MATH 5G/G2/G Test III Fall 26 Name: GTid (9: Instructor: Mitchel T. Keller Teaching Assistant and Section: There are 6 questions on this eam on pages (not counting this coverpage. Answer each question on a separate solution sheet (you may use more than one solution sheet per problem if needed. Be sure to eplain your answers, as answers that are not accompanied by eplanations/work may receive no credit. Place your name, section, and problem number on each solution sheet. Any solution sheet missing any of this information will not be graded. You are to complete this eam completely alone, without the aid of notes, tets, calculators, cellular telephones, personal digital assistants, or any other mechanical or digital calculating device. By signing on the line below, you agree to abide by the Georgia Tech Honor Code, the principles of which are embodied by the Challenge Statement: I commit to uphold the ideals of honor and integrity by refusing to betray the trust bestowed upon me as a member of the Georgia Tech community. Failure to sign this cover page will not be considered evidence of academic misconduct. However, if the cover page is not signed, three points will be deducted from your raw total score on this eam. Student signature: Question Points Score 5 2 5 5 5 5 5 6 5 Total:
. (5 points Find F ( if F ( = e 8 arctan( +π t ln t dt. Solution: We use the chain rule combined with the first part of the Fundamental Theorem of Calculus to compute this derivative after multiplying by in order to put the limit of integration that contains our variable as the upper limit. When we do so, we have F ( = (8 arctan( + π ln(8 arctan( + π 8 + ( + π 2 2. 2. (5 points Find / if y = y. Solution: We use logarithmic differentiation and implicit differentiation to solve this problem. We first take the logarithm of both sides, using absolute values to ensure that the logarithm is defined, and then differentiate. We have y = y y ln = ln y y + ln = y ( ln = ln y y y = ln y y ln. y + ln y. (5 points Find the area of the region bounded by the curves y = 2, y = /, y =, and = 2. Solution: The region is shown in the figure below. From this, we see that it is bounded above by y = 2 from to and then by y = / from to 2. Page of
.25.75 y =.5 y = 2.25.25.5.75.25.5.75 2 -.25 Thus, the area of the region is 2 + = + ln 2 = + ln 2 ln = + ln 2.. (5 points A particle moving along the -ais has acceleration a(t = 2t + units per second per second. At time t =, it is moving left at units per second. Find the average speed (not average velocity of the particle from time t = to time t =. Solution: Since we have a value for the velocity function and the acceleration function, we can find the velocity function by first finding the indefinite integral of the acceleration function. We have v(t = a(t dt = (2t + dt = t 2 + t + C. Since v( = (the particle is moving left at units per second at time t =, we have = v( = + + C = 2 + C C = 6, and therefore v(t = t 2 + t 6 = (t + (t 2. The average value of a continuous b function f on the interval [a, b] is f(t dt, so to find the average speed, we need b a a to integrate the speed function, which is v(t. Since we are only concerned with the speed on the interval [, ], we only need to determine v(t on that interval. Considering the way v factors, we have { (t + (2 t < 2; v(t = (t + (t 2 2. Page 2 of
Thus, the average speed on the interval [, ] is given by v(t dt = ( ( t 2 t + 6 dt + (t 2 + t 6 dt 2 ( [ = ] 2 [ ] t t2 t 2 + 6t + + t2 2 6t 2 = 8 2 ( + 2 + + 2 2 2 2 22 2 + 2 = ( 6 + + 6 = ( + 8 = ( + 6 = 5. 5. (5 points Compute ( ( π sin + 2. 2 + Solution: We use the linearity of the integral to compute this definite integral by evaluating two separate integrals. The first is sin ( π = 2 π sin(u du = 2 π 2 π π cos(u = 2 (cos(π cos( π = 2 π ( 2 = π. For the second integral, we use the substitution u = 2, and then du = 2. Thus, the integral is 2 + = du u 2 + = arctan(u = arctan( arctan( = arctan(. Therefore, the integral we were asked to compute is π + arctan(. 6. (5 points Find the indefinite integral + 2. Page of
Solution: We use the substitution u = + 2. Then du = and = u 2. Therefore, the integral becomes u 2 u 2 (u = + 2 du = du = / 2u / du u u/ = u7/ 7/ 2u/ / + C = 7 ( + 27/ 8 ( + 2/ + C. Page of