Question

A sand hopper is emptied through a chute. The amount \(w\) of sand in kilograms \(t\) seconds after the chute is opened is given by \(w(t)=1000-5 t\). The hopper next to it is being filled from a dump truck. The truck's entire load of 125 kilograms of sand is dumped in 30 seconds. Which is moving sand faster, the open chute or the truck while dumping?

Short answer

The chute is moving sand faster.

Step-by-step solution

Define the Rate of Sand Emptying from the Chute

The chute's sand emptying rate is given by the derivative of the function for the amount of sand in the hopper. The given function is: \[w(t) = 1000 - 5t\]. Differentiate this function with respect to time: \[\frac{dw}{dt} = -5\]. This tells us that the chute is emptying sand at a constant rate of 5 kilograms per second.

Calculate the Rate of Sand Dumping from the Truck

The truck dumps 125 kilograms of sand in 30 seconds. To find the rate, divide the amount of sand by the time: \[\text{Rate} = \frac{125 \text{ kg}}{30 \text{ s}} = \frac{125}{30} \approx 4.17\] Therefore, the truck is dumping sand at a rate of approximately 4.17 kilograms per second.

Compare the Rates of Sand Movement

The chute's rate of emptying sand is 5 kilograms per second, and the truck's rate of dumping sand is approximately 4.17 kilograms per second. Since 5 is greater than 4.17, the chute is moving sand faster than the truck.

Key concepts

Differentiation

In calculus, differentiation is a way of finding the rate at which something changes. When we differentiate a function, we are looking for its derivative, which tells us how the function's value grows or shrinks over time, or with respect to some other variable.

For example, in the given problem, we have a function that describes the amount of sand in a hopper: \[w(t)=1000-5t\]

Differentiating this function with respect to time (\(t\)) gives us: \[ \frac{dw}{dt} = -5 \]

This derivative, \( -5 \), indicates that the sand is being emptied from the hopper at a steady rate of 5 kilograms per second. The negative sign shows that the amount of sand is decreasing over time.

Rate of Emptying

The rate of emptying refers to how quickly the sand is being removed from the hopper. Using differentiation, we found that the rate of emptying was \( -5 \, \text{kg/s} \).

This means every second, the amount of sand in the hopper decreases by 5 kilograms. Since the rate is constant, it does not change over time. This is important because it helps us understand and predict how quickly the hopper will be empty.

Constant rates like this simplify calculations and allow us to make straightforward comparisons with other rates, such as the rate of dumping.

Rate of Dumping

The rate of dumping is about how quickly sand is added to the hopper from the dump truck. Here, 125 kilograms of sand is added in 30 seconds. To find the rate, we divide the total amount of sand by the total time: \[ \text{Rate}=\frac{ 125 \, \text{kg} }{ 30 \, \text{s} } \approx 4.17 \, \text{kg/s} \]

So, the truck is dumping sand into the hopper at a rate of approximately 4.17 kilograms per second. Unlike the emptying process, which is represented by a continuous function over time, the dumping rate is derived from a total amount of sand over a distinct period.

Comparison of Rates

To determine which process is moving sand faster, we compare the rates of emptying and dumping.

The chute's rate of emptying sand is 5 kilograms per second, while the truck's rate of dumping sand is approximately 4.17 kilograms per second.

Since 5 is greater than 4.17, it is clear that the open chute is moving sand at a faster rate than the dump truck.

This comparison helps us understand which process has a greater impact in scenarios where emptying or filling rates are crucial, such as in industrial operations.