Parametric Derivatives: Find D²y/dx² For X=sin(3t), Y=cos(3t)

What's up, math whizzes! Today, we're diving deep into the awesome world of parametric equations and tackling a challenge that'll really get your calculus gears grinding. We're going to figure out how to find the second derivative of y with respect to x, denoted as $\frac{d^2y}{dx^2}$, when our equations are given in terms of a third variable, 't'. Our specific mission, should we choose to accept it, is to conquer the problem: given $x = \sin(3t)$ and $y = \cos(3t)$, find $\frac{d^2y}{dx^2}$. This isn't just about crunching numbers; it's about understanding the rate of change of the rate of change in a more complex, interconnected system. Think of it like driving a car – the first derivative tells you your speed, but the second derivative tells you how your speed is changing, whether you're accelerating or braking. That's crucial info, right? So, buckle up, because we're about to break this down step-by-step, making sure you guys grasp every single part of this calculation. We'll start with the basics of parametric differentiation and then build our way up to that elusive second derivative. Get ready to flex those calculus muscles!

Understanding Parametric Differentiation: The First Step

Alright guys, before we can even think about finding the second derivative, $\frac{d^2y}{dx^2}$, we absolutely must get a solid handle on the first derivative, $\frac{dy}{dx}$, in the context of parametric equations. So, what are parametric equations anyway? Simply put, they're a way to describe a curve using a third variable, often called a parameter (usually 't' for time, but it can be anything!). Instead of $y$ being a direct function of $x$ (like $y = f(x)$), both $x$ and $y$ are defined as functions of this parameter 't'. In our case, we have $x = \sin(3t)$ and $y = \cos(3t)$. The magic of parametric differentiation lies in the chain rule. We want to find $\frac{dy}{dx}$, which represents how $y$ changes with respect to $x$. Since both $x$ and $y$ depend on 't', we can use the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. This formula is your golden ticket to navigating parametric curves! It means we can find the slope of the tangent line to the curve at any point by calculating the derivative of $y$ with respect to $t$ and dividing it by the derivative of $x$ with respect to $t$. Sounds straightforward, right? Let's break down the calculation for our specific problem. First, we need to find $\frac{dx}{dt}$. Using our equation $x = \sin(3t)$, we apply the derivative rules. The derivative of $\sin(u)$ is $\cos(u)$, and by the chain rule, the derivative of $\sin(3t)$ is $\cos(3t)$ multiplied by the derivative of $3t$, which is 3. So, $\frac{dx}{dt} = 3\cos(3t)$. Next, we need $\frac{dy}{dt}$. For $y = \cos(3t)$, the derivative of $\cos(u)$ is $-\sin(u)$. Applying the chain rule again, the derivative of $\cos(3t)$ is $-\sin(3t)$ multiplied by the derivative of $3t$, which is 3. Therefore, $\frac{dy}{dt} = -3\sin(3t)$. Now we can plug these into our formula for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-3\sin(3t)}{3\cos(3t)}$. See how the 3s cancel out? That simplifies things nicely! So, we get $\frac{dy}{dx} = \frac{-\sin(3t)}{\cos(3t)}$. And since $\frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta)$, our first derivative simplifies beautifully to $\frac{dy}{dx} = -\tan(3t)$. Boom! We've successfully found the first derivative. This tells us the instantaneous rate of change of $y$ with respect to $x$ at any given value of 't'. Pretty neat, huh? This foundational step is absolutely critical, so make sure you've got this down before we move on to the more challenging part: the second derivative.

Conquering the Second Derivative: The Path Forward

Alright team, we've conquered the first derivative, $\frac{dy}{dx} = -\tan(3t)$, which is a huge win! But our mission isn't over; we need to find $\frac{d^2y}{dx^2}$. Now, this is where things get a little trickier, and it's super important to pay attention. Remember, $\frac{d^2y}{dx^2}$ means the derivative of $\frac{dy}{dx}$ with respect to $x$. But here's the catch: our current expression for $\frac{dy}{dx}$ is in terms of 't', not 'x'. So, we can't just differentiate $- \tan(3t)$ directly with respect to $x$. We need to use our trusty chain rule again, but in a slightly different way. The key insight is that $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$. Since $\frac{dy}{dx}$ is a function of 't', we can write this as $\frac{d}{dt}\left(\frac{dy}{dx}\right) \times \frac{dt}{dx}$. See what's happening here? We're differentiating our expression for $\frac{dy}{dx}$ with respect to 't', and then multiplying by $\frac{dt}{dx}$. This makes sense because we want the change in $y$ with respect to $x$, and we can only easily calculate the change with respect to $t$. We already know $\frac{dx}{dt} = 3\cos(3t)$. Therefore, $\frac{dt}{dx}$ is simply the reciprocal of that: $\frac{dt}{dx} = \frac{1}{3\cos(3t)}$. Now, let's focus on the first part: $\frac{d}{dt}\left(\frac{dy}{dx}\right)$. We need to find the derivative of $- \tan(3t)$ with respect to 't'. The derivative of $\tan(u)$ is $\sec^2(u)$. Applying the chain rule (you knew it would come back, right?), the derivative of $- \tan(3t)$ is $- \sec^2(3t)$ multiplied by the derivative of $3t$, which is 3. So, $\frac{d}{dt}\left(\frac{dy}{dx}\right) = -3\sec^2(3t)$. Phew! We've got both pieces of the puzzle. Now, let's put them together using our chain rule formula for the second derivative: $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \times \frac{dt}{dx} = (-3\sec^2(3t)) \times \left(\frac{1}{3\cos(3t)}\right)$. Look at that! The 3s cancel out again, which is always a good sign. So, we're left with $\frac{d^2y}{dx^2} = -\sec^2(3t) \times \frac{1}{\cos(3t)}$. Since $\sec(\theta) = \frac{1}{\cos(\theta)}$, we can rewrite $\sec^2(3t)$ as $\frac{1}{\cos^2(3t)}$. Plugging this in, we get $\frac{d^2y}{dx^2} = -\frac{1}{\cos^2(3t)} \times \frac{1}{\cos(3t)}$. Multiplying these together gives us $\frac{d^2y}{dx^2} = -\frac{1}{\cos^3(3t)}$. And since $\frac{1}{\cos(\theta)} = \sec(\theta)$, our final answer is $\frac{d^2y}{dx^2} = -\sec^3(3t)$. How awesome is that? We navigated the complexities of parametric differentiation to arrive at our final answer.

