san14 san14
  • 01-12-2017
  • Mathematics
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zizoux07p0ac3m
zizoux07p0ac3m zizoux07p0ac3m
  • 01-12-2017
let the horizontal distance is x and vertical distance is y and the distance  between them is z  
z^2 = x^2 + y^2  
2z(dz/dt) = 2x(dx/dt) + 2y(dy/dt)               divide by 2
z(dz/dt) = x(dx/dt) + y(dy/dt)
when y = 45 , x = 0 because the boy was under the balloon 
after 3 secs: x = 15 * 3 = 45 , y = 5 * 3 = 15 , z = sqr(x^2 +y^2) = 75 
z(dz/dt) = x(dx/dt) + y(dy/dt)
75(dz/dt) = 45(15) + 15(5) = 975
dz/dt = 13 ft/sec



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