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Chapter 1: Problem 30

Water is poured into a container that has a small leak. The mass \(m\) of the water is given as a function of time \(t\) by \(m=5.00 t^{0.8}-3.00 t+20.00,\) with \(t \geq 0, m\) in grams, and \(t\) in seconds. (a) At what time is the water mass greatest, and (b) what is that greatest mass? In kilograms per minute, what is the rate of mass change at \((\mathrm{c}) t=2.00 \mathrm{~s}\) and (d) \(t=5.00 \mathrm{~s} ?\)

Short Answer

Expert verified
The greatest mass is at \( t \approx 1.52 \) s with a mass of \( 21.45 \) g. The rate of mass change is \( 0.0168 \) kg/min at \( t = 2.00 \) s and \(-0.0624 \) kg/min at \( t = 5.00 \) s.

Step by step solution

01

Understand the Function

The mass of the water as a function of time is given by \( m(t) = 5.00 t^{0.8} - 3.00 t + 20.00 \). Our task is to find when the mass is greatest, its value at that time, and the rate of change at specific times.
02

Find the Time of Greatest Mass

To find when the mass is greatest, we need to find the critical points of \( m(t) \). We do this by finding its derivative: \( m'(t) = 4.00 t^{-0.2} - 3.00 \). Set \( m'(t) = 0 \) to find critical points: \[ 4.00 t^{-0.2} - 3.00 = 0 \] \[ 4.00 t^{-0.2} = 3.00 \] \[ t^{-0.2} = \frac{3.00}{4.00} \] \[ t = \left(\frac{4.00}{3.00}\right)^5 \approx 1.52 \]
03

Confirm Maximum with Derivative Test

To confirm that the critical point at \( t \approx 1.52 \) is a maximum, we use the second derivative test. Calculate the second derivative: \( m''(t) = -0.8 \times 4.00 t^{-1.2} = -3.2 t^{-1.2} \).Evaluate at \( t = 1.52 \): \[ m''(1.52) = -3.2 \times (1.52)^{-1.2} \approx -2.40 \] Since \( m''(1.52) < 0 \), the point is a local maximum.
04

Find the Greatest Mass

Evaluate \( m(t) \) at \( t = 1.52 \) to find the greatest mass:\[ m(1.52) = 5.00 \times (1.52)^{0.8} - 3.00 \times 1.52 + 20.00 \approx 21.45 \text{ grams} \]
05

Find Rate of Change at \( t = 2.00 \) s

To find the rate of mass change at \( t = 2.00 \) s, evaluate the first derivative \( m'(t) \) at \( t=2.00 \): \[ m'(2.00) = 4.00 \times (2.00)^{-0.2} - 3.00 \approx 0.28 \text{ grams/second} \]
06

Convert Rate of Change from Grams/Second to Kilograms/Minute at \( t = 2.00 \) s

Convert the rate from grams per second to kilograms per minute:\( 0.28 \text{ g/s} = 0.28 \times 60 \text{ g/min} = 16.8 \text{ g/min} = 0.0168 \text{ kg/min} \)
07

Find Rate of Change at \( t = 5.00 \) s

Evaluate the first derivative \( m'(t) \) at \( t=5.00 \) to find the rate of mass change:\[ m'(5.00) = 4.00\times(5.00)^{-0.2} - 3.00 \approx -1.04 \text{ grams/second} \]
08

Convert Rate of Change from Grams/Second to Kilograms/Minute at \( t = 5.00 \) s

Convert the rate from grams per second to kilograms per minute:\[-1.04 \text{ g/s} = -1.04 \times 60 \text{ g/min} = -62.4 \text{ g/min} = -0.0624 \text{ kg/min} \]

