/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 Suppose that the concentration o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that the concentration of some pollutant in a river as a function of position \(x\) and time \(t\) is given by \(p(x, t)=p_{0}(x-c t) e^{-\mu t}\) for constants \(p_{0}, c\) and \(\mu .\) Show that \(\frac{\partial p}{\partial t}=-c \frac{\partial p}{\partial x}-\mu p .\) Interpret both \(\frac{\partial p}{\partial t}\) and \(\frac{\partial p}{\partial x},\) and explain how this equation relates the change in pollution at a specific location to the current of the river and the rate at which the pollutant decays.

Short Answer

Expert verified
The differential equation \(\frac{\partial p}{\partial t}=-c \frac{\partial p}{\partial x}-\mu p\) demonstrates that the rate of change of the pollutant concentration depends on the movement of pollution due to the river current, represented by the \(-c \frac{\partial p}{\partial x}\) term, and on the rate of decay or breakdown of the pollutant, represented by the \(-u p\) term.

Step by step solution

01

Calculate the partial derivative with respect to time

Start off by differentiating the function \(p(x,t)\) with respect to time \(t\). For this, the chain rule is applied where the derivative of the outside function is multiplied by the derivative of the inside function. This gives: \(\frac{\partial p}{\partial t} = p_{0}(x-ct)e^{-\mu t}.-c - p_{0}(x - ct)e^{-\mu t}.\mu = -c \frac{\partial p}{\partial x} - \mu p \)
02

Interpret the partial derivative with respect to time

\(\frac{\partial p}{\partial t}\) is the rate of change in the concentration of the pollutant over time at a specific location. If it is negative, the pollutant concentration is decreasing with time. This could be due to dilution from water currents or breakdown of the pollutant.
03

Calculate and interpret the partial derivative with respect to position

Next, calculate the partial derivative \(\frac{\partial p}{\partial x}\). This derivative represents how the concentration of the pollutant changes from one position to another at a specific time. The presence of the velocity constant \(c\) indicates that the rate at which the pollutant is carried downstream by the current.
04

Interpret the differential equation

The derived differential equation \(\frac{\partial p}{\partial t}=-c \frac{\partial p}{\partial x}-\mu p\) states that the rate of change of the pollutant concentration at a point in the river is proportional to the current of the river and the rate at which the pollutant can breakdown or decay.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Pollutant Transport
In the study of pollution in rivers, understanding how pollutants move is crucial. The concentration of a pollutant at any given point in the river over time can be represented using a function like \( p(x, t) = p_{0}(x-ct) e^{-\mu t} \). Here, \( p_{0} \) is the initial concentration, \( c \) is the speed of the river current, and \( \mu \) represents the decay rate of the pollutant. This formula helps model how pollutants are carried downstream and how they decrease over time due to natural breakdown processes.
Pollutant transport is affected by various factors such as:
  • River Current: The speed at which the river moves (\(c\)) can shift the location of the highest concentration of pollutants downstream.
  • Decay Rate: Pollution often undergoes chemical reactions that reduce concentration over time, captured by the term \( e^{-\mu t} \).
  • Initial Distribution: The initial concentration \(p_{0}\) dictates how much pollutant is present initially.
Overall, this function captures how pollution disperses both spatially and temporally, highlighting the dynamic nature of environmental pollution challenges.
Rate of Change
The rate of change in the concentration of a pollutant as it flows in a river is crucial to understanding its environmental impact. Mathematically, this is represented by partial derivatives.
  • Partial Derivative with Respect to Time (\( \frac{\partial p}{\partial t} \) ): This derivative measures how the pollutant concentration changes at a specific location as time progresses. A negative value indicates that the concentration is decreasing, possibly due to the decay of the pollutant or dilution through water flow.
  • Partial Derivative with Respect to Position (\( \frac{\partial p}{\partial x} \)): This measures the change in pollutant concentration over distance at a fixed time. It reflects how the concentration gradient spreads, often increased by the river current \(c\).
The differential equation \( \frac{\partial p}{\partial t} = -c \frac{\partial p}{\partial x} - \mu p \) links these partial derivatives, providing insight into how the temporal and spatial changes work together in pollutant transport.
Differential Equation Interpretation
The differential equation \( \frac{\partial p}{\partial t} = -c \frac{\partial p}{\partial x} - \mu p \) synthesizes the concepts of pollutant transport and rate of change. This equation explains how pollutant concentration changes over time and space.
  • Role of the Current \(c\): The term \(-c \frac{\partial p}{\partial x}\) shows how the river's flow impacts the rate of spread of the pollutant downstream. The faster the current, the more significant this effect becomes.
  • Decay Term \(\mu p\): This part represents the natural reduction in concentration due to chemical decay or other breakdown processes. It shows how pollutants decrease in intensity over time.
By understanding this equation, students can visualize how pollutants disperse and diminish in a river. This mathematical model is vital for predicting pollution impacts and crafting strategies for environmental management.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For a rectangle of length \(L\) and perimeter \(P\), show that the area is given by \(A=\frac{1}{2} L P-L^{2} .\) Compute \(\frac{\partial A}{\partial L} .\) A simpler formula for area is \(A=L W,\) where \(W\) is the width of the rectangle. Compute \(\frac{\partial A}{\partial L}\) and show that your answer is not equivalent to the previous derivative. Explain the difference by noting that in one case the width is held constant while \(L\) changes, whereas in the other case the perimeter is held constant while \(L\) changes.

Show that \(f(x, y)=\left\\{\begin{array}{cl}\frac{x^{2} y}{x^{2}+y^{2}}, & \text { if }(x, y) \neq(0,0) \\ 0, & \text { if }(x, y)=(0,0)\end{array}\right.\) is continuous but not differentiable at \((0,0).\)

A well-known college uses the following formula to predict the grade average of prospective students: \(\mathrm{PGA}=0.708(\mathrm{HS})+0.0018(\mathrm{SATV})+0.001(\mathrm{SATM})-1.13\) Here, PGA is the predicted grade average, HS is the student's high school grade average (in core academic courses, on a four point scale), SATV is the student's SAT verbal score and SATM is the student's SAT math score. Use your scores to compute your own predicted grade average. Determine whether it is possible to have a predicted average of \(4.0,\) or a negative predicted grade average. In this formula, the predicted grade average is a function of three variables. State which variable you think is the most important and explain why you think so.

Use the result of exercise 44 to find an equation of the tangent plane to the parametric surface at the indicated point. \(S\) is the cylinder \(x^{2}+y^{2}=1\) with \(0 \leq z \leq 2 ;\) at \((1,0,1).\)

In exercise 33, we specified that you zoom in on the contour plot until the level curves appear linear and equally spaced. To see why the second condition is necessary, sketch a contour plot of \(f(x, y)=e^{x-y}\) with \(-1 \leq x \leq 1\) and \(-1 \leq y \leq 1 .\) Use this plot to estimate \(\frac{\partial f}{\partial x}(0,0)\) and \(\frac{\partial f}{\partial y}(0,0)\) and compare to the exact values. Zoom in until the level curves are equally spaced and estimate again. Explain why this estimate is much better.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.