First the pieces: dz dx 8z dy da dt dy dt Use the chain rule to find dz dt || || || End result (in terms of just t): dz dt where 2 = x²y + xy², x = 3 + t², y = −2+t³

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Using the Chain Rule to Find \(\frac{dz}{dt}\):**

Given the functions:
\[ z = x^2y + xy^2, \quad x = 3 + t^2, \quad y = -2 + t^3 \]

**First, calculate the partial derivatives:**

\[
\frac{\partial z}{\partial x} = 
\]

\[
\frac{\partial z}{\partial y} = 
\]

**Next, find the derivatives with respect to \(t\):**

\[
\frac{dx}{dt} = 
\]

\[
\frac{dy}{dt} = 
\]

**End result (express \(\frac{dz}{dt}\) solely in terms of \(t\)):**

\[
\frac{dz}{dt} = 
\]
Transcribed Image Text:**Using the Chain Rule to Find \(\frac{dz}{dt}\):** Given the functions: \[ z = x^2y + xy^2, \quad x = 3 + t^2, \quad y = -2 + t^3 \] **First, calculate the partial derivatives:** \[ \frac{\partial z}{\partial x} = \] \[ \frac{\partial z}{\partial y} = \] **Next, find the derivatives with respect to \(t\):** \[ \frac{dx}{dt} = \] \[ \frac{dy}{dt} = \] **End result (express \(\frac{dz}{dt}\) solely in terms of \(t\)):** \[ \frac{dz}{dt} = \]
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