Problem 229 - Calculus 3

Question

Let \(u=e^{x} \sin y\) where \(x=t^{2}\) and \(y=\pi t\). Find \(\frac{d u}{d t}\) when \(x=\ln 2\) and \(y=\frac{\pi}{4}\)

Step-by-Step Solution

Answer

\( \frac{du}{dt} = 2\sqrt{2\ln 2} + \pi \sqrt{2} \)

Step 1: Substitute for x and y

First, substitute the expressions for \(x\) and \(y\) into the function \(u = e^x \sin y\). We have:\[ x = t^2 \quad \text{and} \quad y = \pi t \]Thus, the expression for \(u\) becomes:\[ u = e^{t^2} \sin(\pi t) \]

Step 2: Apply the Chain Rule

To find \( \frac{du}{dt} \), use the chain rule for functions of multiple variables. Given \( u = f(x, y) = e^x \sin y \), we need:\[ \frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt} \]

Step 3: Differentiate u with respect to x

Calculate \( \frac{\partial u}{\partial x} \) while treating \(y\) as a constant:\[ \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(e^x \sin y) = e^x \sin y \]

Step 4: Differentiate u with respect to y

Calculate \( \frac{\partial u}{\partial y} \) while treating \(x\) as a constant:\[ \frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(e^x \sin y) = e^x \cos y \]

Step 5: Differentiate x and y with respect to t

Calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):\[ \frac{dx}{dt} = \frac{d(t^2)}{dt} = 2t \]\[ \frac{dy}{dt} = \frac{d(\pi t)}{dt} = \pi \]

Step 6: Substitute into the Chain Rule formula

Substitute all the derivatives into the chain rule formula:\[ \frac{du}{dt} = \left(e^x \sin y\right)(2t) + \left(e^x \cos y\right)(\pi) \]

Step 7: Evaluate the expression at the given values

Use the given \(x = \ln 2\) and \(y = \frac{\pi}{4}\) to calculate \(t\):From \(x = t^2\), \(t = \sqrt{\ln 2}\).Substitute these into the formula:\[ \frac{du}{dt} = \left(e^{\ln 2} \sin \frac{\pi}{4}\right)(2\sqrt{\ln 2}) + \left(e^{\ln 2} \cos \frac{\pi}{4}\right)(\pi) \]

Step 8: Simplify the Expression

Simplify the result from Step 7:\[ \sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \]Thus:\[ \frac{du}{dt} = \left(2 \cdot \frac{\sqrt{2}}{2}\right)(2\sqrt{\ln 2}) + \left(2 \cdot \frac{\sqrt{2}}{2}\right)(\pi) \] Further simplify:\[ \frac{du}{dt} = 2\sqrt{2\ln 2} + \pi \sqrt{2} \]

Key Concepts

Partial Derivatives

In multivariable calculus, partial derivatives play a crucial role in understanding how a function changes with respect to one variable while keeping others constant. Partial derivatives extend the concept of a derivative to functions of multiple variables. For example, if we have a function \( u = f(x, y) \), the partial derivative of \( u \) with respect to \( x \) is denoted by \( \frac{\partial u}{\partial x} \). This measures how the function \( u \) changes as \( x \) changes, with \( y \) held constant.

• Example: To find \( \frac{\partial u}{\partial x} \) for \( u = e^x \sin y \), we partially differentiate with respect to \( x \), treating \( y \) as a constant, yielding \( \frac{\partial u}{\partial x} = e^x \sin y \).
• Similarly, \( \frac{\partial u}{\partial y} \) refers to differentiating with respect to \( y \): \( \frac{\partial u}{\partial y} = e^x \cos y \).

Understanding partial derivatives is essential for applying the chain rule in scenarios involving multiple variables.

Function Differentiation

Differentiation generalizes to functions of multiple variables through methods such as partial differentiation. The chain rule is an invaluable differentiation technique in calculus. It is especially powerful when you have composite functions. If a function \( u \) depends on two intermediate variables, \( x \) and \( y \), which in turn depend on a single variable \( t \), then the derivative of \( u \) with respect to \( t \) can be calculated using the chain rule:

• Start by finding the partial derivatives \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \).
• Then, find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
• Combine these results using the formula: \( \frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt} \).

This allows us to find how \( u \) changes as \( t \) varies, effectively differentiating the composite function.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions with more than one variable. It involves studying functions that can depend on several independent variables and explores how these functions behave and change. One of the essential aspects of multivariable calculus is handling situations where multiple variables interact and change simultaneously.

• For example, functions like \( u = e^x \sin y \) are common in multivariable calculus, where both \( x \) and \( y \) are independently varying.
• This field covers topics like partial derivatives, gradients, limits, and chain rule applications.

Understanding multivariable calculus helps to model complex systems found in physics, engineering, and other scientific fields, and solve problems involving rates of change in such systems.