Question

Find the equation of the curve which has a horizontal tangent at the point \((0,-1)\), and for which the rate of change, with respect to \(\mathrm{x}\), of the slope at any point is equal to \(8 \mathrm{e}^{2 \mathrm{x}}\).

Short answer

The equation of the curve is \(y = \frac{1}{2}e^{2x} - 4x - \frac{1}{2}\).

Step-by-step solution

Integrate the second derivative

To find the equation of the curve, we need to integrate the given second derivative, \(8e^{2x}\), with respect to x. Let's first find the first derivative by integrating the second derivative: \[ \int 8e^{2x} \mathrm{d}x \] Using integration by substitution, let \(u = 2x\). Then, \(\mathrm{d}u = 2\mathrm{d}x\). So, \[ \int 8e^{2x} \mathrm{d}x = \frac{1}{2}\int 8e^{u} \mathrm{d}u \] Now integrate: \[ \frac{1}{2}\int 8e^{u} \mathrm{d}u = 4e^{u} + C_1 = 4e^{2x} + C_1 \] So, the first derivative is: \[ \frac{d}{dx} = 4e^{2x} + C_1 \]

Apply the condition for the horizontal tangent

The curve has a horizontal tangent at the point \((0, -1)\). This means that the first derivative is \(0\) at this point. Let's substitute \(x=0\) and \(y=-1\) to find the value of the constant \(C_1\): \[ 4e^{2(0)} + C_1 = 0 \] Solving for \(C_1\): \[ C_1 = -4 \] Now, the first derivative becomes: \[ \frac{d}{dx} = 4e^{2x} - 4 \]

Integrate the first derivative

To find the equation of the curve, we need to integrate the first derivative with respect to x: \[ y = \int (4e^{2x} - 4) \mathrm{d}x \] Integrate the terms separately: \[ y = 4 \int e^{2x} \mathrm{d}x - 4 \int \mathrm{d}x \] Using the substitution method from Step 1, we have: \[ y = \frac{1}{2}e^{u} - 4x + C_2 = \frac{1}{2}e^{2x} - 4x + C_2 \]

Apply the condition of the point \((0, -1)\)

Now, we apply the condition that the curve passes through the point \((0, -1)\). We substitute \(x=0\) and \(y=-1\) to find the value of the constant \(C_2\): \[ -1 = \frac{1}{2}e^{2(0)} - 4(0) + C_2 \] Solving for \(C_2\): \[ C_2 = -\frac{1}{2} \]

Write the equation of the curve

With the values of constants \(C_1\) and \(C_2\), we can now write the equation of the curve: \[ y = \frac{1}{2}e^{2x} - 4x - \frac{1}{2} \] Thus, the equation of the curve is: \[ y = \frac{1}{2}e^{2x} - 4x - \frac{1}{2} \]

Key concepts

Integration by Substitution

Integration by substitution, often referred to as u-substitution, is a technique used to simplify the process of finding the antiderivative of a function. It's particularly helpful when dealing with composite functions, and it works by transforming a difficult integral into an easier one by changing the variable of integration.

Here's a step-by-step explanation of the substitution method used in the original exercise. We begin with the integral \[\int 8e^{2x} \mathrm{d}x\]. Realizing that the integral involves an exponential function whose exponent is a linear function of \(x\), we let \(u = 2x\), making the derivative \(\mathrm{d}u = 2\mathrm{d}x\). For the substitution to work, we need to express the original integral in terms of \(u\), so we rewrite the differential \(2\mathrm{d}x\) as \(\mathrm{d}u\), noticing the need to balance the equation, hence multiplying by \(\frac{1}{2}\).

The integral simplifies to \[\frac{1}{2}\int 8e^{u} \mathrm{d}u\], which is much easier to evaluate. After finding the antiderivative, we substitute back \(u = 2x\), obtaining \(4e^{2x} + C_1\) as the solution for the first derivative of our curve.

Understanding this method allows students to tackle a wide range of integrals they will encounter in calculus.

Horizontal Tangent

A horizontal tangent occurs when the slope of the curve at a particular point is zero. In calculus, the slope of a curve at any point is given by its first derivative. Finding where this derivative equals zero can tell us a lot about the behavior of the function at that point.

In the exercise, we are given that the curve has a horizontal tangent at the point \((0,-1)\). This means that the first derivative of the curve must be zero when \(x=0\). By setting the expression for the first derivative, \(4e^{2x} + C_1\), equal to zero with \(x=0\), we can solve for the constant \(C_1\) as was done in step 2 of the solution, giving us the value of \(C_1 = -4\).

This piece of information is critical as it provides us with a necessary condition to find the exact equation of our curve. By understanding the concept of horizontal tangents, students can learn to identify points of interest on curves, such as local maxima and minima, or even inflection points.

Second Derivative

The second derivative of a function is the derivative of the derivative. It measures how the rate of change of the function's slope itself changes. In other words, it provides information about the curvature of the function's graph, revealing whether it's concave up or down at a particular point. This can also indicate points of inflection where the curve changes from being concave up to concave down or vice versa.

In the given exercise, the second derivative of the curve is \(8e^{2x}\), which tells us that the slope of the tangent line changes at a rate proportional to the exponential function of \(2x\). We integrated this second derivative to find the first derivative, and using initial conditions, we found the equation of the curve. The step of finding the first derivative is crucial as it is needed to establish the relationship between the curve's slope and the horizontal tangent mentioned earlier. Recognizing the importance of the second derivative can significantly aid students in understanding curve shaping and the analysis of graphs for differentiable functions.