Homework 2: Problem 12 O (1 point) A jogger runs around a circular track of radius 60 ft. Let (x, y) be her coordinates, where the origin is the center of the track. When the jogger's coordinates are (36, 48), her x-coordinate is changing at a rate of 14 + ft/s. Find dy / dt. dy/dt = ft / sHomework 2: Problem 13 O (1 point) An inverted cylindrical cone, 52 ft deep and 26 ft across at the top, is being filled with water at a rate of 11 fts/min. At what rate is the water rising in the tank when the depth of the water is: + 1 foot? Answer= 10 feet? Answer= 51 feet? Answer=Homework 2: Problem 14 (1 point) The altitude of a triangle is increasing at a rate of 1 centimeters/minute while the area of the triangle is increasing at a rate of 2.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 11.5 centimeters and the area is 81 square centimeters? centimeters Note: Just enter a value but not the units.Homework 2: Problem 13 (1 point) An inverted cylindrical cone, 52 ft deep and 26 ft across at the top, is being filled with water at a rate of 11 fts/ min. At what rate is the water rising in the tank when the depth of the water is: 1 foot? Answer= 14.01 1 10 feet? Answer= 0.1400 51 feet? Answer= 0.00538 Note: You can earn partial credit on this problem. Preview My Answers Submit Answers You have attempted this problem 2 times. Your overall recorded score is 0%. You have 3 attempts remaining.Homework 2: Problem 14 (1 point) The altitude of a triangle is increasing at a rate of 1 centimeters/minute while the area of the triangle is increasing at a rate of 2.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 11.5 centimeters and the area is 81 square centimeters? -13.66 centimeters Note: Just enter a value but not the units. Preview My Answers Submit Answers You have attempted this problem 1 time. Your overall recorded score is 0%. You have 4 attempts remaining.Problem .13 11 = TT hah dt h = 52ft 11 x 16 - TT dh d = 26 11 7 X 16 = 28 du h2 7 so radius = 13 13 1 X 16 x7 = dh 52 22 62 = 4 16 x7 = ah 2 x 62 = 4 56 = dh at Volume of Cone= 1 Toh When 1 foot adh = 56 V = 3 = 56ft/Bin When 10 feet, ah = 56 he df 100 3 x 16 0. 56 f+/min = Th3. db x all + 1x 48 dv - T x3h .gh when 51 feet, db = 56. de 512 48 dt = 0.02153 76/min dv al - d 16 di- given that dv = 11\f