Question

2 1/47 LIT/rz-rol Answer: C) ASA = 41(r2 + rîhar Ass=HTTC ritral Ar r-ri = 411rzt) b. The average rate of change in the surface area of a spherical balloon is given by the equation A E 41(r2 +r1 If the radius of the balloon is increasing at 0.25 inches per second, what is the average rate of change in surface are when the radius increases from 3 to 3.25 inches? Express your answer in terms of it. Answer: 6.251 square inches per second c. A spherical balloon is being inflated. The radius of the balloon is increasing at 0.25 inches per second. The rate of change of the surface area is given by the equation = 470(1; +r) Complete the table below. As the radius increases, is ASA increasing or decreasing? -41 Tir2 2 3 3 4 ASA . Answer: ASA is increasing as r increases iz ASA

Answer 1

Solution @ ASA = 4x (6+ri) At Ar = 0.25 inch/ sec. At ASA when Ar= (3.25-3) inch 1 = 0.25 inch We need to find .. At in our case is l. ASA = 4x (3.25 + 3 ) x 0.25 46.25 0.25 = 6.25 n. 1. (c) Ar= 0.25 inch / sec At li = ha (re+r,) ar ДЕ st (i) for r, = 2,02= 3 - Dr= r2-0,=1 & giver Ar = 0.25 inch/ sec At 28 At = 1x1004 = 4 sec Putting the values in (1) ASA = 4(2+3). = 205 (ii) for rai=3 , 12 = 4 Are Yo-r, = 1 ... Putting the values in (1) ASA = 4* (3+4).|| = 285 OSA | Increa] So as radius is increasing Ć I SA is also increasing s do 200 285