/*! 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 12 A curve passes through the point... [FREE SOLUTION] | 91Ó°ÊÓ

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A curve passes through the point \((0,5)\) and has the property that the slope of the curve at every point \(P\) is twice the \(y\) -coordinate of \(P .\) What is the equation of the curve?

Short Answer

Expert verified
The equation of the curve is \( y = 5e^{2x} \).

Step by step solution

01

Understand the Problem

We are given that the slope of the curve at any point \( P(x, y) \) is twice the \( y \)-coordinate of that point. Mathematically, this means the derivative \( \frac{dy}{dx} = 2y \). Also, the curve passes through the point \((0, 5)\).
02

Interpret the Mathematical Model

The provided relationship \( \frac{dy}{dx} = 2y \) suggests that we have a first-order differential equation. We need to solve this differential equation to find \( y(x) \).
03

Solve the Differential Equation

The differential equation \( \frac{dy}{dx} = 2y \) is separable. We can rewrite it as \( \frac{dy}{y} = 2 \, dx \). Integrating both sides, we have: \[ \int \frac{1}{y} \, dy = \int 2 \, dx \] which results in \( \ln |y| = 2x + C \) where \( C \) is a constant of integration.
04

Solve for \( y \)

To solve for \( y \), exponentiate both sides of \( \ln |y| = 2x + C \) to get \( |y| = e^{2x+C} = e^C e^{2x} \). Let \( e^C = k \), so \( y = ke^{2x} \), where \( k \) is a constant.
05

Apply Initial Condition

The curve passes through \((0, 5)\), so substitute \( x = 0 \) into \( y = ke^{2x} \) to get \( 5 = ke^0 = k \). Thus, \( k = 5 \).
06

Write the Final Equation

Substitute \( k = 5 \) into \( y = ke^{2x} \) to obtain the equation of the curve: \( y = 5e^{2x} \). This is the equation that satisfies both the slope condition and the initial point condition.

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Key Concepts

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

Separable Differential Equations
A separable differential equation is a specific type of differential equation that can be separated into two parts: one involving only the dependent variable, and the other involving only the independent variable. In our exercise, the differential equation is \( \frac{dy}{dx} = 2y \). This is separable because we can rearrange it to \( \frac{dy}{y} = 2 \, dx \).
This separation enables us to tackle each side individually, turning the problem into two separate integrals.
\[\int \frac{1}{y} \, dy = \int 2 \, dx\]
Solving these integrals is straightforward. The left-hand side results in \( \ln |y| \), and the right-hand side becomes \( 2x + C \), where \( C \) is the constant of integration.
  • Separable differential equations simplify the process of solving first-order differential equations.
  • They rely on separating and integrating to find a solution.
Understanding how to recognize and solve these equations is vital to handling many applications in mathematics.
Initial Value Problem
An initial value problem in differential equations specifies not only the differential equation but also the value of the function at a particular point.
This extra piece of information allows us to find a unique solution to the problem. In our scenario, the curve passes through the point \((0, 5)\). This serves as our initial condition.
After solving the separable differential equation, we end up with a general solution: \( y = ke^{2x} \).
To find the specific value of \( k \), we use the initial condition by substituting \( x = 0 \) and \( y = 5 \), leading to:
  • \( 5 = ke^{0} \)
  • \( 5 = k \)
A unique solution is given by \( y = 5e^{2x} \), using the initial condition. Initial value problems are crucial because they ensure the solution is fitting for a real-world scenario where initial circumstances are known.
Exponential Functions
Exponential functions are a type of mathematical function that involve the constant \( e \), the base of natural logarithms, raised to a variable power.
In our exercise, solving the equation led us to an exponential function: \( y = 5e^{2x} \).
The base of the exponential function, \( e \approx 2.718 \), is a fundamental constant in calculus and has unique properties:
  • Exponential functions grow (or decay) at rates proportional to their current value.
  • They are used to model phenomena that grow or shrink rapidly, such as population growth or radioactive decay.
Here, the function \( y = 5e^{2x} \) describes a curve passing through the point \((0, 5)\) that grows exponentially as \( x \) increases.
This exemplifies how exponential functions arise naturally in differential equations and are essential in modeling real-world processes.

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Most popular questions from this chapter

Two sides of a triangle are 4 \(\mathrm{m}\) and 5 \(\mathrm{m}\) in length and the angle between them is increasing at a rate of 0.06 \(\mathrm{rad} / \mathrm{s} .\) Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is \(\pi / 3\) .

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