Prom pt You have been hired by a companyI that has recently.r developed a medication designed to reduce the size of benign tumors. Your role is to confirm that the medication does reduce the size of the tumor, given the rateofchange data. There are mam,r factors to consider, and the goal is to determine the total change in the size of the tumor. Using this data, can you confirm that there is a change in the size of the tumor? Table | t in days 0 5 10 15 20 25 30 35 40 45 rft) in mm 0 0.0105 0.02093 0.03134 0.04121 0.05204 0.06234 0.02261 0.08283 0.09303 per dayr Table II tin days 4 5 14 15 24 25 34 35 44 45 :11\") in mm 0.00839 0.0105 0.02926 0.03134 0.04998 0.05204 0.02056 0.02261 0.09099 0.09303 per day A. Calculating average change in the rate of change. Using the data in Table I, calculate the average change in the rate of change data for the following intervals: i. From t = 0 to t = 45 ii. From t = 25 to t = 45 iii. From t = 40 to t = 45 B. Calculating instantaneous change in the rate of change. 1. Using the data in Table II, calculate the instantaneous acceleration at the following intervals: i. t = 5 ii. t =15 iii. t = 25 iv. t = 35 v. t = 45 2. Explain how you used the limit definition of a derivative to calculate the instantaneous rate of change. Use your results to explain why the limit definition of a derivative is true. 3. At what point is the rate of change at a maximum? How is this relevant to the size of the tumor? For each of the questions above, provide supporting solutions and calculus terminology to explain how you arrived at your answers. Also, explain in detail what your answer represents in a real-world context and why this is useful.chang B. calculating instantaneous chang in the rested instantaneous change = Um8 (ath ) - > () h - go 9 ) at 625 instantaneous change = um (8(5) -8(us) - 0.0105 - (- 0-00839) 2-0. 00211 mm pel day ii ) at 2 15 instantaneous change ? lim 8 (15) - 8(14) h - 0. 03134 - (- 0. 02926) - lim Vicob 1 =-0: 0020 8 mm pes dawg ill ) at ( 225 instantaneous changes lim 8(25 ) - Ex(zu) 4-90 lim .0.05204 - (-0.04948) 1 =-6: 00206 mm Red day in) at 6235 instantaneous change - Cing 8(35) - 7(34) - lim - 0-07261- (-0-07056) = 0. 00 205 mm pel day de ) at t=US instantaneous change - lim 8 (us) -8 ( 44) *0:09303 - 0. 09099 - 0.00204 mm per dayIn order to calculate the instantaneous rate " a change of a function at aspecific point we use the limit definition as derivative. the limit delination a derivative is f' (a ) = Lim [fcath) - fla) h where fla) is the derivative as the function Ple). at 2za to calculate instantaneous rate as change t function " (f) at t=s, we used the limit definition of derivative as 8 (5 ) = Lim 8(5thi ) - 8(5) h at (= 5, 8 (t ) =-0.0105 8 ( 5 ) = ling 8 ( 5 th) - (-0.0105) = Gn 8 (5th ) +0.0105 - lim (8 ( 5th ) + 0. 0105 of h to the second term approche's to zero In -00 but it is multiplied sith small value so become zelo) 8 ( 5 ) = 8 (5 ) + 0 =-0. 00211 wim pel day propo.on toat t= 5, the rate of change of turnor is maximum. be cause initially the size as tumor is big so the medication will work effectively and more tumor will reduces. S