Problem 222 - Calculus 3

Question

For the following exercises, find \(\frac{d f}{d t}\) using the chain rule and direct substitution. $$ f(x, y)=\sqrt{x^{2}+y^{2}}, y=t^{2}, x=t $$

Step-by-Step Solution

Answer

\( \frac{d f}{d t} = \frac{1 + 2t^2}{\sqrt{1 + t^2}} \)

Step 1: Identify Variables and Functions

We need to find \( \frac{d f}{d t} \) where the function \( f(x, y) = \sqrt{x^2 + y^2} \). Here, \( x = t \) and \( y = t^2 \). This means \( f \) is a function of \( x \) and \( y \), and both \( x \) and \( y \) are functions of \( t \).

Step 2: Apply the Chain Rule for Partial Derivatives

Using the chain rule for functions of several variables, we have: \( \frac{d f}{d t} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt} \). Identify what needs to be calculated: \( \frac{\partial f}{\partial x} \), \( \frac{\partial f}{\partial y} \), \( \frac{dx}{dt} \), and \( \frac{dy}{dt} \).

Step 3: Calculate the Partial Derivatives

First, calculate the partial derivative of \( f \) with respect to \( x \):\[ \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \]Now, calculate the partial derivative of \( f \) with respect to \( y \):\[ \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} \]

Step 4: Calculate Derivatives with Respect to t

Find the derivative of \( x \) with respect to \( t \):\[ \frac{dx}{dt} = 1 \]Find the derivative of \( y \) with respect to \( t \):\[ \frac{dy}{dt} = 2t \]

Step 5: Substitute into the Chain Rule Formula

Substitute the calculated partial derivatives and derivatives into the chain rule formula:\[ \frac{d f}{d t} = \frac{x}{\sqrt{x^2 + y^2}} \cdot 1 + \frac{y}{\sqrt{x^2 + y^2}} \cdot 2t \]

Step 6: Substitute x and y with Functions of t

Replace \( x \) with \( t \) and \( y \) with \( t^2 \) in the expression from Step 5:\[ \frac{d f}{d t} = \frac{t}{\sqrt{t^2 + (t^2)^2}} + \frac{t^2}{\sqrt{t^2 + (t^2)^2}} \cdot 2t \]

Step 7: Simplify the Expression

Simplify the expression:\[ \sqrt{t^2 + t^4} = \sqrt{t^2(1 + t^2)} = t \sqrt{1 + t^2} \]Thus:\[ \frac{d f}{d t} = \frac{t}{t \sqrt{1 + t^2}} + \frac{t^2 \cdot 2t}{t \sqrt{1 + t^2}} \ = \frac{1}{\sqrt{1 + t^2}} + \frac{2t^2}{\sqrt{1 + t^2}} \ = \frac{1 + 2t^2}{\sqrt{1 + t^2}} \]

Step 8: Write the Final Answer

The derivative of \( f \) with respect to \( t \) is\[ \frac{d f}{d t} = \frac{1 + 2t^2}{\sqrt{1 + t^2}} \]

Key Concepts

Partial Derivatives

In multivariable calculus, partial derivatives allow us to understand how a function changes as one of its input variables changes, keeping all other variables constant. For a function of two variables, such as

• \( f(x, y) = \sqrt{x^2 + y^2} \)

we can find the partial derivative with respect to each variable. Here, we compute:

• The partial derivative of \( f \) with respect to \( x \), \( \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \).
• When computing for \( y \), \( \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} \).

These calculations show how the function \( f \) changes relative to small changes in \( x \) and \( y \) respectively, assuming the other variable is held constant. This concept is essential when understanding the behavior of multivariable functions in various applications such as physics and engineering.

Multivariable Calculus

In multivariable calculus, we extend the concepts of calculus to functions that have more than one input variable. Consider the function \( f(x, y) = \sqrt{x^2 + y^2} \). This is a function of two variables, \( x \) and \( y \), where both are expressed in terms of another variable\( t \):

• \( x = t \)
• \( y = t^2 \)

By expressing \( x \) and \( y \) in terms of \( t \), we enable calculations using the chain rule to find the derivative of \( f \) with respect to \( t \). This process involves analyzing how simultaneous changes in \( x \) and \( y \) affect \( f \).
This kind of analysis is crucial when working with systems that depend on multiple variables, allowing us to track the impact of these dependencies on the overall system behavior. It finds applications in fields such as economics, fluid dynamics, and machine learning, among others.

Derivatives with Respect to a Parameter

Finding derivatives with respect to a parameter is a vital skill in calculus, especially when handling functions of several variables that depend on a single parameter yet involve multiple expressions. The goal is to compute the total change in the function as the parameter varies. Using the example function: \( f(x, y) = \sqrt{x^2 + y^2} \) where \( x = t \) and \( y = t^2 \), we used the chain rule.
The chain rule formula for this scenario is:

• \[ \frac{df}{dt} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt} \]

Here,

• The derivative of \( x \) with respect to \( t \) is \( 1 \), since \( x = t \).
• The derivative of \( y \) with respect to \( t \) is \( 2t \) since \( y = t^2 \).

After substituting and simplifying, we find that \( \frac{df}{dt} \) equates to \( \frac{1 + 2t^2}{\sqrt{1 + t^2}} \). The ability to transition from individual variable changes to overall parameter changes is useful in scientific modeling, optimization problems, and complex system analyses.