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Chapter 6: Problem 44

A conical-shaped tank is being drained. The height of the water level in the tank is decreasing at the rate \(h^{\prime}(t)=-\frac{t}{2}\) inches per minute. Find the decrease in the depth of the water in the tank during the time interval \(2 \leq t \leq 4\).

Short Answer

Expert verified
The depth of the water decreases by 3 inches.

Step by step solution

01

Identify the given rate of change

The water level in the tank is decreasing at a rate given by the function: { h^{ prime}( t)=-frac{( t)}{ 2}
02

Set up the integral to find the total change

To find the total change in the depth of the water, integrate the rate of change function from t = 2 to t= 4: h ( t )= integral_{ 2}^{ 4} -frac( t}{ 2) d t
03

Integrate the function

The integral of -frac{t}{2} is: -frac{t^2}{4} Evaluating this definite integral from t= 2 to t= 4 : [ -frac{4^2}{4} ] - [ -frac{2^2}{4} ]
04

Simplify the expression

Simplify the values: -frac{16}{4} + frac{4}{4} = - 4 + 1= - 3
05

Write the final answer

The decrease in the depth of the water in the tank during the time interval 2 鈮� t 鈮� 4 is 3 inches.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrals
In calculus, an integral is a way of adding slices to find the whole. It is widely used in various fields like physics, engineering, and economics to compute areas, volumes, and other useful quantities. Integrals can be classified as either definite or indefinite.

An indefinite integral represents a family of functions and includes a constant of integration (C) because the process of differentiation loses information about constants. The notation for an indefinite integral of a function f(t) is \(\int f(t)dt\)\.

On the other hand, a definite integral calculates the net area under a curve within a specified interval. For example, the definite integral of a function f(t) from t = a to t = b is written as \(\int_{a}^{b} f(t) dt\)\.

This concept is particularly important when determining quantities like total distance, area, or volume over a certain period or space. The steps involved in computing an integral include setting up the integral based on the function provided, evaluating it, and simplifying the expression.
Definite Integral
A definite integral provides the exact amount of change or the accumulated quantity for a function over a specific interval. Compared to the indefinite integral where we just get a function plus a constant, with definite integrals, we're interested in real numbers.

For instance, in the problem given, we need to determine the change in the water depth in a conical tank. This is done through the definite integral of the rate of change of the water level from t = 2 to t = 4.

To solve, we set up the definite integral of \(-\frac{t}{2}\) over the interval [2, 4]:
\(\int_{2}^{4} -\frac{t}{2} dt\)\.

Next, we integrate the function to find the antiderivative, which is \(-\frac{t^2}{4}\). Finally, we evaluate this expression from t = 2 to t = 4, compute the result and simplify it. This gives us the exact decrease in the water depth during this interval.
Rate of Change
The rate of change in calculus refers to how a quantity changes with respect to another quantity. It is essentially the derivative of a function and is often represented by \( f'(x) \) or \(df(x)/dx\).

In the exercise, the rate of change of the water level in the tank is given by \(h'(t) = -\frac{t}{2}\). This indicates that the depth of the water is decreasing over time, and the negative sign shows the direction of this change.

To find the total decrease in water depth between two time intervals (t = 2 and t = 4), we use the definite integral of this rate of change function. Through this integral, we accumulate the rate of change over the specific time interval to get the total change in depth:
\(\int_{2}^{4} -\frac{t}{2} dt = -4 + 1 = -3\)\.

This result signifies that the water depth decreases by 3 inches over the time period from t = 2 to t = 4 minutes.

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