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In some chemical reactions, the rate at which the amount of a substance changes with time is proportional to the amount present. For the change of \(\delta\) -glucono lactone into gluconic acid, for example, $$\frac{d y}{d t}=-0.6 y$$ when \(t\) is measured in hours. If there are 100 grams of \(\delta\) -glucono lactone present when \(t=0,\) how many grams will be left after the first hour?

Short Answer

Expert verified
Approximately 54.88 grams will be left after one hour.

Step by step solution

01

Understand the Differential Equation

We start with the differential equation \( \frac{dy}{dt} = -0.6y \). This equation denotes that the rate of change of \( y \), which represents the amount of \( \delta \)-glucono lactone, is proportional to its current amount. The negative sign indicates that the substance is decreasing over time.
02

Set Up the Initial Condition

We are given the initial condition \( y(0) = 100 \) grams. This means that at time \( t = 0 \), there are 100 grams of \( \delta \)-glucono lactone.
03

Solve the Differential Equation

To solve the differential equation \( \frac{dy}{dt} = -0.6y \), we separate the variables to obtain \( \frac{dy}{y} = -0.6 dt \). Integrating both sides, we get \( \ln |y| = -0.6t + C \). Solving for \( y \), we exponentiate both sides to obtain \( y = e^{C} e^{-0.6t} = Ce^{-0.6t} \), where \( C = e^{C} \) is a constant.
04

Determine the Constant C

Using the initial condition \( y(0) = 100 \), we have \( 100 = Ce^{0} \). Thus, \( C = 100 \). The solution to the differential equation is then \( y(t) = 100e^{-0.6t} \).
05

Find the Amount Present After One Hour

Substitute \( t = 1 \) into the equation \( y(t) = 100e^{-0.6t} \) to find the amount remaining after 1 hour: \( y(1) = 100e^{-0.6 \times 1} = 100e^{-0.6} \).
06

Calculate the Result

Compute \( 100e^{-0.6} \) using a calculator. This yields \( y(1) \approx 54.88 \) grams after rounding.

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Key Concepts

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

Exponential Growth and Decay
Exponential growth and decay are fundamental concepts in understanding how quantities evolve over time. These processes occur when the rate of change of a quantity is proportional to its current value. In the case of decay, such as with the chemical reaction of \(\delta\)-glucono lactone to gluconic acid, the amount decreases over time.

Key characteristics of exponential decay include:
  • The presence of a constant rate, often represented by a negative coefficient, indicating a decrease.
  • A natural logarithm function that simplifies to an exponential form when solving the differential equation.
  • An initial value condition that allows us to calculate the constant of integration.
Understanding these principles allows us to predict and model various natural processes, from radioactive decay to cooling bodies and chemical reactions. Utilizing the formula \( y(t) = Ce^{-kt} \), where \( k \) is the decay rate, mathematicians and scientists can effectively calculate future amounts.
Separation of Variables
Separation of variables is a technique used to solve differential equations, especially useful for first-order equations. This method simplifies the problem by separating the variables of the equation into different sides to make integration possible.

Here's how separation of variables typically works:
  • First, rearrange the differential equation to isolate all terms containing one variable on one side, and the other variable with the derivative on the opposite side.
  • In the given problem \( \frac{dy}{dt} = -0.6y \), the equation is rewritten as \( \frac{dy}{y} = -0.6 dt \).
  • Next, integrate both sides separately. This process turns the differential equation into an algebraic one, making it easier to solve for the function.
This technique is especially effective when the equation can be divided cleanly, allowing for two separate integrals that can be evaluated independently. It provides a streamlined approach to finding the general solution of the differential equation.
Integration
Integration is a core mathematical process and is integral to solving differential equations. By integrating, we reverse the differentiation process, thus obtaining the original function from its derivative.

With the differential equation \( \frac{dy}{y} = -0.6 dt \):
  • We integrate \( \frac{dy}{y} \) to obtain \( \ln|y| \).
  • The integration of \(-0.6 dt\) yields \(-0.6t + C\), where \(C\) is the constant of integration.
  • Exponentiation of both sides is used to solve for \(y\), resulting in the formula \(y = Ce^{-0.6t}\).
Integration not only helps in solving for unknown functions but also plays a critical role in physics, engineering, and other fields where modeling continuous processes is necessary. The constant \(C\) is evaluated using initial conditions or boundary values, allowing the solution to be specifically tailored to the problem at hand.

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Most popular questions from this chapter

Suppose that electricity is draining from a capacitor at a rate that is proportional to the voltage \(V\) across its terminals and that, if \(t\) is measured in seconds, $$\frac{d V}{d t}=-\frac{1}{40} V$$ Solve this equation for \(V\), using \(V_{0}\) to denote the value of \(V\) when \(t=0 .\) How long will it take the voltage to drop to \(10 \%\) of its original value?

The intensity \(L(x)\) of light \(x\) feet beneath the surface of the ocean satisfies the differential equation $$\frac{d L}{d x}=-k L$$ As a diver, you know from experience that diving to \(18 \mathrm{ft}\) in the Caribbean Sea cuts the intensity in half. You cannot work without artificial light when the intensity falls below one-tenth of the surface value. About how deep can you expect to work without artificial light?

Skydiving If a body of mass \(m\) falling from rest under the action of gravity encounters an air resistance proportional to the square of the velocity, then the body's velocity \(t\) sec into the fall satisfies the differential equation $$m \frac{d v}{d t}=m g-k v^{2}$$ where \(k\) is a constant that depends on the body's aerodynamic properties and the density of the air. (We assume that the fall is short enough so that the variation in the air's density will not affect the outcome significantly.) a. Show that $$ v=\sqrt{\frac{m g}{k}} \tanh (\sqrt{\frac{g k}{m}} t)$$ satisfies the differential equation and the initial condition that \(v=0\) when \(t=0\) b. Find the body's limiting velocity, lim_,-\inftyv. c. For a 160 -lb skydiver \((m g=160),\) with time in seconds and distance in feet, a typical value for \(k\) is \(0.005 .\) What is the diver's limiting velocity?

A decimal representation of \(e\) Find \(e\) to as many decimal places as your calculator allows by solving the equation \(\ln x=1\) using Newton's method in Section 4.7.

The equation \(x^{2}=2^{x}\) has three solutions: \(x=2, x=4,\) and one other. Estimate the third solution as accurately as you can by graphing.

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