Tiêu đề MSTE 17 Solutions.pdf ✅

17 Differential Calculus: Related Rates Solutions ▣ 1. The radius of an expanding sphere changes at the rate of 2 cm per minute. How fast is the surface area of the sphere changing when the radius is 25 cm, in cm^2/min? [SOLUTION] A = 4πr 2 dA dt = 8πr dr dt From the given, r = 25 cm, dr dt = 2 cm minute dA dt = 8π(25 cm) (2 cm minute ) dA dt = 1256.637 cm2 minute ▣ 2. A balloon is rising vertically over a point A on the ground at the rate of 15 ft/s. A point B is on the ground level 30 ft from A. When the balloon is 40 ft from A, at what rate is its distance from B changing? [SOLUTION] • A. Differentiation L 2 = x 2 + 302 2L dL dt = 2x dx dt L dL dt = x dx dt At x = 40, solve for L. L 2 = 402 + 302 L = 50 ft
(50 ft) dL dt = (40 ft) (15 ft s ) dL dt = 12 ft/s • B. Vectors At x = 40, tan θ = 30 40 = 3 4 cos θ = 4 5 = dL dt dx dt dL dt = 4 5 ( dx dt) dL dt = ( 4 5 ) (15 ft s ) = 12 ft s ▣ 3. Two railroad tracks are perpendicular to each other. At 12 PM there is a train at each track approaching the crossing at 50 kph, one being 100 km and the other being 150 km from the crossing. How fast in kph is the distance between the two trains changing at 4 pm? [SOLUTION]
At any time t, x = 150 − 50t y = 100 − 50t L 2 = (150 − 50t) 2 + (100 − 50t) 2 L 2 = 5000t 2 − 25000t + 32500 At t = 4, L = 50√5. 2L dL dt = 10000t − 25000 dL dt = 10000(4) − 25000 2 × 50√5 dL dt = 67.08 kph ▣ 4. A man walks across a bridge at a rate of 5 ft/s as a boat directly beneath him passes at 10 ft/s. If the bridge is 10 ft above the boat, how fast are the man and the boat separating 1 second later? [SOLUTION] L 2 = x 2 + y 2 + 102 After 1 second, x = (5 ft s ) (1 s) = 5 ft y = (10 ft s ) (1 s) = 10 ft L = √5 2 + 102 + 102 L = 15 ft By time derivatives, 2L dL dt = 2x dx dt + 2y dy dt L dL dt = x dx dt + y dy dt 15 ( dL dt) = (5)(5) + (10)(10) dL dt = 8.333 ft/s ▣ 5. A rectangular trough is 10 ft long and 3 ft wide. Find how fast the surface rises, if water flows in at the rate of 12 ft 3/min. [SOLUTION] • A. Time derivatives
V = (10 ft)(3 ft)h V = 30h dV dt = 30 dh dt 12 ft 3 minute = 30 ( dh dt) dh dt = 0.4 ft minute • B. Discharge formula Q = AV 12 ft 3 minute = (10 ft × 3 ft) ( dh dt) dh dt = 0.4 ft minute ▣ 6. A meteorologist is inflating a spherical balloon. If the radius of the balloon is changing at the rate of 1.5 m/s, express the volume V of the balloon as a function of time t (in seconds). [SOLUTION] dr dt = 1.5 m s r = 1.5t From the volume of a sphere V = 4 3 πr 3 V = 4 3 π(1.5t) 3 V = 9 2 πt 3 ▣ 7. A particle is moving along the curve y = √x. As the particle passes through the point (4,2), its x-coordinate increases at the rate of 3 cm/s. How fast is the distance from the particle to the origin changing at this instant? [SOLUTION] Any point in the curve is (x,√x). Its distance from the origin is L 2 = (x − 0) 2 + (√x − 0) 2 L 2 = x 2 + x At x = 4,