Question

Find the length of the curve \(y=a x^{2}\) from \(x=0\) to \(x=10,\) where \(a>0\) is a real number.

Short answer

Answer: \(L=\frac{1}{2a}sinh^{-1}(20a)\)

Step-by-step solution

Find the derivative of y=ax^2

First step is to find the derivative of the given function with respect to x. Given, \(y=ax^2\). Then, \(\frac{dy}{dx} = 2ax\).

Set up the integral

Next, we need to set up the integral to find the length of the curve. The formula to find the length of a curve is: $$L=\int_a^b \sqrt{1+(\frac{dy}{dx})^2} dx$$ Plugging our derivative, \(\frac{dy}{dx}=2ax\), we get: $$L=\int_0^{10} \sqrt{1+(2ax)^2} dx$$

Simplify the integral

Now, let's simplify the equation inside the square root. $$L=\int_0^{10} \sqrt{1+4a^2x^2} dx$$

Find the antiderivative

To find the antiderivative, we will use substitution method. Let \(u=2ax\). Then, \(du=2adx\Rightarrow dx=\frac{du}{2a}\). Substituting, we get: $$L=\int \sqrt{1+u^2} \frac{du}{2a}$$ The limits of integration will change as well: \(u(0)=0\) and \(u(10)=20a\). Incorporating the constant \(2a\) and the new limits, the integral becomes: $$L=\frac{1}{2a}\int_0^{20a} \sqrt{1+u^2} du$$

Evaluate the integral

Unfortunately, this integral cannot be solved directly using elementary functions. However, the integral \(\int \sqrt{1+u^2} du\) represents the arc length of a hyperbolic function, which is expressed as the hyperbolic sine inverse function, written as \(sinh^{-1}(u)+C\). With this knowledge, we can now evaluate the integral: $$L=\frac{1}{2a}[sinh^{-1}(u)\Big|_0^{20a}]$$

Evaluate the limits and find the final length

Now, we will substitute the limits and find the final length of the curve. $$L=\frac{1}{2a}[sinh^{-1}(20a)-sinh^{-1}(0)]$$ Since \(sinh^{-1}(0)=0\), the expression becomes: $$L=\frac{1}{2a}sinh^{-1}(20a)$$ So, the length of the curve \(y=ax^2\) from \(x=0\) to \(x=10\) is given by $$L=\frac{1}{2a}sinh^{-1}(20a)$$

Key concepts

Curve Length

The concept of curve length in calculus refers to finding the total distance along a curve from one point to another. This concept is important because it extends our ability to measure lengths from straight lines to curved paths. The formula for finding the curve length is critical:

• For a function \( y = f(x) \), the length \( L \) of the curve from \( x = a \) to \( x = b \) is given by the integral: \[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]

In the given problem, we are calculating the length of the parabola \( y = ax^2 \) from \( x = 0 \) to \( x = 10 \). By applying the standard curve length formula, you incorporate the derivative of the function and evaluate the integral within the specified limits. This integral provides a clear measurement of how far a point on the curve travels. Understanding this process helps visualize how mathematical principles can quantify real-world phenomena like paths, trajectories, or edges.

Derivative

Derivatives represent a fundamental concept in calculus and are used to find the rate of change or slope of a function. With the exercise, \( y = ax^2 \), the derivative is calculated to understand how \( y \) changes with respect to \( x \). This is an essential step in defining the curve's geometry:

• To derive \( y = ax^2 \), use the power rule: \( \frac{dy}{dx} = 2ax \).

The derivative \( \frac{dy}{dx} = 2ax \) helps us find the slope of the tangent at any point on the parabola. This slope is then used in the curve length formula. The slope impacts the arc's steepness and hence, affects the curve length computation. Computing derivatives forms a basis for more complex calculations like integrals involving length, area, and volume in calculus. Understanding derivatives makes it easier to predict the behavior of curves and their transformations.

Integral

In calculus, integrals help in finding quantities like areas under curves and, in this case, the length of curves. The integral we set up is based on the curve length formula and includes the derivative:

• The integral for curve length \( L \) over the interval \( [0, 10] \) is given as: \[ L = \int_0^{10} \sqrt{1 + (2ax)^2} \, dx \]

This integral involves evaluating the expression under the square root, \( 1 + 4a^2x^2 \). Solving this requires a substitution method to simplify it to a manageable form. Integration accommodates the sum of infinitesimal pieces of our curve, providing a holistic measure of its entire length. While some integrals can be solved using elementary methods, others, like the one in this exercise, require special functions, particularly hyperbolic functions. This shows how calculus evolves from basic operations to sophisticated analyses.

Hyperbolic Functions

Hyperbolic functions, analogous to trigonometric functions, often appear in calculus involving special integrals. When finding the arc length of a hyperbolic curve or expressions involving square roots like \( \sqrt{1+u^2} \), hyperbolic functions provide solutions:

• For example, \( \int \sqrt{1 + u^2} \, du \) leads to \( \sinh^{-1}(u) + C \), the inverse hyperbolic sine.

In the solution, this hyperbolic function helps express the integral's solution that can't be resolved using regular algebraic methods. When we compute hyperbolic functions, it involves evaluating ratios of exponential functions, which handle complex scenarios with greater ease. Knowing hyperbolic functions and their properties makes them valuable for engineers and physicists, dealing with realistic problems involving catenary curves, relativity, and other applications. Understanding their basic definitions, identities, and properties can vastly improve problem-solving capabilities.