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Consider the curve parametrized by x = t +1 and y = t6 + 4t. dy in terms of t. d.x a) Find Preview 1/2(3t^2+10t) 2 - (312 + 10£) Correct. Good Job! dy in terms of t. da? b) Find Preview dy/da = 4t (3t + 5) dy dr? 4t^3(3t+5)

Consider the curve parametrized by x = t +1 and y = t6 + 4t. dy in terms of t. d.x a) Find Preview 1/2(3t^2+10t) 2 - (312 + 10£) Correct. Good Job! dy in terms of t. da? b) Find Preview dy/da = 4t (3t + 5) dy dr? 4t^3(3t+5)

Calculus: Early Transcendentals
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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Part b
Consider the curve parametrized by x = a) Find dy dx in terms of t. 1/2(3t^2+10t) Correct. Good Job! b) Find ď²y dx² dy/da²4t^3(3t+5) in terms of t. t4 + 1 and y = t6 + 4t5. Preview (31²+10t) Preview d²y/dr² = 4t³ (3t+5)
Transcribed Image Text:Consider the curve parametrized by x = a) Find dy dx in terms of t. 1/2(3t^2+10t) Correct. Good Job! b) Find ď²y dx² dy/da²4t^3(3t+5) in terms of t. t4 + 1 and y = t6 + 4t5. Preview (31²+10t) Preview d²y/dr² = 4t³ (3t+5)
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