Problem 3 - Calculus Volume 1

Question

Find the quantities for the given equation. Find \(\frac{d z}{d t}\) at \((x, y)=(1,3)\) and \(z^{2}=x^{2}+y^{2}\) if \(\frac{d x}{d t}=4\) and \(\frac{d y}{d t}=3.\)

Step-by-Step Solution

Answer

\(\frac{dz}{dt} = \frac{13}{\sqrt{10}}\) at \((x, y) = (1, 3)\).

Step 1: Implicit Differentiation Setup

Given the equation \(z^2 = x^2 + y^2\), we need to differentiate both sides with respect to \(t\). This involves implicit differentiation as \(z\), \(x\), and \(y\) are all functions of \(t\).

Step 2: Differentiate the Left Side

Differentiate \(z^2\) with respect to \(t\). Using the chain rule, we have \(\frac{d}{dt}(z^2) = 2z \cdot \frac{dz}{dt}\).

Step 3: Differentiate the Right Side

Differentiate \(x^2 + y^2\) with respect to \(t\). Applying the chain rule to each term, we get \(2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt}\).

Step 4: Set Up the Equation

Equating the differentiated expressions from both sides, we have:\[2z \cdot \frac{dz}{dt} = 2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt}\]

Step 5: Simplify and Solve for \(\frac{dz}{dt}\)

Cancel the factor of 2 from every term to simplify the equation:\[z \cdot \frac{dz}{dt} = x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt}\]Now, solve for \(\frac{dz}{dt}\):\[\frac{dz}{dt} = \frac{x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt}}{z}\]

Step 6: Substitute Known Values

Substitute the known values \(x = 1\), \(y = 3\), \(\frac{dx}{dt} = 4\), and \(\frac{dy}{dt} = 3\) into the equation. Also, calculate \(z\) at \((x, y) = (1, 3)\) using \(z = \sqrt{x^2 + y^2} = \sqrt{1^2 + 3^2} = \sqrt{10}\).

Step 7: Compute \(\frac{dz}{dt}\)

Substitute the values into the formula:\[\frac{dz}{dt} = \frac{1 \cdot 4 + 3 \cdot 3}{\sqrt{10}} = \frac{4 + 9}{\sqrt{10}} = \frac{13}{\sqrt{10}}\]

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus for finding the derivative of composite functions. When a function is defined in terms of another function, the chain rule helps us differentiate it with respect to a different variable. In the original exercise, the variable \(z\) depends on \(x\) and \(y\), which themselves are functions of \(t\). So, to differentiate \(z^2\) with respect to \(t\), we apply the chain rule.

• The derivative of \(z^2\) with respect to \(t\) is given by \(\frac{d}{dt}(z^2) = 2z \cdot \frac{dz}{dt}\).
• This reflects the chain rule, where we differentiate \(z^2\) with respect to \(z\), and then multiply by the derivative of \(z\) with respect to \(t\).

Here, the chain rule is essential as it connects the derivative of \(z\) to the rates of change of \(x\) and \(y\). Using this step ensures that we capture the relationship between these variables as they change over time, making it a vital part of tackling problems involving implicit differentiation.

Related Rates

Related rates problems arise from situations where multiple variables are changing with respect to time, and the rate of change of one variable affects the others. In this context, related rates are used to connect the changes in \(x\), \(y\), and \(z\) as they vary with \(t\).

• The exercise involves finding \(\frac{dz}{dt}\) when \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are given.
• The equation \(z^2 = x^2 + y^2\) means that any changes in \(x\) and \(y\) will affect \(z\), as all are interdependent.

By differentiating the equation with respect to time \(t\) and applying the concept of related rates, we translate the information about how \(x\) and \(y\) change into how \(z\) changes. These types of problems are typical in scenarios like tracking the speed of moving objects or the rates at which shadows lengthen or contract.

Differentiation Techniques

Differentiation techniques encompass a set of methods used to find the derivative of functions. Understanding and applying these techniques are crucial in solving calculus problems efficiently. The original problem uses implicit differentiation, a technique used when a function isn't given in explicit form (where \(y\) is a function of \(x\), for example), but instead, all variables are interrelated.

• Implicit differentiation allows us to differentiate both sides of an equation involving multiple variables with respect to a single variable \(t\).
• In the exercise, we apply implicit differentiation to \(z^2 = x^2 + y^2\) to derive an equation that relates all the rates of change: \(2z \cdot \frac{dz}{dt} = 2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt}\).

This approach is powerful, as it lets us find derivatives without reorganizing equations into explicit forms. In practice, differentiation techniques like these are widely used across many disciplines, including physics and engineering, where understanding the rates of change is crucial in the analysis of systems.