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In the previous step we found the following just before using a direct substitution to find the velocity at a particular time. (16(a + h) — 1.86(a + h)²) – (16a – 1.86a²) h v(a) = lim h→0 To find the velocity at a general time t = a, we need to first simplify as much as possible. (16(a + h) — 1.86(a + h)²) – (16a 1.86a²) h 16a+ 16h 1.86 a² + v(a) = lim h→0 = lim h→0 = lim h→0 16a + 16h 1.86a² 16h- = lim h→0 = lim (16-( h h h ah - 1.86h2 a 1.86h ]× )ah + h²) - - 16a + 1.86a² ah- 1.86h2 16a + 1.86a²

In the previous step we found the following just before using a direct substitution to find the velocity at a particular time. (16(a + h) — 1.86(a + h)²) – (16a – 1.86a²) h v(a) = lim h→0 To find the velocity at a general time t = a, we need to first simplify as much as possible. (16(a + h) — 1.86(a + h)²) – (16a 1.86a²) h 16a+ 16h 1.86 a² + v(a) = lim h→0 = lim h→0 = lim h→0 16a + 16h 1.86a² 16h- = lim h→0 = lim (16-( h h h ah - 1.86h2 a 1.86h ]× )ah + h²) - - 16a + 1.86a² ah- 1.86h2 16a + 1.86a²

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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In the previous step we found the following just before using a direct substitution to find the velocity at a particular time. 16(a + h) 1.86(a + h)²) – (16a – 1.86a²) h To find the velocity at a general time t = a, we need to first simplify as much as possible. (16(a + h) — 1.86(a + h)²) – (16a – 1.86a²) v(a) = v(a) = lim h→0 lim h→0 = lim h→0 = lim h→0 = lim h 16a + 16h - 1.86 a² + 16a + 16h - 1.86a² - 16h- h→0 = lim (16 - ( h h h ah - 1.86h² a - 1.86h × )ah + h²) - - 16a + 1.86a² ah - 1.86h² - 16a + 1.86a²
Transcribed Image Text:In the previous step we found the following just before using a direct substitution to find the velocity at a particular time. 16(a + h) 1.86(a + h)²) – (16a – 1.86a²) h To find the velocity at a general time t = a, we need to first simplify as much as possible. (16(a + h) — 1.86(a + h)²) – (16a – 1.86a²) v(a) = v(a) = lim h→0 lim h→0 = lim h→0 = lim h→0 = lim h 16a + 16h - 1.86 a² + 16a + 16h - 1.86a² - 16h- h→0 = lim (16 - ( h h h ah - 1.86h² a - 1.86h × )ah + h²) - - 16a + 1.86a² ah - 1.86h² - 16a + 1.86a²
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