Question

(Calculator) The slope of a function \(y=f(x)\) at any point \((x, y)\) is \(\frac{y}{2 x+1}\) and \(f(0)=2\). (a) Write an equation of the line tangent to the graph of \(f\) at \(x=0\). (b) Use the tangent in part (a) to find the approximate value of \(f\) (0.1) (c) Find a solution \(y=f(x)\) for the differential equation. (d) Using the result in part (c), find \(f(0.1)\).

Short answer

Answer: The approximate value of \(f(0.1)\) using the tangent line is \(2.2\), and the exact value of \(f(0.1)\) is approximately \(2.19089\).

Step-by-step solution

(a) Finding Tangent Line Equation

To find the equation of the tangent line to the graph of \(f(x)\) at \(x=0\), we need the slope and the point of tangency. The slope of \(f(x)\) at any point is given by \(\frac{y}{2x+1}\). Using the fact that \(f(0)=2\), the point of tangency is \((0,2)\). The slope at this point is \(\frac{2}{2(0)+1}=\frac{2}{1}=2\). Using the point-slope form, the tangent line equation is \(y-2=2(x-0)\). Simplifying, we get: $$y = 2x + 2$$

(b) Approximate Value of \(f(0.1)\) Using the Tangent Line

To find the approximate value of \(f(0.1)\) using the tangent line equation, we substitute \(x=0.1\) in the equation we found in part (a): $$y = 2(0.1) + 2 = 0.2 + 2 = 2.2$$ So, the approximate value of \(f(0.1)\) is \(2.2\).

(c) Solving the Differential Equation for \(f(x)\)

We are given the differential equation: $$\frac{dy}{dx} = \frac{y}{2x+1}$$ This is a first-order linear homogeneous differential equation. We can solve it by dividing both sides by \(y\) and multiplying both sides by \(dx\): $$\frac{dy}{y} = \frac{1}{2x+1}dx$$ To solve for \(y\), we integrate both sides: $$\int\frac{dy}{y} = \int\frac{1}{2x+1}dx$$ The left side is straightforward: $$\ln |y| = \int\frac{1}{2x+1}dx + C_1$$ To integrate the right side, we use the substitution method, let \(u = 2x + 1, du = 2 dx\): $$\ln |y| = \frac{1}{2}\int\frac{du}{u} + C_1$$ $$\ln |y| = \frac{1}{2}\ln |u| + C_1$$ Back-substitute \(u\) with \(2x+1\) and express \(y\) explicitly: $$\ln |y| = \frac{1}{2}\ln |2x+1| + C_1$$ $$y = e^{\frac{1}{2}\ln(2x+1) + C_1}$$ As \(y=f(0)=2\), we can determine the value of \(C_1\): $$2 = e^{\frac{1}{2}\ln(2(0)+1) + C_1}$$ $$2 = e^{0 + C_1}$$ $$C_1 = \ln 2$$ Now, we have the function: $$y = f(x) = e^{\frac{1}{2}\ln(2x+1) + \ln 2}$$ $$f(x)= 2\sqrt{2x+1}$$

(d) Finding the Exact Value of \(f(0.1)\)

Using the result in part (c), we can find the exact value of \(f(0.1)\): $$f(0.1) = 2\sqrt{2(0.1)+1} = 2\sqrt{1.2} \approx 2.19089$$ So, the exact value of \(f(0.1)\) is approximately \(2.19089\).

Key concepts

Differential Equations

Differential equations play a pivotal role in various fields, from physics to finance, as they describe how a particular quantity changes over time or space. They are essentially equations that involve an unknown function and its derivatives. The unknown function represents something we want to solve for, which often depicts a physical quantity, like velocity or population size. Differential equations come in many forms, but they often categorize as either ordinary differential equations (ODEs) which have a single independent variable, or partial differential equations (PDEs) which have multiple independent variables.

In our exercise, we encountered a first-order ODE, since it involved the first derivative of the function and depended on only one independent variable, which is 'x' in this case. The differential equation \(dy/dx = y/(2x+1)\) can be solved using various methods, including separation of variables, integrating factor, or substitution. The process of solving these equations involves finding an expression for the unknown function that makes the equation true.

Tangent Line Equation

The tangent line equation is a linear representation of a function at a particular point and gives us an immediate rate of change, which can be thought of as the function's velocity at that point. It's crucial when approximating values close to the point of tangency or when assessing the behavior of a curve. To find this equation, one needs the slope of the tangent line at a specific point and the coordinates of that point.

The formula for the tangent line using point-slope form is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope of the function at the point \((x_1, y_1)\). In our exercise, we were given \(f(0)=2\) and used the slope formula \(m = y/(2x+1)\) to find that the slope at \(x=0\) is 2. Substituting these values into the point-slope form yielded the tangent line equation \(y = 2x + 2\). This equation was further used to approximate the value of \(f(0.1)\), illustrating its utility in making close approximations.

First-Order Linear Homogeneous Differential Equation

First-order linear homogeneous differential equations are a subset of differential equations in which the highest order of derivative present is one, and there is no constant or 'non-homogeneous' term added to the function. They have a standard form \(dy/dx + P(x)y = 0\), where the absence of the constant term on the right side of the equation characterizes the 'homogeneous' aspect.

To solve such equations, we can use the method of separation of variables, where we move all the 'y' terms to one side of the equation and all the 'x' terms to the other side. After separation, we integrate both sides to find a general solution. In the case of the differential equation from our exercise, \(dy/dx = y/(2x+1)\), we rearranged terms to give \(dy/y = dx/(2x+1)\) and integrated both sides leading to a solution involving a constant, which we determined using the given initial condition \(f(0)=2\). The systematic approach in finding the constant from the initial condition, also known as an 'initial value problem', provides us with the particular solution to the differential equation.