Hey guys! Today, we're diving into a calculus problem where we need to find the second derivative, specifically d²y/dx², given that x = sin(3t) and y = cos(3t). This involves using the chain rule and some trigonometric identities to simplify the expressions. So, grab your notebooks, and let's get started!

Step 1: Find dy/dx

First, we need to find the first derivative, dy/dx. Since both x and y are given in terms of a parameter t, we'll use the chain rule. The chain rule in this context states that:

dy/dx = (dy/dt) / (dx/dt)

So, let's find dy/dt and dx/dt.

Finding dy/dt

We have y = cos(3t). Differentiating y with respect to t, we get:

dy/dt = -3sin(3t)

This is a straightforward application of the chain rule in differentiation. The derivative of cos(u) is -sin(u), and the derivative of 3t is 3, hence we multiply by 3.

Finding dx/dt

Now, we have x = sin(3t). Differentiating x with respect to t, we get:

dx/dt = 3cos(3t)

Similarly, this uses the chain rule. The derivative of sin(u) is cos(u), and the derivative of 3t is 3, so we multiply by 3.

Calculating dy/dx

Now that we have both dy/dt and dx/dt, we can find dy/dx:

dy/dx = (dy/dt) / (dx/dt) = (-3sin(3t)) / (3cos(3t)) = -tan(3t)

So, the first derivative dy/dx is -tan(3t). Keep this result handy, as we'll need it for the next step.

Step 2: Find d²y/dx²

Now, let's find the second derivative, d²y/dx². Remember that d²y/dx² is the derivative of dy/dx with respect to x. Since dy/dx is in terms of t, we'll use the chain rule again:

d²y/dx² = d/dx (dy/dx) = d/dt (dy/dx) * (dt/dx)

We already know dy/dx = -tan(3t). So, we need to find d/dt (-tan(3t)) and (dt/dx).

Finding d/dt (-tan(3t))

Differentiating -tan(3t) with respect to t, we get:

d/dt (-tan(3t)) = -3sec²(3t)

The derivative of tan(u) is sec²(u), and the derivative of 3t is 3, so by the chain rule, we multiply by 3. The negative sign remains.

Finding dt/dx

We know dx/dt = 3cos(3t). Therefore, dt/dx is the reciprocal of dx/dt:

dt/dx = 1 / (dx/dt) = 1 / (3cos(3t)) = (1/3)sec(3t)

Calculating d²y/dx²

Now, we can find d²y/dx²:

d²y/dx² = d/dt (dy/dx) * (dt/dx) = (-3sec²(3t)) * ((1/3)sec(3t)) = -sec³(3t)

So, the second derivative d²y/dx² is -sec³(3t). That's our final answer!

Summary and Key Points

To recap, here’s what we did:

1. Found dy/dx: Using the chain rule, we found dy/dt and dx/dt, then calculated dy/dx = (dy/dt) / (dx/dt) = -tan(3t).
2. Found d²y/dx²: We used the chain rule again to find d²y/dx² = d/dt (dy/dx) * (dt/dx). We found d/dt (-tan(3t)) = -3sec²(3t) and dt/dx = (1/3)sec(3t). Multiplying these gave us d²y/dx² = -sec³(3t).

Key points to remember:

• The chain rule is essential when dealing with parametric equations.
• dy/dx = (dy/dt) / (dx/dt)
• d²y/dx² = d/dt (dy/dx) * (dt/dx)
• Basic trigonometric derivatives are crucial (e.g., derivative of sin(u), cos(u), tan(u)).

Additional Tips for Solving Similar Problems

When tackling problems like this, here are a few tips that might help:

• Write everything down: Keep track of each derivative and substitution. This will help you avoid making mistakes.
• Simplify as you go: Simplify expressions as soon as possible to make subsequent steps easier.
• Double-check your derivatives: Make sure you've correctly differentiated each function. A small mistake can throw off the entire solution.
• Know your trigonometric identities: Sometimes, simplifying trigonometric expressions can lead to a more manageable form.
• Practice, practice, practice: The more you practice, the more comfortable you'll become with these types of problems.

Common Mistakes to Avoid

• Forgetting the chain rule: The chain rule is crucial in these types of problems. Forgetting to apply it correctly is a common mistake.
• Incorrect trigonometric derivatives: Make sure you know the derivatives of basic trigonometric functions.
• Algebraic errors: Be careful with your algebra. A small mistake can lead to an incorrect final answer.
• Not simplifying: Failing to simplify expressions can make the problem more difficult than it needs to be.

Practice Problems

To reinforce your understanding, try these practice problems:

1. Given x = cos(2t) and y = sin(2t), find d²y/dx².
2. Given x = t² and y = 2t, find d²y/dx².
3. Given x = e^t and y = e^(-t), find d²y/dx².

Work through these problems, and you'll become much more confident in finding second derivatives with parametric equations. Compare your solutions with available resources or ask for help if needed.

Conclusion

Finding the second derivative d²y/dx² when x = sin(3t) and y = cos(3t) involves a few steps, but it’s manageable if you take it one step at a time. By using the chain rule and carefully differentiating each function, you can arrive at the correct answer: d²y/dx² = -sec³(3t).

Remember, practice makes perfect. Keep working on similar problems, and you’ll become a pro in no time. Good luck, and happy calculating!

I hope this helps you guys understand how to solve this type of problem. If you have any questions, feel free to ask. Keep practicing, and you'll get the hang of it!