Find the equation for the line tangent to the parametric curve: at the points where t = 4 and t = -4. For t = 4, the tangent line (in form y = mx + b) is y For t = -4, the tangent line is x = t³ - 16t Y 16t² - 14 =
Find the equation for the line tangent to the parametric curve: at the points where t = 4 and t = -4. For t = 4, the tangent line (in form y = mx + b) is y For t = -4, the tangent line is x = t³ - 16t Y 16t² - 14 =
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter6: Applications Of The Derivative
Section6.CR: Chapter 6 Review
Problem 31CR
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Question
![Find the equation for the line tangent to the parametric curve:
at the points where t
For t = 4, the tangent line (in form y = mx + b) is
0
y = =
For t
y =
=
-
= 4 and t = -4.
-4, the tangent line is
0
X
Y
=
=
t³ - 16t
16t²t4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feffa94ab-9dff-49df-b80f-e51acf479d61%2Fe4fec96b-88fe-4527-9182-ad4b98817f08%2F63kqo2s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Find the equation for the line tangent to the parametric curve:
at the points where t
For t = 4, the tangent line (in form y = mx + b) is
0
y = =
For t
y =
=
-
= 4 and t = -4.
-4, the tangent line is
0
X
Y
=
=
t³ - 16t
16t²t4
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