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Charylopos
Answers (4)
Pythagorous Theorem, z^2 = y^2 + x^2, where y = 2.5 cm and x is the legth of the pull. substituting z^2 = 2.5^2 +x^2, z =sqrt(x^2 +6.25) dz/dx (2z) = 2x dz/dx = (x/z) ......... (a) but dz/dt = dz/dx * dx/dt Thus, dz/dt = (x/z) * dx/dt = x/z * 3 when z = 12
A. x = 2√[ y^2 + (2.5)^2 ] cm (using Pythagorus theorem) B. x^2 = 4y^2 + 25 => 2x dx/dt = 8y dy/dt => dx/dt = 4 (y/x) dy/dt When x = 12 cm, y = (1/2)√[ 144 - 25 ] = 2.727 cm => dx/dt = 4 (2.727/12) x 3 cm/s = 2.727 cm/s.
A. x = 2√[ y^2 + (2.5)^2 ] cm (using Pythagorus theorem) B. x^2 = 4y^2 + 25 => 2x dx/dt = 8y dy/dt => dx/dt = 4 (y/x) dy/dt When x = 12 cm, y = (1/2)√[ 144 - 25 ] = 2.727 cm => dx/dt = 4 (2.727/12) x 3 cm/s = 2.727 cm/s.
each side of the elastic is the hypotenuse of a triangle with legs 5/2 and y, so x = 2√(2.5^2 + y^2) dx/dt = 2y/√(2.5^2+y^2) dy/dt when x=12, y=5.45, so dx/dt = 2(5.45)/6 (-3) = -1/5.45 = -0.18 cm/s
