Arc Length Of Parametric Curve X = E^t, Y = E^{-t}, Z = \sqrt{2}t
Hey guys! Today, we're diving into the fascinating world of parametric curves and learning how to calculate their arc length. Specifically, we'll be tackling the curve defined by the parametric equations x = e^t, y = e^{-t}, and z = \sqrt{2}t, over the interval 0 ≤ t ≤ 2. This might seem a bit daunting at first, but trust me, with a little calculus magic, we'll unravel this problem together. We'll break down each step, making sure you grasp the underlying concepts and can confidently apply them to other similar problems. So, grab your thinking caps, and let's embark on this mathematical adventure!
Understanding Parametric Curves and Arc Length
Before we jump into the calculations, let's take a moment to understand what parametric curves are and what arc length represents. Imagine a point moving in three-dimensional space. Instead of describing its position directly as a function of x, y, and z, we use a parameter, usually denoted as 't', to define the coordinates. Think of 't' as time – as time changes, the point traces out a curve in space. Our given equations, x = e^t, y = e^{-t}, and z = \sqrt{2}t, are examples of this. Each value of 't' corresponds to a specific point (x, y, z) on the curve. So, the parametric equations serve as a set of instructions, dictating where the point should be in space at any given moment in time. Now, the arc length is simply the distance the point travels along this curve over a specific interval of 't'. It's like measuring the length of a winding road on a map. We're not just interested in the straight-line distance between the starting and ending points, but the actual distance covered along the curvy path. In this case, we want to find the arc length of the curve traced out by our equations as 't' varies from 0 to 2. This concept has practical applications in various fields, such as physics (calculating the distance traveled by a particle), computer graphics (determining the length of a curve for rendering), and engineering (designing smooth curves for roads and bridges). Understanding parametric curves and arc length isn't just an academic exercise; it's a powerful tool for describing and analyzing motion and shapes in the real world.
The Arc Length Formula: Your New Best Friend
Now that we have a solid understanding of what we're trying to find, let's introduce the arc length formula. This formula is the key to unlocking the solution to our problem. For a parametric curve defined by x = f(t), y = g(t), and z = h(t), where 't' varies from 'a' to 'b', the arc length (L) is given by the following integral:
L = ∫[a, b] √[(dx/dt)² + (dy/dt)² + (dz/dt)²] dt
Don't let this formula intimidate you! Let's break it down piece by piece. The integral symbol (∫) tells us we're summing up infinitesimally small segments of the curve's length. The limits of integration, 'a' and 'b', define the interval over which we're calculating the arc length (in our case, 0 and 2). Inside the integral, we have a square root containing three terms: (dx/dt)², (dy/dt)², and (dz/dt)². These terms represent the squares of the derivatives of our parametric equations with respect to 't'. In simpler terms, we're finding how quickly each coordinate (x, y, and z) is changing as 't' changes. The square root combines these rates of change to give us the overall rate of change of the curve's position, which is related to its speed. Finally, multiplying this rate of change by dt (an infinitesimally small change in 't') gives us an infinitesimally small segment of the arc length. Integrating over the interval [a, b] then sums up all these tiny segments to give us the total arc length. So, the formula is essentially a way of adding up tiny pieces of the curve's length to find the total length. It's a powerful tool that allows us to calculate the length of even complex curves, as long as we can find the derivatives of the parametric equations and evaluate the integral. This formula will be our workhorse in solving our specific problem, so it's crucial to understand each part of it.
Calculating the Derivatives: The First Step to Success
Alright, guys, the arc length formula is our roadmap, and now it's time to start our journey! The first crucial step in calculating the arc length is finding the derivatives of our parametric equations with respect to 't'. Remember, our equations are: x = e^t, y = e^-t}, and z = \sqrt{2}t. We need to find dx/dt, dy/dt, and dz/dt. Let's tackle them one by one. Starting with x = e^t, the derivative dx/dt is simply e^t. This is a fundamental derivative that you'll encounter frequently in calculus. The exponential function e^t has the unique property that its derivative is itself. Moving on to y = e^{-t}, we need to apply the chain rule. The chain rule states that if we have a composite function, the derivative is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In this case, the outer function is e^u and the inner function is u = -t. The derivative of e^u is e^u, and the derivative of -t is -1. So, applying the chain rule, we get dy/dt = e^{-t} * (-1) = -e^{-t}. Finally, for z = \sqrt{2}t, the derivative dz/dt is simply \sqrt{2}. This is because the derivative of a constant times 't' is just the constant. So, we've successfully found our derivatives, and dz/dt = \sqrt{2}. These derivatives represent the instantaneous rates of change of the x, y, and z coordinates with respect to 't'. They are the building blocks we need to plug into the arc length formula and get one step closer to our final answer. Now that we have these derivatives, the next step is to square them, which will prepare us for the square root in the formula.