The Big Picture: Why Does This Matter?

So, we've crunched the numbers and found that for $x = \sin(3t)$ and $y = \cos(3t)$, the second derivative $\frac{d^2y}{dx^2}$ is $- \sec^3(3t)$. But why should you guys care about this? What's the real-world significance of finding the second derivative of parametric equations? Well, think about it. The first derivative, $\frac{dy}{dx}$, tells us the slope of the tangent line to the curve at any given point. It describes the instantaneous rate of change of $y$ with respect to $x$. This is super useful for understanding how a system is behaving right now. For instance, if $x$ and $y$ represent positions in physics, $\frac{dy}{dx}$ could tell you about the direction of motion. But the second derivative, $\frac{d^2y}{dx^2}$, tells us how that rate of change is itself changing. It's the rate of change of the slope. This concept is absolutely fundamental in understanding the curvature of a path. If $\frac{d^2y}{dx^2}$ is positive, the curve is concave up (like a smiley face 😃), meaning the slope is increasing. If it's negative, the curve is concave down (like a frowny face ☹️), meaning the slope is decreasing. In our specific case, $x = \sin(3t)$ and $y = \cos(3t)$, we can actually recognize this as a circle! If you square both equations and add them, you get $x^2 + y^2 = \sin^2(3t) + \cos^2(3t) = 1$. This is the equation of a unit circle centered at the origin. As 't' changes, the point $(x, y)$ moves around this circle. The second derivative helps us understand how the direction of movement is changing as the point traverses the circle. For example, on the top half of the circle, the slope is generally negative and becoming less negative (increasing), so the second derivative should be positive. On the bottom half, the slope is positive and becoming less positive (decreasing), so the second derivative should be negative. Our result, $- \sec^3(3t)$, directly reflects this changing curvature. Parametric equations and their derivatives are not just abstract math concepts; they are the language used to describe complex motion, trajectories, and shapes in fields like engineering, physics, computer graphics, and economics. Whether you're designing a roller coaster track, plotting the path of a satellite, or analyzing the dynamics of a financial market, understanding these rates of change is paramount. So, while this problem might seem like just another calculus exercise, it's actually a building block for understanding the dynamic world around us!

Final Answer Recap and Key Takeaways

Alright guys, let's wrap this up with a quick recap and some key takeaways. We started with the parametric equations $x = \sin(3t)$ and $y = \cos(3t)$ and our goal was to find the second derivative, $\frac{d^2y}{dx^2}$.

Key Steps and Results:

1. Find the First Derivative ($\frac{dy}{dx}$):

   • We calculated $\frac{dx}{dt} = 3\cos(3t)$.
   • We calculated $\frac{dy}{dt} = -3\sin(3t)$.
   • Using the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, we found $\frac{dy}{dx} = \frac{-3\sin(3t)}{3\cos(3t)} = -\tan(3t)$.

2. Find the Second Derivative ($\frac{d^2y}{dx^2}$):

   • We used the chain rule formula: $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \times \frac{dt}{dx}$.
   • We found $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}(-\tan(3t)) = -3\sec^2(3t)$.
   • We found $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{3\cos(3t)}$.
   • Multiplying these together: $\frac{d^2y}{dx^2} = (-3\sec^2(3t)) \times \left(\frac{1}{3\cos(3t)}\right)$.
   • After simplification, we arrived at the final answer: $\boxed{\frac{d^2y}{dx^2} = -\sec^3(3t)}$.

Key Takeaways:

• Parametric Differentiation: Always remember that $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. This is your fundamental tool.
• Second Derivative Chain Rule: To find $\frac{d^2y}{dx^2}$, you'll typically use $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \times \frac{dt}{dx}$. It involves differentiating the first derivative with respect to 't' and then multiplying by the reciprocal of $\frac{dx}{dt}$.
• Understanding Notation: $\frac{d^2y}{dx^2}$ means differentiating the result of $\frac{dy}{dx}$ with respect to $x$. Don't get tricked into differentiating directly with respect to 't' without accounting for the change of variables.
• Application: The second derivative is crucial for understanding curvature, concavity, and the acceleration of motion described by parametric equations.

Keep practicing these types of problems, guys! The more you work through them, the more intuitive parametric differentiation will become. You've got this!