Question

Revlew the Explore It, then use it to complete the exercise below, Explore It EXPLORE & TEST CONCEPT EXAMPLE VIDEO WHEN WOULDI USE THIS YOU WILL LEARN ABOUT original function is not known. used to estimate the rate of change and the derivative when How rise over run can i Zacarias da Matalus fotolia.com In real world applications, measurements of certain quantisies such a are taken at specific points in time. The rate of change of these quantities can be estimated using the rise over run formula for slope. This may also be used to estimate the derivative of a function fthat is not position, temperature, number of individuals, and concentrations fiziAz)-fie slope derivative f' () lim r)-Me) Click here to access the Explore It in a new window. Select Function 3, N(t) = 100,000e--10)-/8, under the Explore Test section of the Explore It. The secant line can be used approximate the derlvative at a particular polnt on the graph by placing either the right or the left side of the secant line on that point. (Round your answer to the nearest integer.) Consider the point o ap dtt 10 dae approximatk line at this point. Set the other end of the secant line at the point that is 1 day earler. What Is the graph at I lope using t cases/day (b) Which of the following best describes the meaning of this slope part (a)? The number of cases is growing between days 9 and 10. The number of cases is declining between days 9 and 10. The number of cases is steady between days 9 and 10. There are no new cases between days 9 and 10. (c) Consider again the point on the graph at t 10 days. Set one end of the secant line at t 10 days and the other end at the point that is 1 day later. What is the approximation the slope of the graph at t 10 days using this secant line? cases/day (d) Which of the following best describes the meaning of this slope part (c)? The number of cases is growing between days 10 and 11. The number of cases is declining between days 10 and 11. The number of cases is steady between days 10 and 11. There are no new cases between days 10 and 11. is also possible to approximate the slope of a graph using points day 11. What is the approximation of the slope of the graph e) the right and the left of the position of interest. Set one end of the secant line to day 9 and the other end day using this secant line? cases/day r) Which of the following best describes the meaning of this slope part (e)? The number of cases is growing between days 9 and 11 The number of cases is declining between days 9 and 11. The number of cases Is steady between days 9 and 11. There are no new cases between days 9 and 11. (g) Set the t-coordinate of the fixed point to 10. Drag the adjustable point toward the fixed point. Use the zoom feature to see if the secant line overlaps the tangent line. Using the slope of the tangent line, what is the actual growth rate of the disease on day 10? cases/day

Answer 1

(a) slope =

f(x) f(x AT) x- (x - Ar)

in this case x= 10 and Δx=1 so

slope=

(10-10) 1000000e -(S -1000000e9-10)

         = cases/day

1000000(1 e) = 117503.0974

(b) Since slope is positive so the number of cases at t=10 is more than at t=9 days. Therefore, the number of cases are growing between 9 and 10 days. So first option is correct.

(c)       slope =

Γα+ Δυ)-) f ί+ Δα) π+Δr-

=
(10-10)2 (11-10) - 1000000e 1000000e
          =
1000000(e 1) 117503.0974
cases/day

(d) Since slope is negative so the number of cases at t=10 is more than at t=11 days. Therefore, the number of cases are decreasing between 10 and 11 days. So second option is correct.

(e) slope =

f(xAr) - f(x - Ar) Ar-+ Ax

               =

(11-10) 1000000e -(S 1000000e9-10

               = 0 cases/day

(f)   Since slope is zero so the number of cases should be constant i.e. the number of cases at t=9 is same as that at t=11 days so option third is correct.

(g) Actual rate =

d 1000000-10)2 1000000(-2(r- 10)/8 )e-(r-10)2/8

at x=10 its value is
|1000000(-2(10 10)/8)e-(10-10)/8
= cases/day