i did this problem and it isn't working out, so i think i'm either making a mistake or misunderstanding what it's asking.
A particle moves along the x axis so that its velocity at any time t greater than or equal to 0 is given by v(t) = 1  sin(2pi*t).
a. Acceleration a(t) of the particle at any time t?
i said that acceleration was the derivative of velocity, so i took the derivative of the velocity function.
v(t) = 1  sin(2pi*t)
v'(t) = cos(2pi*t) x 2pi
a(t) = 2pi*cos(2pi*t)
b. Values of t between 0 and 2 where the particle is at rest?
i said it was at rest when velocity = 0.
0 = 1  sin(2pi*t)
1 = sin(2pi*t)
sin(pi/2) = 1, so...
2pi*t = pi/2
4pi*t = pi
4t = 1
t = 1/4
c. Position x(t) of the particle at any time t if x(0) = 0?
i didn't understand the part about x(0) = 0. i kind of ignored it and it didn't work out. i figured that the position function was the antiderivative of the velocity function and did:
antiderivative of 1  sin(2pi*t)
1/2pi x antiderivative of 1  sin(2pi*t)
1/2pi (t + cos(2pi*t)) = x(t)
but that doesn't equal 0 when t = 0. help?
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3 answers

in b) you were right to find
t = 1/4
but remember the period of 1  sin(2pi*t)
is 2pi/2pi = 1
so another solution would be 1/4 + 1 or
t = 5/4
so the times between 0 and 2 when the object is at rest is
t = 1/4 and t = 5/4
for c) if
v(t) = 1  sin(2pi*t)
then x(t) = t + (1/2pi)cos(2pi*t) + c
but you were given x(0) = 0
0 = 0 + (1/2pi)cos(0) + c
0 = )1/2pi) + c
c = 1/(2pi)
so x(t) = t + (1/2pi)cos(2pi*t)  1/(2pi) 👍
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answered by Reiny 
omg i can't believe i just forgot c in part c. it's even called part c!
and thank you for the help with part b. i completely forgot about that! 👍
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answered by jane 
Let R be the region bounded by the yaxis and the graph of y=xcubed divided by 1+xsquared and y=42x,as shown inthe figure above.
find the area of R
find the volume of the solid generated when R is revolved about the xaxis 👍
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answered by Anonymous
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