2d(dd/dt) = 2A(dA/dt) + 2T(dT/dt)
d(dd/dt) = A(dA/dt) + T(dT/dt)
100(dd/dt) = 60(- 30) + 80(40)
100(dd/dt) = - 1800 + 3200
100(dd/dt) = 1400
dd/dt = 1400 / 100
dd/dt = 14
The distance between the two vehicles is increasing at a rate of 14 ft./sec.
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2.) Height, y = 4 ft.
Distance from Dock, x = 12 ft.
Rope Length = z
dz/dt = 3 ft./sec
Find dx/dt:
z² = x² + y²
z² = (12)² + (4)²
z² = 144 + 16
z² = 160
z = √160
z = 12.65 (Negative root disregarded)
z² - x² = y²
z² - x² = (4)²
z² - x² = 16
Differentiate Implicitly Over Time:
Since y² is a constant, dy/dt = 0, so
2z(dz/dt) - 2x(dx/dt) = 0
z(dz/dt) - x(dx/dt) = 0
12.65(3) - 12(dx/dt) = 0
37.95 - 12(dx/dt) = 0
12(dx/dt) = 37.95
dx/dt = 37.95 / 12
dx/dt = 3.162
The boat is moving at a rate of 3.162 ft./sec.
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