6. Two ships are sailing toward a very small island. One ship, the Pinta is east
ID: 2870570 • Letter: 6
Question
6. Two ships are sailing toward a very small island. One ship, the Pinta is east of the island and is sailing due west at 15 mi/h. The other ship, the Nina is north of the island and is sailing due south at 20 mi/h. At a certain time the Pinta is 30 mi from the island and the Nina is 40 mi from the island. How rapidly is the distance between the ships decreasing at that time. 7. A large garden clock has an hour hand that is 1.5 meters. And a minute hand that is 2.75 meters long. Find the rate at which the distance between the tips of the hands are changing at 12:20. 8. A spherical tank of radius 2 meters has water in it. Y is the height of the water in the tank. Water is being drained from the tank at a rate of 100 liters/mm. A) Find the rate at which the depth of the water is decreasing when the height of the water is 0.8 meters. V = pi(rh^2 - 1/3 h^3) (Volume of a sphere filled to height, h with radius r.)Explanation / Answer
Post question seperately, i have solved the first problem in complete detailed manner. Thanks
6) Since Pinta is moving in the x west direction and nina is moving in the south direction, the resultant movement will be 45 degree south of west
Pinto to island = x
Nina to island = y
Distance between pinta and nina = L = sqrt(x^2+y^2)
We need to find the derivative at the value of x=30 mi and y = 40 mi in order to find the decreasing distance
L = (x^2 + y^2)^(1/2)
dL/dt = 1/2*(x^2 + y^2)^(1/2-1) * (2x dx/dt + 2ydy/dt)
=> 1/2 * [2*30*15 + 2*40*20] * 1/(900 + 1600)^(1/2)
=> 1/2 * (2500)/50
=> 25
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