Problem 13 - A Course in Calculus and Real Analysis

Question

If \(x:=\int_{0}^{y} \frac{d t}{\sqrt{1+t^{2}}}\), find \(\frac{d^{2} y}{d x^{2}}\)

Step-by-Step Solution

Answer

The short answer is: \(\frac{d^2 y}{dx^2} = y\).

Step 1: Find the derivative of x with respect to y#

We are given that: \[x = \int_{0}^{y} \frac{dt}{\sqrt{1+t^2}}\] To find the derivative, use the Fundamental Theorem of Calculus and the given integral expression: \[\frac{dx}{dy}=\frac{1}{\sqrt{1+y^2}}\]

Step 2: Find the derivative of y with respect to x#

Now, we have the derivative of x with respect to y. To get the derivative of y with respect to x, we just need to take the reciprocal of this value: \[\frac{dy}{dx}=\sqrt{1+y^2}\]

Step 3: Find the second derivative of y with respect to x#

Finally, we'll find the second derivative by differentiating the expression for the first derivative, \(\frac{dy}{dx}\), with respect to x: \[\frac{d^2 y}{dx^2} = \frac{d}{dx}\left(\sqrt{1+y^2}\right)\] Note that we have a function of y, so we need to apply the chain rule for differentiation, which states that if we have a composite function h(x) = g(f(x)), then h'(x) = g'(f(x)) * f'(x). In this case, our composite function is \(\sqrt{1+y^2}\) and our inner function is y(x). So we have: \[\frac{d^2 y}{dx^2} = \frac{d}{dy}\left(\sqrt{1+y^2}\right) \cdot \frac{dy}{dx}\] Next, differentiate the outer function with respect to y: \[\frac{d}{dy}(\sqrt{1+y^2}) = \frac{1}{2}(1+y^2)^{-\frac{1}{2}} \cdot 2y\] Which simplifies to: \[\frac{y}{\sqrt{1+y^2}}\] So, the second derivative of y with respect to x is: \[\frac{d^2 y}{dx^2} = \frac{y}{\sqrt{1+y^2}} \cdot \sqrt{1+y^2}\] This simplifies to: \[\frac{d^2 y}{dx^2} = y\]

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on finding the accumulation of quantities. In particular, it focuses on determining the total change, area, or volume from a rate of change. The exercise here starts with an integral, \(x = \int_{0}^{y} \frac{dt}{\sqrt{1+t^2}}\), which represents the accumulation of changes from 0 to \(y\) for the function \(\frac{1}{\sqrt{1+t^2}}\). This specific integral involves a trigonometric identity since it is similar to the derivative of arcsinh or the inverse hyperbolic sine function.

• In integral calculus, the goal is often to compute definite integrals, which reflects the total accumulation between two points.
• It helps in solving real-world problems involving areas under curves and accumulated values over an interval.

With this integral, you essentially get a net effect from starting point 0 to an upper limit \(y\), which is crucial for understanding changes over intervals in many scientific fields.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with that of integration, serving as a critical component in calculus. In our exercise, we apply it to determine \(\frac{dx}{dy}\). This theorem ensures that the derivative of an integral function equipped with a variable upper limit returns the original integrand.

• Part 1 of the theorem states that if a function \(f\) is continuous over an interval \([a, b]\), then the function \(F\), defined as \(F(x) = \int_{a}^{x} f(t) \, dt\), is continuous over \([a, b]\), differentiable over \((a, b)\), and \(F'(x) = f(x)\).
• This exercise utilizes this concept when switching from the integral to the differential, finding \(\frac{dx}{dy} = \frac{1}{\sqrt{1+y^2}}\).

It's this powerful connection that allows us to easily move between the processes of integration and differentiation, proving crucial in solving the problem at hand.

Chain Rule

The chain rule is a fundamental method in calculus for finding the derivative of composite functions. This rule is vital in our solution, particularly when deriving the second derivative, \(\frac{d^2 y}{dx^2}\), and handling compositions such as \(\sqrt{1+y^2}\).
The chain rule states that if you have two functions combined, \(h(x) = g(f(x))\), then the derivative is \(h'(x) = g'(f(x)) \cdot f'(x)\). We apply this principle when finding the second derivative by differentiating \(\sqrt{1+y^2}\) with respect to \(y\) and multiplying it by \(\frac{dy}{dx}\).

• This is crucial because it allows us to differentiate more complex functions that are built up from simpler ones.
• In the given case, differentiating \(\sqrt{1+y^2}\) involves calculating \(\frac{d}{dy}\left(\sqrt{1+y^2}\right)\), which simplifies due to the chain rule to \(\frac{y}{\sqrt{1+y^2}}\).

The chain rule helps manage the complexity by giving a reliable method for breaking down and solving derivatives of these composite functions. This results in accurately determining \(\frac{d^2 y}{dx^2} = y\), the ultimate goal of the problem.