course Mth 173
quiz #1 Version 1
If y = .015t^2 - 1.7t + 93. You want to find the average rate of depth change between t = 13.9 and t = 27.8. You first have to find the depth at these two points.
So the depth when t = 13.9 is .015(13.9)^2 - 1.7(13.5) + 93 = 72.26815.
The depth when t = 27.8 is .015(27.8)^2 - 1.7 (27.8) + 93 = 57.3326.
So you would have the following two order pairs (13.9, 72.26815) (27.8, 57.3326).
Now the average rate of depth change between these two points is -14.93555 / 13.9 = -.93.
Right calculation but it looks like you divided backwards. This should come out identical to the -1.0745 you get below; it does come our -1.07; you can check the rest of the significant figures.
The function that represents rate at clock time is r = 2at + b. The clock time that is halfway between t = 13.9 and t = 27.8 is t = 20.85. The rate at t = 20.85 is r = 2(.015)(20.85) + -1.7 = .6255 + -1.7 = -1.0745.
When r(t) = .193t + - 2.1, in order to find the depth change between t = 13.9 and t = 27.8. You must first find the rate a both points.
So when t = 13.9 the rate is .193(13.9) + -2.1 = .5827 cm / s.
When t = 27.8 the rate is .193(27.8) + -2.1 = 3.2654 cm / s.
Then you find the average between them two rates which is (.5827 + 3.2645) / 2 = 1.92405. In order to find the depth change you multiply the average change of rates by the time that has passed between the two points. Then depth change is 1.92405 * 13.9 = 26.74 cm. The depth function is y = .0965t^2 + -2.1. If t = 0 and the depth is 130. The depth function would be y = a(0)^2 + b(0) + 130.
Excellent. See my note and let me know if you have questions.