A particle travels along the x-axis so that its velocity is given by v(t)=cos3x for 0<(or equal to)t<(or equal to)5. When t=0, the particle is at x=5.
a. What is the smallest x-coordinate of the particle?
b. What is the total distance traveled?
c. At what t is the particle moving right?
This is from my antiderivative chapter. Pleas help!
To solve this problem, we need to use the information given in the question and apply the concepts of velocity, displacement, and integrating the velocity function.
a) The smallest x-coordinate of the particle can be found by determining when the velocity function is zero. This occurs when cos(3x) = 0. Since the cosine function is zero at π/2, 3π/2, 5π/2, etc., we can find the values of x that satisfy these conditions.
π/2 = 3x => x = π/6
3π/2 = 3x => x = π/2
5π/2 = 3x => x = 5π/6
We have found two critical points - π/6 and 5π/6. To determine which one is the smallest x-coordinate, we need to substitute these values into the given velocity function and see which one yields the smallest result. The smaller value is the one we are looking for.
v(π/6) = cos(3π/6) = cos(π/2) = 0
v(5π/6) = cos(15π/6) = cos(5π/2) = 0
Both values are equal to zero, so both critical points give us the same x-coordinate. Therefore, the smallest x-coordinate of the particle is π/6.
b) The total distance traveled is given by finding the absolute value of the displacement. To find the displacement, we need to integrate the velocity function over the given time interval.
The displacement, Δx, can be obtained by evaluating the integral of the velocity function v(t) from t = 0 to t = 5:
Δx = ∫[0 to 5] v(t) dt = ∫[0 to 5] cos(3x) dx
Since we are given that when t = 0, x = 5, we need to express the bounds of the integral in terms of x.
When t = 0, x = 5; therefore, the initial x-coordinate is 5. To find the final x-coordinate, we need to determine x when t = 5:
t = 5 => x = 5 - ∫[0 to 5] cos(3x) dx
Now, we can evaluate the integral and find the displacement:
Δx = 5 - ∫[0 to 5] cos(3x) dx
To find this integral, we need to use the anti-derivative of the cosine function. The anti-derivative of cos(3x) is sin(3x)/3. We can now evaluate the integral:
Δx = 5 - [sin(3x)/3] evaluated from 0 to 5
Δx = 5 - [sin(15)/3] - [sin(0)/3]
Finally, we can calculate the total distance traveled by taking the absolute value of the displacement:
Distance traveled = |Δx|
c) To determine when the particle is moving right, we need to consider the sign of the velocity function v(t). If v(t) > 0, then the particle is moving right.
In this case, the velocity function is v(t) = cos(3x). We need to find when cos(3x) > 0. Since the cosine function is positive in the first and second quadrants, we can set up the following inequality:
cos(3x) > 0
To solve this inequality, we need to consider the critical points where cos(3x) changes sign. These points occur when 3x is equal to π/2, 3π/2, 5π/2, etc. We have already found these values in part a).
For x = π/6, cos(3x) = cos(π/2) = 0
For x = 5π/6, cos(3x) = cos(5π/2) = 0
Since both values are equal to zero, they are not solutions of the inequality. Therefore, there are no values of x for which the particle is moving right during the given time interval.