Consider the curve parametrized by x = t4 + 1 and y = tº + 4t5. a) Find dy dx da in terms of t. 3/2t^2+5t Correct. Good Job! d'y b) Find in terms of t. dx² d²y/dx² = (3t+5)/(4t^3) Correct. Good Job! 33 Preview = 1² + 5t d²y/dx² = Preview 3t+5 4t3 c) For what intervals of this the curve concave up? Enter your answer as intervals separated by a comma.

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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Consider the curve parametrized by x = t4 + 1 and y = t6 + 4t5.
dy
a) Find
da
dx
3/2t^2+5t
in terms of t.
Correct. Good Job!
d²y
b) Find in terms of t.
dx²
d²y/dx²
-
(3t+5)/(4t^3)
Correct. Good Job!
dy
dx
Preview
1²+5t
d²y/dx²
=
Preview
3t+5
4t3
c) For what intervals of t is the curve concave up? Enter your answer as intervals
separated by a comma.
Transcribed Image Text:Consider the curve parametrized by x = t4 + 1 and y = t6 + 4t5. dy a) Find da dx 3/2t^2+5t in terms of t. Correct. Good Job! d²y b) Find in terms of t. dx² d²y/dx² - (3t+5)/(4t^3) Correct. Good Job! dy dx Preview 1²+5t d²y/dx² = Preview 3t+5 4t3 c) For what intervals of t is the curve concave up? Enter your answer as intervals separated by a comma.
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