Calculating Arc Length Parametric Equations X T 3t³ And Y T 10t²
Hey guys! Let's dive into a cool math problem where we'll figure out the length of a curve defined by parametric equations. We've got x(t) = 3t³ and y(t) = 10t², and we're interested in the part of the curve that exists when t is between 0 and 2. Get ready, because we're going to use some calculus magic to solve this!
Understanding Parametric Equations and Arc Length
Before we jump into the calculations, let's take a step back and make sure we're all on the same page about what parametric equations are and how they relate to arc length.
Parametric equations are a way of defining a curve using a third variable, often called a parameter (in our case, it's 't'). Instead of directly relating 'y' to 'x', we express both 'x' and 'y' as functions of 't'. Think of 't' as time, and the parametric equations as describing the position of a point moving along the curve as time changes. In our example, as 't' varies, the point (x(t), y(t)) traces out a specific curve in the xy-plane. Visualizing this can be super helpful – imagine a tiny robot following these equations to create a path!
Now, arc length is simply the distance along the curve between two points. Imagine you're walking along the curve; the arc length is the total distance you'd travel. For a standard function y = f(x), we have a formula to calculate arc length using an integral. But when we're dealing with parametric equations, we need a slightly different approach, but it’s derived from the same fundamental ideas. The core idea involves chopping up the curve into tiny little segments, approximating each segment with a straight line, and then adding up the lengths of those lines. This is where calculus comes to the rescue, allowing us to make those segments infinitely small and get a precise answer. Remember your Pythagorean theorem? It’s the backbone of this calculation, helping us find the length of those tiny straight lines.
The Arc Length Formula for Parametric Curves
Here's the star of the show: the formula for calculating the arc length (L) of a parametric curve defined by x(t) and y(t) over an interval [a, b]:
L = ∫[a to b] √[(dx/dt)² + (dy/dt)²] dt
Don't let the formula intimidate you! Let's break it down:
- ∫[a to b]… dt: This is the integral, which essentially means we're summing up an infinite number of tiny pieces. We're integrating with respect to 't', and 'a' and 'b' are the limits of our interval (the starting and ending values of 't').
- √(…): This is the square root, a familiar friend from geometry.
- (dx/dt) and (dy/dt): These are the derivatives of x(t) and y(t) with respect to 't'. Remember, derivatives tell us the rate of change. In this context, dx/dt represents the horizontal speed of our point moving along the curve, and dy/dt represents the vertical speed.
- (dx/dt)² + (dy/dt)²: This part is crucial. Think of it like the Pythagorean theorem in action. (dx/dt)² and (dy/dt)² are like the squares of the sides of a tiny right triangle, and their sum is the square of the hypotenuse. The hypotenuse represents the length of a tiny segment of the curve.
- √[(dx/dt)² + (dy/dt)²]: This whole term inside the integral is the length of an infinitesimally small segment of the curve. We're summing up all these tiny lengths to get the total arc length.
So, the formula is really just a fancy way of saying: “Add up the lengths of all the tiny pieces of the curve.” Calculus makes this possible!
Applying the Formula to Our Specific Problem
Now, let’s get our hands dirty and apply this formula to our problem where x(t) = 3t³ and y(t) = 10t², with 0 ≤ t ≤ 2. This is where the fun really begins!
Step 1: Find the Derivatives
The first thing we need to do is find the derivatives of x(t) and y(t) with respect to t. Remember the power rule for differentiation? It's going to be our best friend here. The power rule states that if f(t) = at^n, then f'(t) = nat^(n-1). Let's apply it:
- dx/dt = d(3t³)/dt = 3 * 3t² = 9t²
- dy/dt = d(10t²)/dt = 10 * 2t = 20t
Piece of cake, right? We've found the rates of change of x and y with respect to t. These derivatives tell us how quickly the x and y coordinates are changing as t changes.
Step 2: Square the Derivatives
Next, we need to square these derivatives. This is a straightforward step, but it's important to get the algebra right:
- (dx/dt)² = (9t²)² = 81t⁴
- (dy/dt)² = (20t)² = 400t²
Now we have the squared rates of change, which are essential for the next step in the arc length formula.
