Question

The function \(G(t)\) is defined by $$ G(t)=F(x, y)=x^{2}+y^{2}+3 x y $$ where \(x(t)=a t^{2}\) and \(y(t)=2 a t\). Use the chain rule to find the values of \((x, y)\) at which \(G(t)\) has stationary values as a function of \(t\). Do any of them correspond, to the stationary points of \(F(x, y)\) as a function of \(x\) and \(y\) ?

Short answer

The point (0,0) is the common stationary point for both G(t) and F(x,y).

Step-by-step solution

Write out the given functions

The function is defined as follows: \[ G(t) = F(x, y) = x^2 + y^2 + 3xy \] where \[ x(t) = a t^2 \] and \[ y(t) = 2a t \].

Express G(t) in terms of t

Substitute \(x(t)\) and \(y(t)\) into \(G(t)\): \[ G(t) = (a t^2)^2 + (2a t)^2 + 3(a t^2)(2a t) \] Simplifying, we get \[ G(t) = a^2 t^4 + 4a^2 t^2 + 6a^2 t^3 \].

Differentiate G(t) with respect to t

To find the stationary values, compute \(\frac{dG}{dt}\): \[ \frac{dG}{dt} = 4a^2 t^3 + 8a^2 t + 18a^2 t^2 \].

Solve \(\frac{dG}{dt} = 0\)

Set the derivative equal to zero and solve for \(t\): \[ 4a^2 t^3 + 18a^2 t^2 + 8a^2 t = 0 \]. Factor out \(4a^2 t\): \[ 4a^2 t ( t^2 + 4.5 t + 2) = 0 \]. This gives the solutions: \[ t = 0 \] and solving the quadratic equation \[ t^2 + 4.5 t + 2 = 0 \], we find: \[ t = \frac{-4.5 \pm \sqrt{4.5^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{-4.5 \pm 3.605}{2} \] \[ t = -0.447, -4.05 \].

Calculate the corresponding (x, y) points

Using \( x(t) = a t^2 \) and \( y(t) = 2a t \), find the \((x,y)\) pairs for \( t = 0, -0.447, -4.05 \): For \( t = 0 \), \( x(0) = 0 \), \( y(0) = 0 \). For \( t = -0.447 \), \( x(-0.447) = a ( -0.447)^2 \approx 0.2a \), \( y(-0.447) = 2a (-0.447) \approx -0.894a \). For \( t = -4.05 \), \( x(-4.05) = a (-4.05)^2 \approx 16.4a \), \( y(-4.05) = 2a (-4.05) \approx -8.1a \).

Check for stationary points of F(x, y)

To find the stationary points of \( F(x,y) \) as a function of \( x \) and \( y \), find the critical points by solving \( \frac{\partial F}{\partial x} = 0 \) and \( \frac{\partial F}{\partial y} = 0 \): \[ \frac{\partial F}{\partial x} = 2x + 3y = 0 \], \[ \frac{\partial F}{\partial y} = 2y + 3x = 0 \].

Solve the system of equations

Solve the system of linear equations: From the first equation: \( x = \frac{-3y}{2} \). Substitute into the second equation: \[ 2y + 3\left( \frac{-3y}{2} \right) = 0 \] \[ 2y - \frac{9y}{2} = 0 \] \[ \left( 2 - \frac{9}{2} \right)y = 0 \] \[ y(2 - 4.5) = 0 \] \[ y = 0 \]. Substitute \( y = 0 \) into \( x = \frac{-3y}{2} \): \[ x = 0 \]. So the stationary point for \( F(x,y) \) is \( (0,0) \).

Compare stationary points

Compare the stationary points obtained from \( G(t) \) with \( F(x,y) \): The stationary points from \( G(t) \) were \((0,0)\), \((0.2a, -0.894a)\), and \((16.4a, -8.1a)\). The stationary point from \( F(x,y) \) is \((0,0)\). Only the point \((0,0)\) is common in both cases.

Key concepts

Stationary Values

In calculus, stationary values are points where the derivative of a function is zero. These points indicate where the function may have local maxima, local minima, or saddle points. To find them, we usually set the derivative equal to zero and solve for the variable in question. For the function in the exercise: G(t) = F(x, y) = x^2 + y^2 + 3xy To find the stationary values of G(t) with respect to t, we first need to express G(t) solely in terms of t and then differentiate it with respect to t. Setting this derivative equal to zero provides the stationary points.

Partial Derivatives

When dealing with functions of multiple variables, such as F(x, y) in this exercise, partial derivatives are essential. A partial derivative of a function with respect to one variable is the derivative of that function while keeping the other variables constant. For example, the partial derivatives of F(x, y) = x^2 + y^2 + 3xy are: ∂F/∂x = 2x + 3y ∂F/∂y = 2y + 3x

System of Equations

A system of equations is a collection of two or more equations with a common set of variables. In the context of finding stationary values or critical points, solving a system of equations means finding where all the derivatives equal zero simultaneously. For our function F(x, y), we have the system of linear equations: 2x + 3y = 0 2y + 3x = 0 By solving this, we determine where both partial derivatives are zero. From this example, substituting x and y respectively from one equation into the other simplifies this system, helping us find critical points more efficiently.

Differentiation

Differentiation is a core principle in calculus used to find the rate at which a function changes. It's essential for determining gradients, slopes, and more importantly for identifying stationary points or critical points in functions. In this exercise, we used differentiation twice: first, to find the derivative of G(t) with respect to t, and then to find the partial derivatives of F(x, y) with respect to x and y. ∂F/∂x gives the slope of F in the x-direction, while ∂F/∂y gives the slope in the y-direction. Together they help find where the function levels out.