Problem 3 - Calculus Early Transcendentals

Question

\(1-6\) Use the Chain Rule to find \(d z / d t\) or \(d w / d t\) $$z=\sqrt{1+x^{2}+y^{2}}, \quad x=\ln t, \quad y=\cos t$$

Step-by-Step Solution

Answer

\(\frac{dz}{dt} = \frac{\ln t}{t \sqrt{1 + (\ln t)^2 + \cos^2 t}} - \frac{\cos t \sin t}{\sqrt{1 + (\ln t)^2 + \cos^2 t}}\)

Step 1: Understand the Problem

We have a function \(z\) defined as \(z = \sqrt{1 + x^2 + y^2}\), where \(x = \ln t\) and \(y = \cos t\). We need to find \(\frac{dz}{dt}\) using the Chain Rule.

Step 2: Apply the Chain Rule

The Chain Rule states: if a variable \(z\) depends on \(x\) and \(y\), both of which depend on \(t\), then \(\frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}\).

Step 3: Find Partial Derivatives of z

Calculate \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\): - \(\frac{\partial z}{\partial x} = \frac{x}{\sqrt{1 + x^2 + y^2}}\)- \(\frac{\partial z}{\partial y} = \frac{y}{\sqrt{1 + x^2 + y^2}}\)

Step 4: Find Derivatives of x and y with respect to t

Calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\): - \(\frac{dx}{dt} = \frac{1}{t}\) (since \(x = \ln t\)) - \(\frac{dy}{dt} = -\sin t\) (since \(y = \cos t\))

Step 5: Substitute and Simplify

Substitute all derivatives back into the Chain Rule formula:\[\frac{dz}{dt} = \left( \frac{x}{\sqrt{1 + x^2 + y^2}} \right) \cdot \frac{1}{t} + \left( \frac{y}{\sqrt{1 + x^2 + y^2}} \right) \cdot (-\sin t)\]Substitute \(x = \ln t\) and \(y = \cos t\) into the expression and simplify:\[\frac{dz}{dt} = \frac{\ln t}{t \sqrt{1 + (\ln t)^2 + \cos^2 t}} - \frac{\cos t \sin t}{\sqrt{1 + (\ln t)^2 + \cos^2 t}}\]This is the derivative of \(z\) with respect to \(t\).

Key Concepts

Partial Derivatives

In calculus, when you deal with functions of several variables, understanding partial derivatives becomes important. Partial derivatives help us determine how a function changes with respect to just one of its variables, while keeping the others constant. This is crucial in many applications, like evaluating the slope in a multi-dimensional space.

When you see the notation \(\frac{\partial z}{\partial x}\) or \(\frac{\partial z}{\partial y}\), it means we're only considering the change in \(z\) with respect to \(x\) or \(y\), ignoring any dependency on other variables at that moment. In our problem:

• \(\frac{\partial z}{\partial x} = \frac{x}{\sqrt{1 + x^2 + y^2}}\)
• \(\frac{\partial z}{\partial y} = \frac{y}{\sqrt{1 + x^2 + y^2}}\)

Each expression shows how the function \(z\) changes as \(x\) or \(y\) changes individually, while the denominator \(\sqrt{1 + x^2 + y^2}\) scales this change appropriately.

Derivatives of Composite Functions

The derivative of composite functions extends the idea of differentiation when functions are nestled inside each other, like layers in an onion. The Chain Rule is the tool we use to handle this complexity.

Consider a composite function where \(z\) depends on \(x\) and \(y\), while both \(x\) and \(y\) depend on \(t\). The Chain Rule formula allows us to "peel back the layers" and compute the total rate of change of \(z\) with respect to \(t\):

• \(\frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}\)

By applying this formula, each component of the function's change is considered individually, allowing for a combined assessment of the rate of change across all variables involved.

Multivariable Calculus

Multivariable calculus expands upon single-variable calculus by exploring functions that depend on more than one variable. This is the realm where calculus meets dimensions, and our simple slopes and rates of change extend into all directions a function can vary.

In this context, functions are not just lines or curves but surfaces or even higher-dimensional shapes. This gives rise to a beautiful complexity where we use tools like partial derivatives and the Chain Rule to understand how each component of a function shifts.

• In our exercise, we used multivariable calculus concepts to explore \(z = \sqrt{1 + x^2 + y^2}\).
• Both \(x\) and \(y\) interact with \(t\) leading to more intricate dependencies.

As you advance, these concepts will help you explore areas from physical sciences to economics, wherever multiple changing variables are at play.

Implicit Differentiation

Often in calculus, not all functions are given explicitly in terms of a single variable. Instead, they are intertwined, which is where implicit differentiation shines. It allows us to find derivatives even when functions are not "solved" for a specific variable.

Implicit differentiation involves differentiating both sides of an equation with respect to a variable. Although our exercise didn't require implicit differentiation directly, understanding it is essential for dealing with complex equations, like \(x = \ln t\) and \(y = \cos t\), when they are part of a larger context.

• It is a powerful method when you cannot easily isolate one variable.
• Implicit differentiation is useful when the relationship between variables is defined but not straightforwardly separated.

This technique builds on the core idea of the Chain Rule, where multiple interconnected derivatives are calculated simultaneously.