Position from velocity Consider an object moving along a line with the given velocity v and initial position a. Determine the position function, for t ≥ 0, using the antiderivative method b. Determine the position function, for t ≥ 0, using the Fundamental Theorem of Calculus ( Theorem 6.1 ). Check for agreement with the answer to part (a). 20. v ( t ) = 3 sin π t on [ 0 , 4 ] ; s ( 0 ) = 1
Position from velocity Consider an object moving along a line with the given velocity v and initial position a. Determine the position function, for t ≥ 0, using the antiderivative method b. Determine the position function, for t ≥ 0, using the Fundamental Theorem of Calculus ( Theorem 6.1 ). Check for agreement with the answer to part (a). 20. v ( t ) = 3 sin π t on [ 0 , 4 ] ; s ( 0 ) = 1
Solution Summary: The author explains the position function of the given velocity by anti-derivative method: s(t)=-3pi mathrmcos
Position from velocity Consider an object moving along a line with the given velocity v and initial position
a. Determine the position function, for t ≥ 0, using the antiderivative method
b. Determine the position function, for t ≥ 0, using the Fundamental Theorem of Calculus (Theorem 6.1). Check for agreement with the answer to part (a).
20.
v
(
t
)
=
3
sin
π
t
on
[
0
,
4
]
;
s
(
0
)
=
1
a
->
f(x) = f(x) = [x] show that whether f is continuous function or not(by using theorem)
Muslim_maths
Use Green's Theorem to evaluate F. dr, where
F = (√+4y, 2x + √√)
and C consists of the arc of the curve y = 4x - x² from (0,0) to (4,0) and the line segment from (4,0) to
(0,0).
Evaluate
F. dr where F(x, y, z) = (2yz cos(xyz), 2xzcos(xyz), 2xy cos(xyz)) and C is the line
π 1
1
segment starting at the point (8,
'
and ending at the point (3,
2
3'6
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.
Fundamental Theorem of Calculus 1 | Geometric Idea + Chain Rule Example; Author: Dr. Trefor Bazett;https://www.youtube.com/watch?v=hAfpl8jLFOs;License: Standard YouTube License, CC-BY