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Key Concepts

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

Derivatives
Derivatives are a fundamental concept in calculus. They measure how a function changes as its input changes. When you hear the word "derivative," think "rate of change," because this is essentially what a derivative tells you.
A function's derivative at a point gives you the slope of the tangent line to the function's graph at that point. This means it tells us how steeply the graph rises or falls.
  • If a derivative is positive, the function is increasing at that point.
  • If it's negative, the function is decreasing.
  • When the derivative is zero, the function could be at a peak, a trough, or a flat spot.
For example, if we take the derivative of the mass function over time, we calculate how quickly the mass is changing at any given time.
Critical Points
Critical points are where the derivative of a function is either zero or undefined. These points help identify where something special happens with the function, such as a maximum, a minimum, or a point of inflection.
In our example with the mass of water, critical points are used to determine when the mass is greatest. To find a critical point, we set the derivative of the mass function to zero:
  • The equation transformed and solved gives us the critical point.
  • In this exercise, the critical point for the maximum mass is found when the derivative equals zero.
Finding critical points is an important step when analyzing any function, as they often mark significant changes in behavior.
Rate of Change
The rate of change is a phrase you'll hear a lot in calculus. It refers to how quickly a quantity changes over time. In practical terms, the rate of change tells you how fast something is happening.
For this exercise, the rate of change is calculated at specific times:
  • At time = 2 seconds, we evaluate the derivative, which tells us how fast the mass is changing at that precise moment.
  • The same calculation is done at time = 5 seconds to see how the rate of change has shifted.
Additionally, to make this data more meaningful, it's often necessary to convert the units, such as from grams per second to kilograms per minute, providing clearer insights for understanding the problem.
Second Derivative Test
The second derivative test is a handy tool to confirm whether a critical point is a maximum or a minimum. This involves taking the derivative of the derivative, which is called the second derivative.
Once we have the second derivative, we plug in our critical point's value:
  • If the second derivative is negative at a critical point, the function has a local maximum there.
  • If it's positive, there's a local minimum.
  • If it's zero, the test is inconclusive, and other methods need to be used.
In the water mass problem, the second derivative test helped us confirm that the critical point was indeed a point of maximum mass, enhancing our understanding and providing a clear conclusion to the problem.

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Most popular questions from this chapter

The description for a certain brand of house paint claims a coverage of \(460 \mathrm{ft}^{2} /\) gal. (a) Express this quantity in square meters per liter. (b) Express this quantity in an SI unit (see Appendices A and D. (c) What is the inverse of the original quantity, and (d) what is its physical significance?

A lecture period ( \(50 \mathrm{~min}\) ) is close to 1 microcentury. (a) How long is a microcentury in minutes? (b) Using percentage difference \(=\left(\frac{\text { actual }-\text { pproximation }}{\text { actual }}\right) 100\) find the percentage difference from the approximation.

(a) A unit of time sometimes used in microscopic physics is the shake. One shake equals \(10^{-8} \mathrm{~s}\). Are there more shakes in a second than there are seconds in a year? (b) Humans have existed for about \(10^{6}\) years, whereas the universe is about \(10^{10}\) years old. If the age of the universe is defined as 1 "universe day," where a universe day consists of "universe seconds" as a normal day consists of normal seconds, how many universe seconds have humans existed?

The cubit is an ancient unit of length based on the distance between the elbow and the tip of the middle finger of the measurer. Assume that the distance ranged from 43 to \(53 \mathrm{~cm},\) and suppose that ancient drawings indicate that a cylindrical pillar was to have a length of 9 cubits and a diameter of 2 cubits. For the stated range, what are the lower value and the upper value, respectively, for (a) the cylinder's length in meters, (b) the cylinder's length in millimeters, and (c) the cylinder's volume in cubic meters?

A vertical container with base area measuring \(14.0 \mathrm{~cm}\) by \(17.0 \mathrm{~cm}\) is being filled with identical pieces of candy, each with a volume of \(50.0 \mathrm{~mm}^{3}\) and a mass of \(0.0200 \mathrm{~g}\). Assume that the volume of the empty spaces between the candies is negligible. If the height of the candies in the container increases at the rate of \(0.250 \mathrm{~cm} / \mathrm{s}\) at what rate (kilograms per minute) does the mass of the candies in the container increase?
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