Question
2. Coffee is draining out of a conical filter at a rate of 4.5cm^3/min . If the cone is 5 cm tall and has a radius of 4 cm, how fast is the coffee level dropping when the coffee is 5 cm deep?
Solution
4.4
(265 Votos)
Pablo
Profissional · Tutor por 6 anos
Resposta
To solve this problem, we need to find the rate at which the coffee level is dropping when the coffee is 5 cm deep.Given information:- The coffee is draining out of a conical filter at a rate of 4.5 cm³/min.- The cone has a height of 5 cm and a radius of 4 cm.Let's define the variables:- V(t) = the volume of coffee in the cone at time t- h(t) = the height of the coffee level at time tWe can use the formula for the volume of a cone to relate the volume and the height:V(t) = (1/3)πr²hSince the radius of the cone is 4 cm, we can express the radius in terms of the height:r = (4/5)hSubstituting this into the volume formula, we get:V(t) = (1/3)π((4/5)h)²hV(t) = (1/3)π(16/25)h³Now, we can differentiate this equation with respect to time to find the rate of change of the volume:dV/dt = (1/3)π(16/25) * 3h² * dh/dtdV/dt = (16/25)πh² * dh/dtWe are given that dV/dt = -4.5 cm³/min (the negative sign indicates that the volume is decreasing).Now, we can substitute the given values and solve for dh/dt:-4.5 = (16/25)π(5)² * dh/dt-4.5 = (16/25)π(25) * dh/dt-4.5 = 16π * dh/dtdh/dt = -4.5 / (16π)dh/dt ≈ -0.089 cm/minTherefore, the coffee level is dropping at a rate of approximately 0.089 cm/min when the coffee is 5 cm deep.