Final answer:
The question is related to the topic of related rates in calculus, which involves finding the rate of change in the height of water in an inverted cone-shaped tank as it is being filled. The student needs to derive a relationship between the volume and height of water, and then use differentiation with respect to time to solve for the rate at which the water level is rising when it is 4 meters deep.
Step-by-step explanation:
The student's question on related rates involves calculus, and it is asking about how fast the water level in an inverted cone-shaped water tank is rising when the water is at a certain depth. To solve this, we must relate the volume V of water in the tank at any time t to the height h of the water at that time. Since the tank is an inverted cone, the volume of the water can be expressed as V = (1/3)πr^2h, where r is the radius of the water's surface at height h. Because the radius and the height of the cone are proportional (due to similar triangles), we can express r in terms of h and the tank's dimensions.
When water is pumped in at a constant rate, we can apply the chain rule to find the rate of change of the height with respect to time, dh/dt, given the rate at which volume is changing, dV/dt. To find the rate dh/dt when the water is 4 m deep, we need to substitute h = 4 m into the expression we have for dV/dt and solve for dh/dt using the given values for the dimensions of the tank and the rate at which water is being pumped.