Problem 221 - Calculus Volume 3

Question

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

Step-by-Step Solution

Answer

Using both direct differentiation and chain rule confirms \(\frac{d f}{d t} = 2t + 4t^3\).

Step 1: Express f in terms of t

First, substitute the expressions for \(x\) and \(y\) into the function \(f(x, y) = x^2 + y^2\). This means substituting \(x = t\) and \(y = t^2\) into \(f\). The function becomes \(f(t) = t^2 + (t^2)^2\). Simplify this to get \(f(t) = t^2 + t^4\).

Step 2: Differentiate using direct substitution

Differentiate the function \(f(t) = t^2 + t^4\) with respect to \(t\). The derivative \(\frac{d}{dt}(t^2) = 2t\) and \(\frac{d}{dt}(t^4) = 4t^3\). Therefore, \(\frac{d f}{d t} = 2t + 4t^3\).

Step 3: Verify with the chain rule

To use the chain rule, differentiate \(f\) first with respect to \(x\) and \(y\). Compute \(\frac{\partial f}{\partial x} = 2x\) and \(\frac{\partial f}{\partial y} = 2y\). Now apply the chain rule, using \(\frac{d x}{d t} = 1\) and \(\frac{d y}{d t} = 2t\).The chain rule gives: \[ \frac{d f}{d t} = \frac{\partial f}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial f}{\partial y} \cdot \frac{d y}{d t} = 2x \cdot 1 + 2y \cdot (2t) \].Substitute back the expressions for \(x\) and \(y\) into the equation:\[ 2(t) \cdot 1 + 2(t^2) \cdot 2t = 2t + 4t^3 \].This confirms the direct differentiation method.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, providing a way to understand how a function changes at any given point. In this exercise, we are given the function \( f(x, y) = x^2 + y^2 \), which depends on both \( x \) and \( y \). Here's how we approach differentiation in this context:

• First, we make use of direct substitution to express the function in terms of a single variable, \( t \). In the exercise, \( x = t \) and \( y = t^2 \) are substituted into \( f \), transforming it to \( f(t) = t^2 + t^4 \).
• Next, we differentiate the resulting function with respect to \( t \). Differentiation involves calculating the derivative, which indicates the function's rate of change. For example, the derivative of \( t^2 \) is \( 2t \), and for \( t^4 \), it is \( 4t^3 \). Therefore, the derivative of \( f(t) \) is \( \frac{d f}{d t} = 2t + 4t^3 \).

Breaking down complex expressions into simpler ones using differentiation can help unravel how variables and functions interplay, all through the lens of calculus.

Partial Derivatives

Partial derivatives are central to multivariable calculus, offering a glimpse into how a function changes with respect to one variable alone while the others are held constant. In the exercise, we identify how \( f(x, y) = x^2 + y^2 \) changes individually with \( x \) and \( y \).

• Finding the partial derivative of \( f \) with respect to \( x \), written as \( \frac{\partial f}{\partial x} \), involves treating \( y \) as a constant. In this context, the expression \( 2x \) indicates the sensitivity of \( f \) to alterations in \( x \) alone.
• Similarly, when determining \( \frac{\partial f}{\partial y} \), \( x \) is fixed and \( y \) is the dynamic variable. Thus, the derivative here, \( 2y \), demonstrates how \( f \) responds to changes in \( y \).

Utilizing partial derivatives, we capture the isolated contributions of each variable to the overall rate of change, forming the backbone of the chain rule application in the exercise.

Direct Substitution

Direct substitution is an intuitive method used to simplify complex expressions by replacing variables with given values or expressions. Here is how it was used in the given problem:

• The exercise involves substituting \( x = t \) and \( y = t^2 \). This turns the expression in \( f(x, y) \) into a simpler function \( f(t) = t^2 + t^4 \).
• By reducing \( f(x, y) = x^2 + y^2 \) to a single-variable expression \( f(t) \), we can compute the function's derivative directly without handling multiple variables simultaneously.

Direct substitution provides a straightforward pathway to transforming and solving complex equations, enabling us to leverage simpler calculus rules like the standard derivative handling in a single-variable context.