i did this problem and it isn't working out, so i think i'm either making a dumb 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 anti-derivative of the velocity function and did:

anti-derivative of 1 - sin(2pi*t)
1/2pi x anti-derivative of 1 - sin(2pi*t)
1/2pi (t + cos(2pi*t)) = x(t)

but that doesn't equal 0 when t = 0. help?

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!

Let R be the region bounded by the y-axis and the graph of y=xcubed divided by 1+xsquared and y=4-2x,as shown inthe figure above.

find the area of R

find the volume of the solid generated when R is revolved about the x-axis

Let's go through each part of the problem step by step:

a. To find the acceleration of the particle at any time t, you correctly used the fact that acceleration is the derivative of velocity. The derivative of the velocity function v(t) = 1 - sin(2πt) is v'(t) = -cos(2πt) * 2π. Therefore, the acceleration function a(t) is a(t) = -2πcos(2πt).

b. To find the values of t between 0 and 2 where the particle is at rest (meaning its velocity is zero), you set the velocity function equal to zero: 0 = 1 - sin(2πt). Then you solved for t by taking the inverse sine of both sides: sin(2πt) = 1. This is where the misunderstanding occurred. Instead of using sin(π/2) = 1, you should have used sin(0) = 0 since it's asking for where the particle is at rest.

Setting sin(2πt) = 0, you have 2πt = 0. Solving for t, you get t = 0. So the particle is at rest at t = 0.

c. Now let's move on to finding the position function x(t) of the particle at any time t, given that x(0) = 0. The position function is the antiderivative (indefinite integral) of the velocity function. So you were on the right track with finding the antiderivative, but you missed a constant term in the antiderivative which accounts for the initial condition x(0) = 0.

Starting with the velocity function v(t) = 1 - sin(2πt), integrate it with respect to t:

∫ v(t) dt = ∫ (1 - sin(2πt)) dt

The integral of 1 with respect to t is t. The integral of sin(2πt) with respect to t is -1/(2π)cos(2πt). So the antiderivative of v(t) is:

x(t) = t - (1/(2π)cos(2πt)) + C

Now, to find the constant term C, we use the initial condition x(0) = 0. Plug in t = 0 and x(t) = 0 into the equation:

0 = 0 - (1/(2π)cos(0)) + C

Simplifying, 0 = 0 + (1/(2π)) + C. Subtracting (1/(2π)) from both sides, we find C = -(1/(2π)).

Therefore, the position function x(t) is:

x(t) = t - (1/(2π)cos(2πt)) - (1/(2π))

Make sure to check your work by evaluating x(0) to see if it equals 0. When t = 0, we have x(0) = 0 - (1/(2π)cos(2π*0)) - (1/(2π)) = 0 - (1/(2π)*1) - (1/(2π)) = -(2/(2π)) = -1/π, which is not equal to 0. So the position function x(t) you found is not correct.

To correctly account for the initial condition x(0) = 0, the position function should be:

x(t) = t - (1/(2π)cos(2πt))

I hope this explanation helps you understand the problem and the steps involved in solving it. If you have any further questions, feel free to ask!

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)