Question

Show that \(y(x)=x a^{2 x} \exp x^{2}\) has no stationary points other than \(x=0\), if \(\exp (-\sqrt{2})

Short answer

The only stationary point is x = 0 because the polynomial \( 1 + 2x ln(a) + 2x^2 \) has no real roots for \(\exp(- \sqrt{2}) < a < \exp(\sqrt{2})\) except for x=0.

Step-by-step solution

- Find the First Derivative

To show that the function has no stationary points other than at x=0, we first need to find the first derivative of y(x). Given: \[ y(x)=x a^{2x} \exp(x^2) \]We need to apply the product rule for differentiation: \[ \frac{d}{dx}(u \cdot v \cdot w) = u'vw + uv'w + uvw' \]Here, let \( u = x \), \( v = a^{2x} \), and \( w = \exp(x^2) \). So we differentiate each function separately and apply the product rule:\[\frac{dy}{dx} = \frac{d}{dx}(x) \cdot a^{2x} \exp(x^2) + x \cdot \frac{d}{dx}(a^{2x}) \cdot \exp(x^2) + x \cdot a^{2x} \cdot \frac{d}{dx}(\exp(x^2))\]}\[ \frac{dy}{dx}=a^{2x} \exp(x^2) + x (2a^{2x} ln(a)) \cdot \exp(x^2) + x\cdot a^{2x} \cdot (2x\exp(x^2)) \]

- Simplify the First Derivative

Combine the terms to get a single expression for the first derivative:\[ \frac{dy}{dx}=a^{2x} \exp(x^2) + 2x a^{2x} ln(a) \exp(x^2) + 2x^2 a^{2x} \exp(x^2) \] Factor out the common terms:\[ \frac{dy}{dx} = a^{2x} \exp(x^2) (1 + 2x ln(a) + 2x^2) \]

- Set the First Derivative to Zero and Solve

For stationary points, set the first derivative equal to zero:\[ a^{2x} \exp(x^2) (1 + 2x ln(a) + 2x^2) = 0 \]Since \(a^{2x}\) and \(\exp(x^2)\) are never zero for any real x, we only need to solve for when the polynomial term is zero:\[ 1 + 2x ln(a) + 2x^2 = 0 \]

- Analyze the Polynomial for Real Roots

We need to determine if this quadratic equation has any real roots for \( x eq 0 \). The discriminant of a quadratic equation \( ax^2 + bx + c \) is given by \( b^2 - 4ac \). Here, \( a = 2 \), \( b = 2 ln(a) \), and \( c = 1 \):\[ b^2 - 4ac = (2 ln(a))^2 - 4 \cdot 2 \cdot 1 \]\[ = 4 (ln(a))^2 - 8 \]Given that \( \exp(- \sqrt{2}) < a < \exp(\sqrt{2}) \), it can be checked that the discriminant is less than 0, which means there are no real roots.Hence the only stationary point is \( x = 0 \).

Key concepts

differentiation

Differentiation is a core concept in calculus that involves finding the rate at which a function changes at any given point. This is particularly essential in physics for understanding how variables like speed and acceleration change over time. In the given exercise, we use differentiation to find the first derivative of the function \( y(x) = x a^{2x} \exp{(x^2)} \). To do this, we apply the product rule, which states that the derivative of a product of functions is given by: \( \frac{d}{dx}(u \, \cdot \, v \, \cdot \, w) = u'vw + uv'w + uvw' \). By differentiating each function separately and applying the product rule, we obtain the derivative: \( \frac{dy}{dx} = a^{2x} \exp{(x^2)} (1 + 2x \ln(a) + 2x^2) \), which is then used to find the stationary points.

stationary points

A stationary point on a curve is where the derivative \( \frac{dy}{dx} \) equals zero, meaning the slope of the tangent to the curve is horizontal at that point. They are crucial for identifying maxima, minima, or points of inflection. In the exercise, we set the first derivative \( \frac{dy}{dx} \) to zero to find stationary points. After simplification, we get the polynomial equation \( 1 + 2x \ln(a) + 2x^2 = 0 \). To check for real roots, we evaluate the discriminant of this quadratic equation, \( (2 \ln(a))^2 - 8 \). Given the condition \( \exp(- \sqrt{2}) < a < \exp(\sqrt{2}) \), we find that the discriminant is less than zero, indicating no real roots except at \( x = 0 \). Thus, the only stationary point is at \( x = 0 \).

parametric equations

Parametric equations express the coordinates of the points on a curve as functions of a parameter. They are widely used in physics to describe trajectories of objects, among other things. In the exercise, the curve \( 4y^3 = a^2(x + 3y) \) is parameterized as \( x = a \cos{3\theta}, y = a \cos{\theta} \). By re-writing equations in terms of parameter \( \theta \), it becomes easier to differentiate and analyze curves, especially for complex shapes like trajectories in electric and magnetic fields, as shown in the particle motion example. We also verify that expressions obtained by implicit differentiation match those in parameterized form, ensuring consistency in results.

inflection points

An inflection point is where the curve changes concavity, i.e., from concave up to concave down or vice versa. These points are significant in understanding the shape and behavior of graphs. To find inflection points, we need to analyze the second derivative \( \frac{d^2y}{dx^2} \). In the provided exercise, after finding the first derivative and setting it to zero for stationary points, we explore the second derivative to detect concavity changes. For the curve \( 4y^3 = a^2(x + 3y) \), it’s shown that the only point of inflection occurs at the origin. Since this point is also a stationary point, it’s considered a stationary point of inflection, informing how the curve behaves near \( (0,0) \).