#higher derivatives

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Higher Derivatives and Motion

Since the derivative of f(x) is itself a function, f'(x).   We can also take its derivative, f''(x) or d²y/dx². This function is called the second derivative of f(x).

Find the second derivative: f(x) = x³ f(x) = x³ f'(x) = 3x² f''(x) = 6x

Rectilinear Motion The position of a particle moving on a line is given by the equation: y(x) = 2x³ - 21x² + 60x where x ≥ 0, y(x) is measured in meters and x is measured in seconds. 1. Find the velocity and acceleration in factored form. 2. Construct sign diagrams with descriptors for all three. 3. Create a trajectory describing the motion of the particle. 4. Describe the acceleration of the particle in interval notation. 5. Find the velocity after 3 seconds. 6. Find when the particle is at rest. 7. Find the total distance traveled during the first 6 seconds. Question One The reason we factor each function is to determine the zeros, or x-intercepts. First, we factor the original function. y(x) = 2x³ - 21x² + 60x y(x) = x(2x² - 21x + 60) The velocity is the first derivative of the motion function. v(x) = 6x² - 42x + 60 v(x) = 6(x² - 7x + 10) v(x) = 6(x - 5)(x - 2) The acceleration is the second derivative of the motion function. a(x) = 12x - 42 a(x) = 6(2x - 7) Question Two Now we must create sign diagrams and put the zeros on the number line, where x = seconds. Then we indicate what is positive or negative by plugging in a number between each zero to each function. *Note: Since all factored terms have an odd exponent, the positive and negative signs will fluctuate. *Note: Relate acceleration with its force. y(x) = x(2x² - 21x + 60)

v(x) = 6(x - 5)(x - 2)

a(x) = 6(2x - 7)

Question Three Imagine the trajectory line as the y-axis and the line's movement is the graph of y(x). Things to keep in mind: *y starts at zero, and it is the only zero for y(x), therefore the motion of the particle will never return to the origin. *To know where the point in time when the particle changes directions: plug in the velocity's zeros into the motion function. *At x = 2 seconds, y = 52 meters, at x = 5 seconds, y = 25 meters.

Question Four We describe acceleration with its force, to determine when the particle is slowing down or speeding up. The force changes direction at acceleration's zero, which is 3.5. So, when x < 3.5, the force is moving left (negative), therefore the particle is slowing down during from 0 seconds to 2 seconds, due to the particle changing direction at 2 seconds. When the particle changes direction at x = 2, the particle is moving left, and since 2 < 3.5, the force is going with the particle, speeding the particle up. However, when x > 3.5, the force is going right (positive), so while the particle is moving left until x = 5 seconds, the force is moving right, therefore slowing the particle down. When x = 5, the particle starts moving right, the same direction as its force, therefore it speeds up to infinity. We can describe this relationship in interval notation:

Question Five Use the velocity's function to solve this problem. v(x) = 6x² - 42x + 60 v(x = 3) = 6x² - 42x + 60 v(3) = 6(3)² - 42(3) + 60 v(3) = -12 m/s Question Six The particle is at rest when the velocity is equal to zero, or the zeros we plotted on the velocity's sign diagram. Therefore, the particle is at rest when x = 2 and 5. Question Seven We first find the distance from 0 to 2 seconds. |y(0) - y(2)| = |52 - 0| = 52 m Next, we find the distance from 2 to 5 seconds. |y(2) - y(5)| = |25 - 52| = 27 m Now we subtract this distance with the distance of the particle at 6 seconds. |y(6) - y(5)| = |36 - 25| = 11 m Finally, we add all three distances together to get the total distance until 6 seconds. 52 + 27 + 11 = 90 m