4 Chapter Review #44
(a) The mouse changes direction at approximately t=17, 23, 27. To determine this, we look at where the graph changes from positive to negative, or vice versa. At t=17, the graph goes from positive to negative, or in terms of position, the mouse changes from moving right to moving left. At t=23, the mouse changes from moving left to moving right, and at t=27, the mouse changes from moving right to moving left. (b) The mouse is moving most rapidly to the right where there is a maximum on the graph, so at approximately t=10. He is moving most rapidly to the left where there is a minimum, so around t=40. (c) From looking at the graph, we can determine when the mouse is farthest to the right of center where the area under the curve is the greatest positive value. In this case the greatest area under the curve is found on the interval (0,17), so the greatest area occurs when t=17. Therefore at t=17, the mouse is the farthest to the right of center. To find the time at which the mouse is farthest to the left of center, we must find where the greatest negative area under the curve occurs. Since this is found on the interval (27,40), the greatest area occurs at t=40, which means mouse is farthest to the left of center at t=40. (d) (10,17), (20,23), (24,27) ??? (e) To estimate where the mouse is at the center of the tunnel, we had to find the area under the curve to determine at what time the mouse's displacement was zero. We started by taking the left and right approximations at n=2.5 intervals. At each interval, we averaged the right and left value. We then added them together and multiplied our sum by our interval 2.5 to yield an of answer -138.75. We then worked backwards, adding the intervals back to find an estimate what values of t a displacement of 0 fell between. At t=35, the displacement was 1.25 and at t=37.5, it equaled -67.5. From this we could determine that was in this interval, but closer to 35. So we narrowed our interval in increments, by cutting each interval and each approximation in half, until we reached a very small interval of (35.078125, 35.0390625). Table of Values!
t (in seconds) 2.5 5 7.5 10 12.5 15 17.5 20 22.5 25
approximation 3.5 10 15.5 19 19 14 1.5 -13.5 -13.5 0
27.5 30 32.5 35 37.5 40
2 -10 -21 -26 -27.5 -28.5
If the displacement (area under the curve) at t=35 is 1.25, we needed a smaller interval to give us an area closer to 0. We did this by added smaller and smaller rectangles to 1.25 to narrow it down substantially. We took smaller increments, starting at t=35.625 which we determined to have an average approximation of -27.5. We then added 1.25 and .625(-27.5) to get a total area under the curve of -15.93. From this we repeatedly divided the number after the decimal by 2. So t=35.3125, 35.15625 and so on and did the same to -27.5 to yield -27.125, -27.0625, etc. We finally narrowed down our interval to 35.0390625<t<35.078125. At t=35.625~ -27.5 1.25 + (.625)(-27.5)~ -15.93 At t=35.3125~ -27.125 1.25 + (.3125)(-27.125)~ -7.22 At t=35.15625~ -27.0625 1.25 + (.15625)(-27.0625)~ -2.97 At t=35.078125~ -27.03125 1.25 + (.078125)( -27.03125)~ -.86182 At t=35.0390625~ -27.015625 1.25 + (.0390625)( -27.015625)~ .193
(getting closer...)
So, the area under the curve is zero somewhere on the interval of 35.0390625<t<35.078125.