Chapter 13: Problem 12
Suppose that $$ w=x \sin y z^{2} ; \quad x=\cos t, \quad y=t^{2}, \quad z=e^{t} $$ Find the rate of change of \(w\) with respect to \(t\) at \(t=0\) by using the chain rule, and then check your work by expressing \(w\) as a function of \(t\) and differentiating.
Short Answer
Step by step solution
Define the Functions
Apply Chain Rule for Differentiation
Calculate Partial Derivatives
Differentiate with Respect to t
Evaluate at \(t = 0\)
Verify by Differentiating \(w(t)\) Directly
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
- x represents one dimension, and the other two dimensions are represented by y and z.
- This is where partial derivatives become important.
Here are the three partial derivatives computed in this exercise:
- \( \frac{\partial w}{\partial x} \) measures how \(w\) changes with changes in \(x\) while keeping \(y \) and \(z\) constant. It's given as \( \sin(yz^2) \).
- \( \frac{\partial w}{\partial y} \) looks at change in \(w\) with respect to \(y\), producing \( xz^2 \cos(yz^2) \).
- \( \frac{\partial w}{\partial z} \) considers changes in \(w\) due to \(z\), leading to \( 2xyz \cos(yz^2) \).
Differentiation Techniques
- Begin by defining \(w = \cos t \sin(t^2 e^{2t})\) and differentiating it using chain and product rules.
- The chain rule here is used to link the rate of change of \( w \) with \(x, y, z\) and combines them elegantly into a single expression in terms of \( t \).
This solidifies our connection between partial derivatives and the total derivative via the chain rule.This process highlights how powerful differentiation techniques aid in solving complex problems beyond simple one-variable functions.
Rate of Change
- We used the partial derivatives with respect to each variable and their rates of change concerning \(t\).
- Then, these rates are compounded into one by the chain rule to provide \( \frac{dw}{dt} \), the total rate of change in \(w\) as \(t\) varies.