Unformatted text preview: SECTIONLZ MATHEMATICALMODELS El 13 14. (a) Using d in place of m and C in place of y, we ﬁnd the (c) slope to be 02—01 460—380V80 1 dg—di _ 800—480 _%_4
Soalinear equation isC—460= % (d—800) <=>
0—460=§d—200 e C=§d+260 (b) Letting d = 1500 we get 0 = i (1500) + 260 : 635. 50°
The cost of driving 1500 miles is $635. The slope of the line represents the cost per mile, $0.25.
(d) The y-intercept represents the ﬁxed cost, $260. (e) A linear function gives a suitable model in this situation because you have ﬁxed monthly costs such as insurance and car payments, as well as costs that increase as you drive, such as gasoline, oil, and tires, and the cost of these
for each additional mile driven is a constant. 15. (a) The data appear to be periodic and a sine or cosine function would make the best model. A model of the form
f (at) = a cos(bm) + c seems appropriate. (b) The data appear to be decreasing in a linear fashion. A model of the form f (as) = mx + b seems appropriate.
16. (a) The data appear to be increasing exponentially. A model of the form f (x) = a - bf or f (x) = (1 ~ b“ + c seems
(b) The data appear to be decreasing similarly to the values of the reciprocal function. A model of the form f (2:) : a/x seems appropriate. Some values are given to many decimal places. These are the results given by several computer algebra systems—rounding is left to the reader.
17. (a) 15 (b) Using the points (4000, 14.1) and (60,000, 8.2), we obtain
82 — 14.1
— 14.1 = E — '
y 60,000 _ 4000 (x 4000) or, equlvalently,
y m —0.000105357$ + 14.521429.
0 61,000 A linear model does seem appropriate. 61,000 ...
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- Spring '10