Question

For the curve $y=\sqrt{x},$ between $x=0$ and $x=2,$ find: The arc length.

Short answer

The arc length is 4.

Step-by-step solution

Understand the arc length formula

For a curve described by the function y = f(x) the arc length from x = a to x = b is given by the integral: $$L={\int }_a^b\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$$

Find the derivative of the function

For the curve y = \sqrt{x} the derivative is: $$\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$$

Substitute the derivative into the arc length formula

Substitute $\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$ into the arc length formula: $$L={\int }_0^2\sqrt{1+{\left(\frac{1}{2\sqrt{x}}\right)}^2}dx$$ This simplifies to: $$L={\int }_0^2\sqrt{1+\frac{1}{4x}}dx$$

Simplify the integrand expression

Combine the terms inside the square root: $$L={\int }_0^2\sqrt{\frac{4x+1}{4x}}dx$$ Which can be simplified to: $$L={\int }_0^2\sqrt{\frac{4x+1}{4x}}={\int }_0^2\frac{\sqrt{4x+1}}{2\sqrt{x}}dx$$

Perform a u-substitution

Let $u=4x+1$. Then $du=4dx$, and $dx=\frac{du}{4}$. Also, when $x=0$, $u=1$ and when $x=2$, $u=9$. Substituting these values into the integral gives: $$L={\int }_1^9\frac{\sqrt{u}}{2\sqrt{\frac{u-1}{4}}}\cdot \frac{du}{4}={\int }_1^9\frac{\sqrt{u}}{\frac{2\sqrt{u-1}}{2}}\cdot \frac{du}{4}={\int }_1^9\frac{\sqrt{u}}{\sqrt{u-1}}\cdot \frac{du}{4}={\int }_1^9\frac{du}{2}$$

Integrate and evaluate the definite integral

Integrate and evaluate: $$L={\int }_1^9\frac{du}{2}=\frac{1}{2}(u){|}_1^9=\frac{1}{2}(9-1)=4$$

Key concepts

arc length formula

To find the length of a curve, we use the arc length formula. The arc length $L$ of a function $y=f(x)$ from $x=a$ to $x=b$ is given by:
$$L={\int }_a^b\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$$ This formula provides the total length of the curve by integrating the square root of the sum of 1 and the square of the derivative of the function.
It's a way to measure the distance along the curve rather than a straight line. First, you need to compute the derivative $\frac{dy}{dx}$ and then square it to substitute back into the formula.

• Arc length calculates the actual 'distance' along the curve.
• Derivatives play a crucial role in this formula.
• Understanding how to manipulate and simplify the integrand is essential.

u-substitution

U-substitution is a method for simplifying integrals by substituting a part of the integrand with a new variable, usually denoted as u. This technique is very handy in integrals that involve complicated expressions.
Let's take an example from the arc length problem:

We had an integral \int_0^2 \frac{\sqrt{4x + 1}}{2\sqrt{x}} \, dx\. To simplify, we set $u=4x+1$. Then, our differential $dx$ changes as well: $du=4dx$ or $dx=\frac{du}{4}$.
When $x=0$, $u=1$, and when $x=2,u=9$. This transforms the integral into:

• The variable limits change according to the new substitution.
• The integral becomes simpler and often more manageable to solve.
• Finally, involve back-substitution to finish the integration.

definite integral evaluation

Definite integrals are used to find the total value of the function over a specific interval. These integrals have upper and lower limits and are noted as follows: \int_{a}^{b} f(x) \, dx\. They give the net area between the function graph and the x-axis.
Using our example problem, after performing u-substitution, we got the integral: $$L={\int }_1^9\frac{du}{2}$$ This is much simpler to evaluate. The antiderivative of $\frac{1}{2}u$ is straightforward:
$$\frac{1}{2}(u){|}_1^9$$ Finally, substitute the limits of integration into the antiderivative:
$$\frac{1}{2}(9)-\frac{1}{2}(1)=4$$ The definite integral evaluates to 4, which gives us the arc length.

• Evaluation turns the integral function value into tangible results.
• Consistent simplification and back-substitution are key.