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?

To solve this problem, we need to find the rate at which the coffee level is dropping when the coffee is 5 cm deep.<br /><br />Given information:<br />- The coffee is draining out of a conical filter at a rate of 4.5 cm³/min.<br />- The cone has a height of 5 cm and a radius of 4 cm.<br /><br />Let's define the variables:<br />- V(t) = the volume of coffee in the cone at time t<br />- h(t) = the height of the coffee level at time t<br /><br />We can use the formula for the volume of a cone to relate the volume and the height:<br />V(t) = (1/3)πr²h<br /><br />Since the radius of the cone is 4 cm, we can express the radius in terms of the height:<br />r = (4/5)h<br /><br />Substituting this into the volume formula, we get:<br />V(t) = (1/3)π((4/5)h)²h<br />V(t) = (1/3)π(16/25)h³<br /><br />Now, we can differentiate this equation with respect to time to find the rate of change of the volume:<br />dV/dt = (1/3)π(16/25) * 3h² * dh/dt<br />dV/dt = (16/25)πh² * dh/dt<br /><br />We are given that dV/dt = -4.5 cm³/min (the negative sign indicates that the volume is decreasing).<br /><br />Now, we can substitute the given values and solve for dh/dt:<br />-4.5 = (16/25)π(5)² * dh/dt<br />-4.5 = (16/25)π(25) * dh/dt<br />-4.5 = 16π * dh/dt<br />dh/dt = -4.5 / (16π)<br />dh/dt ≈ -0.089 cm/min<br /><br />Therefore, the coffee level is dropping at a rate of approximately 0.089 cm/min when the coffee is 5 cm deep.