Calculating The Arc Length Of A Curve Defined By An Integral

Hey there, math enthusiasts! Today, we're diving into a cool problem that combines calculus and trigonometry to find the length of a curve. Buckle up, because we're going to tackle this step-by-step, making sure everything is crystal clear.

Problem Statement

We're given a curve defined by the equation:

$$x = \int_0^y \sqrt{10 \sec^4 t - 1} \, dt$$

Our mission, should we choose to accept it, is to find the length of this curve over the interval $-\frac{\pi}{4} \leq y \leq \frac{\pi}{4}$.

Understanding Arc Length

Before we jump into the nitty-gritty, let's take a moment to refresh our understanding of arc length. Arc length is essentially the distance along a curve. Imagine you're walking along a winding path; the arc length is the total distance you've traveled. In calculus, we have a handy formula to calculate this.

For a curve defined by $x = f(y)$, the arc length $L$ over the interval $a \leq y \leq b$ is given by:

$$L = \int_a^b \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy$$

This formula might look a bit intimidating, but don't worry, we'll break it down. The key is to find the derivative $\frac{dx}{dy}$ and then plug it into the formula. Let's get started!

Step 1: Finding ${\frac{dx}{dy}}$

The first crucial step in finding the arc length is to determine the derivative of $x$ with respect to $y$, denoted as $\frac{dx}{dy}$. We are given the equation:

$$x = \int_0^y \sqrt{10 \sec^4 t - 1} \, dt$$

Notice that $x$ is defined as an integral with respect to $t$, and the upper limit of integration is $y$. This is where the Fundamental Theorem of Calculus comes to our rescue. The Fundamental Theorem of Calculus (Part 1) states that if we have a function defined as an integral with a variable upper limit, the derivative of that function is simply the integrand evaluated at the upper limit. In simpler terms, it helps us to differentiate an integral.

Applying the Fundamental Theorem of Calculus, we get:

$$\frac{dx}{dy} = \frac{d}{dy} \int_0^y \sqrt{10 \sec^4 t - 1} \, dt = \sqrt{10 \sec^4 y - 1}$$

That wasn't so bad, was it? The derivative $\frac{dx}{dy}$ is simply the expression inside the square root, but with $t$ replaced by $y$. This is a direct application of the Fundamental Theorem of Calculus, which is a cornerstone concept in calculus. This theorem provides a powerful link between differentiation and integration.

Step 2: Plugging into the Arc Length Formula

Now that we have $\frac{dx}{dy}$, we can plug it into the arc length formula we discussed earlier. Remember, the arc length formula for a curve defined by $x = f(y)$ over the interval $a \leq y \leq b$ is:

$$L = \int_a^b \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy$$

In our case, $\frac{dx}{dy} = \sqrt{10 \sec^4 y - 1}$, and the interval is $-\frac{\pi}{4} \leq y \leq \frac{\pi}{4}$. Substituting these values into the formula, we get:

$$L = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sqrt{1 + \left(\sqrt{10 \sec^4 y - 1}\right)^2} \, dy$$

Let's simplify the expression inside the integral. Squaring the square root eliminates the radical, giving us:

$$L = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sqrt{1 + (10 \sec^4 y - 1)} \, dy$$

The $1$ and $-1$ cancel out, further simplifying the integral:

$$L = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sqrt{10 \sec^4 y} \, dy$$

We can take the square root of $10$ and $\sec^4 y$ separately:

$$L = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sqrt{10} \cdot \sqrt{\sec^4 y} \, dy$$

\sqrt{\sec^4 y}$ simplifies to $\sec^2 y$, so we have: $L = \sqrt{10} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2 y \, dy

Step 3: Evaluating the Integral

Now we need to evaluate the integral of $\sec^2 y$ with respect to $y$. Recall that the derivative of the tangent function, $\tan y$, is $\sec^2 y$. Therefore, the antiderivative of $\sec^2 y$ is $\tan y$. This is a fundamental integral in calculus that's crucial to remember.

So, we have:

$$L = \sqrt{10} \left[ \tan y \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}$$

To evaluate the definite integral, we subtract the value of the antiderivative at the lower limit from its value at the upper limit:

$$L = \sqrt{10} \left( \tan \left(\frac{\pi}{4}\right) - \tan \left(-\frac{\pi}{4}\right) \right)$$

We know that $\tan(\frac{\pi}{4}) = 1$ and $\tan(-\frac{\pi}{4}) = -1$. Substituting these values, we get:

$$L = \sqrt{10} (1 - (-1))$$

$$L = \sqrt{10} (1 + 1)$$

$$L = 2\sqrt{10}$$

Final Answer

Therefore, the length of the curve is $2\sqrt{10}$.

Conclusion

And there you have it! We successfully found the length of the curve using the arc length formula and the Fundamental Theorem of Calculus. This problem showcases how different concepts in calculus can come together to solve interesting problems. The key takeaways from this exercise are the application of the arc length formula, the use of the Fundamental Theorem of Calculus, and the importance of recognizing basic trigonometric integrals. Keep practicing, and you'll become a calculus whiz in no time!

Remember, understanding the underlying concepts is just as important as memorizing formulas. So, keep exploring, keep questioning, and keep learning!