Squaring the Derivatives: Preparing for the Square Root
Great job on finding those derivatives! Now, before we can plug them into the arc length formula, we need to square each of them. This step is crucial because the formula involves the sum of the squares of the derivatives under a square root. So, let's take our derivatives – dx/dt = e^t, dy/dt = -e^-t}, and dz/dt = \sqrt{2} – and square them. Squaring dx/dt = e^t, we get (dx/dt)² = (e^t)² = e^(2t). Remember the rule of exponents, we get (dy/dt)² = (-e^{-t})² = e^(-2t). Here, the negative sign disappears because squaring a negative number results in a positive number. Again, we use the rule of exponents to simplify. Squaring dz/dt = \sqrt{2}, we get (dz/dt)² = (\sqrt{2})² = 2. This is straightforward – the square of the square root of 2 is simply 2. So, we now have the squares of our derivatives: (dx/dt)² = e^(2t), (dy/dt)² = e^(-2t), and (dz/dt)² = 2. These squared terms represent the squares of the rates of change in each direction. Squaring them ensures that we're dealing with positive quantities, which is necessary when we take the square root in the arc length formula. These squared derivatives are the ingredients we need to add together under the square root, which will give us a measure of the overall rate of change of the curve's position. With these squares in hand, we're ready to move on to the next step: plugging everything into the arc length formula and simplifying.
Plugging into the Formula and Simplifying: Almost There!
Okay, team, we've done the hard work of finding the derivatives and squaring them. Now comes the exciting part – plugging everything into the arc length formula and simplifying! Remember our formula:
L = ∫[a, b] √[(dx/dt)² + (dy/dt)² + (dz/dt)²] dt
We know our limits of integration are a = 0 and b = 2. We also know (dx/dt)² = e^(2t), (dy/dt)² = e^(-2t), and (dz/dt)² = 2. Let's substitute these values into the formula:
L = ∫[0, 2] √[e^(2t) + e^(-2t) + 2] dt
Now, the expression under the square root looks a bit messy, but don't worry, there's a clever simplification we can make. Notice that e^(2t) + e^(-2t) + 2 looks suspiciously like a perfect square. In fact, it is! We can rewrite it as (e^t + e^{-t})². To see why, let's expand (e^t + e^{-t})²:
(e^t + e^{-t})² = (e^t)² + 2(et)(e{-t}) + (e^{-t})² = e^(2t) + 2(e^(t-t)) + e^(-2t) = e^(2t) + 2(e^0) + e^(-2t) = e^(2t) + 2 + e^(-2t)
This is exactly the expression we have under the square root! So, we can rewrite our integral as:
L = ∫[0, 2] √[(e^t + e^{-t})²] dt
Now, the square root and the square cancel each other out, leaving us with:
L = ∫[0, 2] (e^t + e^{-t}) dt
Wow, that's much simpler! We've successfully simplified the integral, and we're now ready to evaluate it. This simplification was key to making the integral solvable, and it highlights the importance of looking for patterns and algebraic tricks when dealing with calculus problems. We're on the home stretch now; all that's left is to find the antiderivative and evaluate it at the limits of integration.
Evaluating the Integral: The Final Calculation
Fantastic work, guys! We've simplified the integral to L = ∫[0, 2] (e^t + e^-t}) dt. Now, it's time to evaluate this integral and find the arc length. To do this, we need to find the antiderivative of (e^t + e^{-t}). The antiderivative of e^t is simply e^t. The antiderivative of e^{-t} is -e^{-t}. Remember, the derivative of -e^{-t} is -(-e^{-t}) = e^{-t}, so this checks out. Therefore, the antiderivative of (e^t + e^{-t}) is (e^t - e^{-t}). Now, we need to evaluate this antiderivative at the limits of integration, 0 and 2. This means we need to find (e^2 - e^{-2}) and (e^0 - e^{-0}), and then subtract the second value from the first. Let's start with the upper limit, t = 2) is approximately 7.389 - 0.135, which is about 7.254. Now, let's evaluate at the lower limit, t = 0: (e^0 - e^-0}) = (1 - 1) = 0. Remember that any number raised to the power of 0 is 1. Finally, we subtract the value at the lower limit from the value at the upper limit) - (e^0 - e^{-0}) = 7.254 - 0 = 7.254. So, the arc length of the parametric curve x = e^t, y = e^{-t}, z = \sqrt{2}t over the interval 0 ≤ t ≤ 2 is approximately 7.254 units. We've successfully navigated through all the steps, from understanding the concept of arc length to applying the formula, simplifying the integral, and finally, evaluating it. This is a great accomplishment! You've demonstrated a strong understanding of calculus and problem-solving skills. This method can be applied to a wide variety of parametric curves, so you now have a valuable tool in your mathematical toolkit. Congratulations on reaching the finish line!
In this article, we've journeyed through the process of finding the arc length of a parametric curve. We started by understanding the concepts of parametric curves and arc length, then introduced the arc length formula. We meticulously calculated the derivatives of our parametric equations, squared them, and plugged them into the formula. We then used a clever simplification technique to make the integral manageable and evaluated it to find the final arc length. This journey highlights the power of calculus in solving real-world problems and demonstrates the importance of breaking down complex problems into smaller, manageable steps. By understanding the underlying concepts and mastering the techniques, you can confidently tackle similar challenges in the future. So, keep practicing, keep exploring, and keep pushing your mathematical boundaries!
