Question

Show the differential equation of the system of parabolas \(\mathrm{y}^{2}=4 \mathrm{a}(\mathrm{x}-\mathrm{b})\) is given by \(y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=0\).

Short answer

Question: Show that the system of parabolas represented by the equation \(y^2 = 4a(x-b)\) can be represented by the differential equation \(y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0\). Answer: By finding the first derivative (\(\frac{dy}{dx} = \frac{2a}{y}\)) and second derivative (\(\frac{d^2y}{dx^2} = -\frac{4a^2}{y^3}\)) of the given equation, we showed that when these expressions are substituted into the provided differential equation, the equation holds true. Therefore, the system of parabolas \(y^2 = 4a(x-b)\) can indeed be represented by the differential equation \(y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0\).

Step-by-step solution

Write the equation of the parabola.

The provided equation for the system of parabolas is \(y^2 = 4a(x-b)\).

Differentiate y with respect to x. (First derivative)

Differentiate both sides of the equation with respect to x. Using chain rule on left-hand side and product rule on the right-hand side, we have: \(\frac{d}{dx}(y^2) = \frac{d}{dx}(4a(x-b))\) Or, \(2y \frac{dy}{dx} = 4a(1)\) Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{2a}{y}\)

Differentiate y' with respect to x. (Second derivative)

Differentiate the expression for \(\frac{dy}{dx}\) with respect to x: \(\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{2a}{y}\right)\) This expression requires the use of the quotient rule, \([\frac{d}{dx}(u/v) = \frac{vu' - uv'}{v^2}]\). In this case, \(u = 2a\) and \(v = y\). To find \(u'\) and \(v'\), we again differentiate, this time with respect to \(x\). Note that \(a\) is a constant, so \(u' = 0\). And we already found \(y' = \frac{dy}{dx} = \frac{2a}{y}\) in step 2. Therefore, applying the quotient rule, we get: \(\frac{d^2y}{dx^2} = \frac{y(0) - (2a)(\frac{2a}{y})}{y^2}\) Simplifying, \(\frac{d^2y}{dx^2} = -\frac{4a^2}{y^3}\)

Check if the given differential equation is satisfied.

We have derived expressions for \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\). Now we will be substituting their expressions into the provided differential equation, \(y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0\). If it holds true, our derivation is correct. Substituting and simplifying, we get: \(y\left(-\frac{4a^2}{y^3}\right) + \left(\frac{2a}{y}\right)^2 = 0\) \(-\frac{4a^2}{y^2} + \frac{4a^2}{y^2} = 0\) The given differential equation holds true, so our derivation is correct. Indeed, the system of parabolas \(y^2 = 4a(x-b)\) can be represented by the differential equation \(y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0\).

Key concepts

Derivative of Parametric Functions

In calculus, parametric functions are a way of defining a function using parameters. For the equation of a parabola given as \(y^2 = 4a(x-b)\), we see that both \(x\) and \(y\) are expressed in terms of another variable, which could be a parameter. To find the derivative of \(y\) with respect to \(x\) in this context, we first express \(y\) implicitly in terms of \(x\) and then differentiate.
When differentiating parametric functions, one useful approach is to differentiate each function separately with respect to the parameter and then use the derivative of one function with respect to the other. This is how we found \(\frac{dy}{dx} = \frac{2a}{y}\). It provides a way to understand how two related rates of change, in this case, the rates of change of \(x\) and \(y\), work together within a system.

Chain Rule in Differentiation

The chain rule is a fundamental tool in calculus for differentiating composite functions. In the context of our parabola \(y^2 = 4a(x-b)\), the chain rule allows us to differentiate a function of a function. For example, \(y^2\) is a function of \(y\), which in turn is a function of \(x\).
The chain rule states that if \(h(x) = f(g(x))\), then \(h'(x) = f'(g(x))\cdot g'(x)\). When we differentiated the equation of the parabola, we applied the chain rule to \(y^2\), treating \(y\) as a function of \(x\), and \(y^2\) as a composite function. This gives us the first derivative \(2y\frac{dy}{dx}\) as part of our step-by-step solution.

Quotient Rule in Differentiation

The quotient rule is used in differentiation when we have a function that is the quotient of two other functions. If we have a function \(\frac{u(x)}{v(x)}\), its derivative with respect to \(x\) is given by \(\frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}\).
In the solution, this rule is applied when finding the second derivative \(\frac{d^2y}{dx^2}\). Since we have the derivative of \(y\) already found, \(\frac{dy}{dx} = \frac{2a}{y}\), and by identifying \(u = 2a\) and \(v = y\), we could apply the quotient rule to obtain the second derivative. This highlights the value of understanding and correctly applying the quotient rule to physical problems or, in this case, the shape of a parabola described in a parametric form.

Integral Calculus Applications

While not directly used in solving the given problem, integral calculus is an important counterpart to differential calculus, providing tools for finding areas, volumes, and central points among other things. The area under a curve, for instance, is found by integrating the function describing the curve.
If we had the equation of the parabola and needed to find the area enclosed by the parabola and the x-axis, we would set up an integral with the appropriate limits. Similarly, if we want to revert back from the velocity function (which is a derivative of the position function) to the position function, we would integrate the velocity function. Understanding integrals is crucial as they often represent the 'accumulated effect', which has many practical applications in physics, engineering, economics, and beyond. Integral calculus is used to solve a wide range of problems where we aim to find the whole from the change or the rate of change.