Problem 18 - Calculus with Concepts in Calculus

Question

Compute \(d w / d t\) $$ w=\ln \left(x^{2}+y^{2}+z^{2}\right) ; x=\sin t, y=\cos t, z=e^{-t^{2}} $$

Step-by-Step Solution

Answer

The derivative \( \frac{dw}{dt} = \frac{-4t(e^{-t^2})^2}{1 + e^{-2t^2}} \).

Step 1: Understand the Derivatives Needed

To find \( \frac{dw}{dt} \), we apply the chain rule. This requires finding the partial derivatives \( \frac{\partial w}{\partial x} \), \( \frac{\partial w}{\partial y} \), and \( \frac{\partial w}{\partial z} \) as well as the derivatives of \( x \), \( y \), and \( z \) with respect to \( t \).

Step 2: Compute Partial Derivatives of w

Given \( w = \ln(x^2 + y^2 + z^2) \), find the partial derivatives:\[ \frac{\partial w}{\partial x} = \frac{2x}{x^2 + y^2 + z^2} \]\[ \frac{\partial w}{\partial y} = \frac{2y}{x^2 + y^2 + z^2} \]\[ \frac{\partial w}{\partial z} = \frac{2z}{x^2 + y^2 + z^2} \]

Step 3: Compute Derivatives of x, y, and z

Find the derivatives of each variable with respect to \( t \):\[ \frac{dx}{dt} = \cos t \]\[ \frac{dy}{dt} = -\sin t \]\[ \frac{dz}{dt} = -2te^{-t^2} \]

Step 4: Apply Chain Rule to Find dw/dt

Combine the results using the chain rule to determine \( \frac{dw}{dt} \):\[ \frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} + \frac{\partial w}{\partial z} \cdot \frac{dz}{dt} \]This simplifies to:\[ \frac{dw}{dt} = \frac{2x}{x^2 + y^2 + z^2} \cdot \cos t + \frac{2y}{x^2 + y^2 + z^2} \cdot (-\sin t) + \frac{2z}{x^2 + y^2 + z^2} \cdot (-2t e^{-t^2}) \]

Step 5: Simplify the Expression

Substitute back the expressions for \( x = \sin t \), \( y = \cos t \), and \( z = e^{-t^2} \) into the equations to express everything in terms of \( t \). Simplification leads to:\[ \frac{dw}{dt} = \frac{2 \sin t \cdot \cos t - 2 \cos t \cdot \sin t - 4t \cdot e^{-t^2} \cdot e^{-t^2}}{\sin^2 t + \cos^2 t + (e^{-t^2})^2} \]\[ \frac{dw}{dt} = \frac{-4t (e^{-t^2})^2}{1 + e^{-2t^2}} \]The terms \(2 \sin t \cdot \cos t - 2 \cos t \cdot \sin t\) cancel each other out because they are equal and opposite.

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus used to find the derivative of composite functions. It allows us to differentiate functions that are made up of other functions. In this context, when we want to find the derivative of a function like \(w\) that depends on \(x\), \(y\), and \(z\) which in turn depend on \(t\), the chain rule helps us measure how \(w\) changes as \(t\) changes.
For a function where \(w\) is defined by variables \(x\), \(y\), and \(z\) (all functions of \(t\)), the chain rule is expressed as:

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

Here, each term shows the influence of a variable and its rate of change with respect to \(t\). This complete expression gives us the total rate of change of \(w\) with respect to \(t\), effectively capturing the contributions from all dependent functions.

Partial Derivatives

Partial derivatives are used when dealing with functions of multiple variables. They measure how a function changes as one specific variable changes while keeping other variables constant. If \(w\) is a function of \(x\), \(y\), and \(z\), then the partial derivative with respect to \(x\), for example, is how \(w\) changes as \(x\) changes.

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

These expressions show the gradient or slope of the function \(w\) in the direction of each of these axes, effectively showing how sensitive \(w\) is to small changes in \(x, y,\) and \(z\). Understanding partial derivatives is crucial for using the chain rule to derive \(\frac{dw}{dt}\).
They decompose the effect of each variable, forming the backbone of multivariable differentiation.

Derivative of Exponential Functions

The derivative of exponential functions involves understanding how expressions involving the constant \(e\) change. Exponential functions arise frequently in problems involving growth and decay, like compound interest or radioactive decay.
For the function \(z = e^{-t^2}\), the derivative with respect to \(t\) is

• \(\frac{dz}{dt} = -2te^{-t^2}\)

This derives from the chain rule and the derivative of its exponent \(-t^2\). When faced with an exponential function, the derivative can often be retained in the same exponential form, shifted by a coefficient from the differentiation of the exponent.
Considering the entire process of finding \(\frac{dw}{dt}\), these derivatives play a key role in the chain rule. They contribute to the understanding of how \(w\), as a more complex function of \(t\), changes when the independent variable shifts. Delving into derivatives of exponentials teaches the subtle transformations that these functions experience in calculus.