Saturday

April 19, 2014

April 19, 2014

Posted by **jonh** on Tuesday, February 19, 2013 at 2:10am.

- math -
**Reiny**, Tuesday, February 19, 2013 at 9:19amYour situation can be described by a right-angled triangle, where the base is along the water.

let the boat be x ft from the dock, let the length of rope from the man to the boat be y ft

then y^2 = x^2 + 7^2

2y dy/dt = 2x dx/dt

y dy/dt = x dx/dt

case #1, dx/dt = -3 ft/sec, x = 25

y^2 = 25^2 + 7^ = 674 , ----> y = √674

√674 dy/dt = 25(-3)

dy/dt = -75/√674 = appr -2.89 ft/sec

case 2, when x = 10, dx/dt = -3

y^2 = 100+49 = 149 , ---> y = √149

√149 dy/dt = 10(-3)

dy/dt = --2.46 ft/s

case3 , when x = 28 , dy/dt = -3

y^2 = 784+49 = 833 , -----> y = √833

√833(-3 = 28 dx/dt

dx/dt = √833(-3)/28 = -3.09 ft/s

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