AP Calculus AB hard particle motion question?

Question

particle X moves along the positive x-axis so that its position at time t≥0 is given by x(t)=5t³-9t²+7.

(A) Is particle X moving toward the right or toward the left at time t=1? Give a reason for your answer.
(B) At what time t≥0 is particle X farthest from the left? Justify your answer.
(C) A second particle, Y, moves along the positive y-axis so that its position at time t is given by y(t) = 7t +3. At any time t, t≥0, the origin and the positions of the particles X and Y are the vertices of a triangle in the first quadrant. Find the rate of change of the area of the triangle at time t=1. Show the work that leads to your answer.

Answer 1

.



(A)

x(t) = 5t³ - 9t² + 7

x’(t) = 15t² - 18t

===>at t = 1; x’(1) = 15*1² - 18*1 = -3

The negative speed at t = 1 implies the particle is moving to the left.



(B)

x’(t) = 15t² - 18t

critical points

15t² - 18t = 0

3t(5t - 6) = 0

t = 0, t = 1.2



intervals: 0 < t < 1.2 and t > 1.2

===> test any point on 0 < t < 1.2; Use t = 1; x’(1) = -3

===> test any point on t > 1.2; Use t = 2; x’(2) = 24

∴ t = 1.2 gives a minimum position that is farthest to the left



(C)

A second particle, Y, moves along the positive y-axis so that its position at time t is given by y(t) = 7t +3. At any time t, t≥0, the origin and the positions of the particles X and Y are the vertices of a triangle in the first quadrant. Find the rate of change of the area of the triangle at time t=1. Show the work that leads to your answer.



A = ½ (5t³ - 9t² + 7)(7t + 3)

A = 17.5t⁴ - 24t³ - 13.5t² + 24.5t + 10.5

A’(t) = 70t³ - 72t² - 27t + 24.5

A’(1) = -4.5

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Answer 2

A graphing calculator can provide easy answers.



a) At t=1, the position curve has a downward slope. This means the particle is moving toward the left.



b) On the interval t = [0, 1.8), the particle is farthest from the origin at t=0. After t=1.8, the particle gets farther and farther to the right of the origin. The problem statement does not define the domain of t, so we must conclude ...

.. the particle is farthest from the left when t approaches infinity.



c) The area (A) and its rate of change (R) are shown on the graph. At t=1, R=-4.5.

.. The area is changing at the rate of -4.5 square units per time unit at t=1.



_____

Click on any curve on the graph to see its intercepts and critical points. Click on those to see their coordinates.

Answer 3

A) x'(t) = 15t^2-18t

x'(1) = -18

Particle is moving left

B) 15t^2-18t = 0

3t(5t-6) = 0

t = 0 (max) or 6/5 (min)

C) A = Area of triangle = (1/2)(5t^3-9t^2+7)(7t+3)

A'(t) = (1/2)[(7t+3)(15t^2-18t)+7(5t^3-9t^2+7)]

A'(1) = (1/2)[-30+7*3]

= -9/2

Answer 4

dx/dt = 15 t² - 18 t....(A) to the left ; (B) t = 0 has d²x / dt² < 0 ===> max x(t) ;



( C ) ; A = (1/2) xy ===> dA/dt = (1/2) { y dx/dt + x dy/dt}...evaluate when t = 1

Answer 5

b