Step 3: Plug into the Arc Length Formula and Simplify
Now comes the exciting part: plugging everything into our arc length formula! Remember, L = ∫[a to b] √[(dx/dt)² + (dy/dt)²] dt. Our limits of integration are a = 0 and b = 2, and we've calculated (dx/dt)² and (dy/dt)². Let's substitute:
L = ∫[0 to 2] √[81t⁴ + 400t²] dt
Okay, the integral looks a bit intimidating, but don't worry, we can simplify it. Notice that both terms inside the square root have a common factor of t². Let's factor that out:
L = ∫[0 to 2] √[t²(81t² + 400)] dt
Now, we can use the property √(ab) = √a * √b to separate the square root:
L = ∫[0 to 2] √t² * √(81t² + 400) dt
Since √(t²) = |t|, and t is non-negative in our interval [0, 2], we can simplify further:
L = ∫[0 to 2] t * √(81t² + 400) dt
Our integral is looking much friendlier now!
Step 4: Evaluate the Integral
To evaluate this integral, we'll use a u-substitution. This is a classic technique for dealing with integrals involving composite functions. Let's let u = 81t² + 400. Then, we need to find du:
du = d(81t² + 400)/dt = 162t dt
Notice that we have 't dt' in our integral, which is great! We can solve for 't dt':
t dt = du / 162
Now we need to change our limits of integration to be in terms of 'u'. When t = 0:
u = 81(0)² + 400 = 400
And when t = 2:
u = 81(2)² + 400 = 81 * 4 + 400 = 324 + 400 = 724
So our new limits of integration are 400 and 724. Let's substitute everything into the integral:
L = ∫[400 to 724] √(u) * (du / 162)
We can pull the constant 1/162 out of the integral:
L = (1/162) ∫[400 to 724] √u du
Now we can rewrite √u as u^(1/2) and use the power rule for integration, which states that ∫x^n dx = (x^(n+1))/(n+1) + C:
L = (1/162) [ (u^(3/2)) / (3/2) ] evaluated from 400 to 724
Simplify the fraction:
L = (1/162) * (2/3) [ u^(3/2) ] evaluated from 400 to 724
L = (1/243) [ u^(3/2) ] evaluated from 400 to 724
Now we evaluate at the upper and lower limits:
L = (1/243) [ (724)^(3/2) - (400)^(3/2) ]
Let's simplify the powers: 724^(3/2) = (√724)³ and 400^(3/2) = (√400)³ = 20³ = 8000
L = (1/243) [ (√724)³ - 8000 ]
We can simplify √724 as √(4 * 181) = 2√181, so (√724)³ = (2√181)³ = 8 * 181√181 = 1448√181
L = (1/243) [ 1448√181 - 8000 ]
This is the exact length of the curve! What a journey!
The Grand Finale: The Exact Arc Length
So, after all that calculus wizardry, we've arrived at the exact length of the curve represented by the parametric equations x(t) = 3t³ and y(t) = 10t² over the interval 0 ≤ t ≤ 2:
L = (1/243) [1448√181 - 8000]
This is the precise answer, expressed in exact form using symbolic notation and fractions. We didn't resort to decimal approximations, which is the hallmark of a true calculus enthusiast! This answer might look a bit complex, but it's a testament to the power of calculus to provide exact solutions to geometric problems. The ability to calculate arc length has significant applications in various fields, including physics, engineering, and computer graphics. Think about designing a roller coaster – you'd need to know the exact length of the track!
Key Takeaways and Final Thoughts
Let's recap the key steps we took to solve this problem:
- We understood the concept of parametric equations and arc length.
- We recalled the arc length formula for parametric curves: L = ∫[a to b] √[(dx/dt)² + (dy/dt)²] dt.
- We found the derivatives of x(t) and y(t) with respect to t.
- We squared the derivatives.
- We plugged everything into the formula and simplified the integral.
- We used u-substitution to evaluate the integral.
- We arrived at the exact arc length.
This problem beautifully illustrates how calculus can be used to solve geometric problems. The arc length formula might seem daunting at first, but by breaking it down step-by-step, we can tackle even complex calculations. Remember, the key is to understand the underlying concepts and practice, practice, practice! Whether you're a student tackling calculus for the first time or a seasoned mathematician, these techniques and problem-solving approaches are fundamental.
I hope you enjoyed this exploration of arc length! Keep those calculus skills sharp, and you'll be amazed at what you can accomplish.
