A man starts walking north at 3 ft/s from point P. Five minutes later, a woman starts walking south at 4 ft/s from a point 500 ft due east of P. At what rate are the people moving apart 15 minutes after the woman starts walking? (Round your answer to two decimal places.) ft/s At noon, ship A is 70 km west of ship B. Ship A is sailing south at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4:00 PM? (Round your answer to one decimal place.) km/h
The Correct Answer and Explanation is:
Problem 1: Man and Woman Walking Apart
Let the coordinate system place point P at the origin (0,0)(0, 0)(0,0).
The man walks north at 333 ft/s from point P, so his position at time ttt seconds after the woman starts walking is:(x1,y1)=(0,3(t+300))(x_1, y_1) = (0, 3(t + 300))(x1,y1)=(0,3(t+300))
(The man starts 5 minutes = 300 seconds earlier.)
The woman starts 500 ft east of P and walks south at 444 ft/s:(x2,y2)=(500,−4t)(x_2, y_2) = (500, -4t)(x2,y2)=(500,−4t)
The distance DDD between the two people is:D=(x2−x1)2+(y2−y1)2=(500)2+(3t+300+4t)2=250000+(7t+300)2D = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} = \sqrt{(500)^2 + (3t + 300 + 4t)^2} = \sqrt{250000 + (7t + 300)^2}D=(x2−x1)2+(y2−y1)2=(500)2+(3t+300+4t)2=250000+(7t+300)2
Differentiate with respect to ttt:dDdt=12⋅(250000+(7t+300)2)−1/2⋅2(7t+300)⋅7=7(7t+300)250000+(7t+300)2\frac{dD}{dt} = \frac{1}{2} \cdot \left(250000 + (7t + 300)^2\right)^{-1/2} \cdot 2(7t + 300) \cdot 7 = \frac{7(7t + 300)}{\sqrt{250000 + (7t + 300)^2}}dtdD=21⋅(250000+(7t+300)2)−1/2⋅2(7t+300)⋅7=250000+(7t+300)27(7t+300)
At t=900t = 900t=900 seconds (15 minutes):dDdt=7(7⋅900+300)250000+(7⋅900+300)2=7(6300+300)250000+66002=7⋅6600250000+43560000=4620043810000≈462006617.41≈6.98 ft/s\frac{dD}{dt} = \frac{7(7 \cdot 900 + 300)}{\sqrt{250000 + (7 \cdot 900 + 300)^2}} = \frac{7(6300 + 300)}{\sqrt{250000 + 6600^2}} = \frac{7 \cdot 6600}{\sqrt{250000 + 43560000}} = \frac{46200}{\sqrt{43810000}} ≈ \frac{46200}{6617.41} ≈ 6.98 \text{ ft/s}dtdD=250000+(7⋅900+300)27(7⋅900+300)=250000+660027(6300+300)=250000+435600007⋅6600=4381000046200≈6617.4146200≈6.98 ft/s
Answer: 6.98 ft/s
Problem 2: Two Ships
At noon:
- Ship A is 70 km west of Ship B.
- Let Ship B be at origin (0, 0).
- Ship A starts at (-70, 0).
Ship A sails south at 35 km/h
Ship B sails north at 25 km/h
Let t=4t = 4t=4 hours after noon.
Position of Ship A: (−70,−35t)=(−70,−140)(-70, -35t) = (-70, -140)(−70,−35t)=(−70,−140)
Position of Ship B: (0,25t)=(0,100)(0, 25t) = (0, 100)(0,25t)=(0,100)
Distance between ships:D=(−70)2+(−140−100)2=4900+57600=62500=250 kmD = \sqrt{(-70)^2 + (-140 – 100)^2} = \sqrt{4900 + 57600} = \sqrt{62500} = 250 \text{ km}D=(−70)2+(−140−100)2=4900+57600=62500=250 km
Differentiate:D2=x2+y2D^2 = x^2 + y^2D2=x2+y22D⋅dDdt=2x⋅dxdt+2y⋅dydt2D \cdot \frac{dD}{dt} = 2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt}2D⋅dtdD=2x⋅dtdx+2y⋅dtdy
At t=4t = 4t=4:
- x=−70x = -70x=−70, dx/dt=0dx/dt = 0dx/dt=0
- y=−140−100=−240y = -140 – 100 = -240y=−140−100=−240
- dy/dt=−35−25=−60dy/dt = -35 – 25 = -60dy/dt=−35−25=−60 km/h
Substitute:2(250)⋅dDdt=2(−240)(−60)⇒500⋅dDdt=28800⇒dDdt=28800500=57.6 km/h2(250) \cdot \frac{dD}{dt} = 2(-240)(-60) \Rightarrow 500 \cdot \frac{dD}{dt} = 28800 \Rightarrow \frac{dD}{dt} = \frac{28800}{500} = 57.6 \text{ km/h}2(250)⋅dtdD=2(−240)(−60)⇒500⋅dtdD=28800⇒dtdD=50028800=57.6 km/h
Answer: 57.6 km/h
