Problem 221 - Calculus 3

Question

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

Step-by-Step Solution

Answer

\(\frac{df}{dt} = 2t + 4t^3\).

Step 1: Identify Variables and Functions

Given the function \(f(x, y) = x^2 + y^2\) and the substitutions \(x = t\) and \(y = t^2\), we want to find \(\frac{df}{dt}\) by using the chain rule.

Step 2: Apply Chain Rule

The chain rule for a function \(f(x, y)\), where \(x = g(t)\) and \(y = h(t)\), states that \(\frac{df}{dt} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt}\).

Step 3: Compute Partial Derivatives

Compute the partial derivatives: \(\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2 + y^2) = 2x\) and \(\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 + y^2) = 2y\).

Step 4: Compute Derivatives with Respect to \(t\)

Compute the derivatives of \(x\) and \(y\) with respect to \(t\): \(\frac{dx}{dt} = \frac{d}{dt}(t) = 1\) and \(\frac{dy}{dt} = \frac{d}{dt}(t^2) = 2t\).

Step 5: Substitute in Chain Rule Expression

Substitute the partial derivatives and the derivatives with respect to \(t\) into the chain rule expression: \(\frac{df}{dt} = (2x) \cdot 1 + (2y) \cdot 2t\).

Step 6: Substitute \(x = t\) and \(y = t^2\)

Substitute \(x = t\) and \(y = t^2\) into the expression: \(\frac{df}{dt} = 2t + 4t(t^2)\).

Step 7: Simplify the Expression

Simplify \(\frac{df}{dt} = 2t + 4t^3\). Combine terms to get the final expression: \(\frac{df}{dt} = 2t + 4t^3\).

Key Concepts

Partial Derivatives

In calculus, partial derivatives are essential when working with functions of several variables. These derivatives depict how a function changes as each particular variable changes while keeping other variables constant.
For instance, in the function \( f(x, y) = x^2 + y^2 \), the partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), gives us an idea of how the function \( f \) changes as \( x \) varies, while \( y \) is kept fixed. Here, \( \frac{\partial f}{\partial x} = 2x \).

• Partial Derivative with Respect to \( x \): Treat \( y \) as a constant and differentiate \( x \)'s terms.
• Partial Derivative with Respect to \( y \): Treat \( x \) as a constant and focus on the \( y \) terms, yielding \( \frac{\partial f}{\partial y} = 2y \).

Partial derivatives enable us to dissect how distinct variables influence a multi-variable function. This dissection is foundational to solving problems involving the chain rule in functions with more than one variable.

Substitution Method

The substitution method simplifies solving calculus problems by temporarily changing the variables to something more manageable. In many cases, it involves expressing the variables in terms of another single parameter.
In this exercise, given \( x = t \) and \( y = t^2 \), we substitute these expressions back into the original function \( f(x, y) = x^2 + y^2 \). This substitution forms the new function \( f(t) = t^2 + (t^2)^2 \), which then simplifies our calculations significantly.

• Step 1: Identify which variables can be expressed in terms of a new single parameter (in this case, \( t \)).
• Step 2: Substitute the expressions for \( x \) and \( y \) into the function \( f \) to create a single-variable expression.

Using this method breaks down complex problems involving multiple variables into simpler computations. It's crucial for finding solutions quickly and efficiently.

Derivatives with Respect to a Parameter

When a function's variables are expressions of another parameter, we calculate derivatives with respect to that parameter, commonly \( t \). This approach helps us understand how the entire function changes concerning this parameter. By doing so, we derive the rate of change of the original function based on how \( x \) and \( y \) change with \( t \).
For our function \( f(x,y) = x^2 + y^2 \), and substitutions \( x = t \) and \( y = t^2 \):

• \( \frac{dx}{dt} \): Derivative of \( x = t \) with respect to \( t \) is \( 1 \).
• \( \frac{dy}{dt} \): Derivative of \( y = t^2 \) with respect to \( t \) is \( 2t \).

Inserting these into the chain rule, we find \( \frac{df}{dt} = (2x) \times 1 + (2y) \times 2t \). After substituting \( x = t \) and \( y = t^2 \), we get the final result \( \frac{df}{dt} = 2t + 4t^3 \). This showcases the changes in \( f \) directly related to \( t \), providing deeper insight into the function's behavior.