Given the equation z=f(x, y) = ln(x² + 2y² +1) with the change of variables x=3-1² and y = 4t-1³, use the chain rule to find the derivative at t = 1: dz dt Now find the equation of the tangent line at t = 1, then graph: (hint: use velocity vector)
Given the equation z=f(x, y) = ln(x² + 2y² +1) with the change of variables x=3-1² and y = 4t-1³, use the chain rule to find the derivative at t = 1: dz dt Now find the equation of the tangent line at t = 1, then graph: (hint: use velocity vector)
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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![**Mathematics Education: Calculus - Chain Rule and Tangent Lines**
**Topic: Calculating Derivatives Using the Chain Rule**
Given the equation \( z = f(x, y) = \ln(x^2 + 2y^2 + 1) \) with the change of variables \( x = 3 - t^2 \) and \( y = 4t - t^3 \), use the chain rule to find the derivative \( \frac{dz}{dt} \) at \( t = 1 \).
\[ \frac{dz}{dt} = \]
**Solution Steps:**
1. Begin by expressing \( z \) in terms of \( t \) using the given transformations for \( x \) and \( y \).
2. Apply the chain rule:
\[ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \]
3. Determine the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).
4. Compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
5. Evaluate all the derivatives at \( t=1 \).
**Equation of the Tangent Line:**
Now find the equation of the tangent line at \( t = 1 \), then graph (hint: use the velocity vector).
---
**Graphical Representation:**
The image below shows a 3D graph of the function \( z = \ln(x^2 + 2y^2 + 1) \).
- The graph is plotted in a 3D coordinate system with axes labeled \( x \), \( y \), and \( z \).
- The surface appears to be a smooth curve rising toward the center.
- A tangent line or a vector could be visualized on the surface, indicating the slope at a specific point.
---
**Further Insight:**
The problem highlights the application of the chain rule in multivariable calculus to find the rate of change of a composite function. Additionally, it integrates the concept of tangent lines in three-dimensional space, aiding in visualizing how derivatives describe the behavior of functions locally.
For interactive visualizations and more examples, visit our calculus section](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3d2f6440-8765-40b0-9188-246cb7d5292a%2F5ab2a319-06d5-406a-9ffd-0e391778d9bc%2F2bamrhn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Mathematics Education: Calculus - Chain Rule and Tangent Lines**
**Topic: Calculating Derivatives Using the Chain Rule**
Given the equation \( z = f(x, y) = \ln(x^2 + 2y^2 + 1) \) with the change of variables \( x = 3 - t^2 \) and \( y = 4t - t^3 \), use the chain rule to find the derivative \( \frac{dz}{dt} \) at \( t = 1 \).
\[ \frac{dz}{dt} = \]
**Solution Steps:**
1. Begin by expressing \( z \) in terms of \( t \) using the given transformations for \( x \) and \( y \).
2. Apply the chain rule:
\[ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \]
3. Determine the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).
4. Compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
5. Evaluate all the derivatives at \( t=1 \).
**Equation of the Tangent Line:**
Now find the equation of the tangent line at \( t = 1 \), then graph (hint: use the velocity vector).
---
**Graphical Representation:**
The image below shows a 3D graph of the function \( z = \ln(x^2 + 2y^2 + 1) \).
- The graph is plotted in a 3D coordinate system with axes labeled \( x \), \( y \), and \( z \).
- The surface appears to be a smooth curve rising toward the center.
- A tangent line or a vector could be visualized on the surface, indicating the slope at a specific point.
---
**Further Insight:**
The problem highlights the application of the chain rule in multivariable calculus to find the rate of change of a composite function. Additionally, it integrates the concept of tangent lines in three-dimensional space, aiding in visualizing how derivatives describe the behavior of functions locally.
For interactive visualizations and more examples, visit our calculus section
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