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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![The equation given is:
\[ x = 8 + 3t, \quad y = -8t \]
You are asked to find the derivative:
\[ \frac{dy}{dx} = \]
To solve for \(\frac{dy}{dx}\):
1. Find \(\frac{dx}{dt} = 3\).
2. Find \(\frac{dy}{dt} = -8\).
Use the chain rule to find \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{-8}{3}
\]
Therefore, \(\frac{dy}{dx} = \frac{-8}{3}\).
No graphs or diagrams are included in this image.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F941eb469-a862-4ae1-815d-a6698b94400a%2F1a444eb6-d9b3-4233-8bd2-f2176ac07434%2Fk7kst5y_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The equation given is:
\[ x = 8 + 3t, \quad y = -8t \]
You are asked to find the derivative:
\[ \frac{dy}{dx} = \]
To solve for \(\frac{dy}{dx}\):
1. Find \(\frac{dx}{dt} = 3\).
2. Find \(\frac{dy}{dt} = -8\).
Use the chain rule to find \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{-8}{3}
\]
Therefore, \(\frac{dy}{dx} = \frac{-8}{3}\).
No graphs or diagrams are included in this image.
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