorough insights into how changing each input impacts the function as a whole.
• Chain rule usage helps in combining these individual effects into a comprehensive answer.
• This process simulates real-world scenarios where multiple factors intervene to influence results, such as in physics, engineering, and economics.

The challenge in multivariable calculus is mastering how to juggle these variables and correctly apply mathematical principles such as the chain rule to untangle their interdependencies and solve complex problems.