Hey guys! Today, we're diving into a fun calculus problem where we need to find the second derivative d²y/dx² given that x = sin(3t) and y = cos(3t). This involves using parametric equations and a bit of chain rule magic. So, grab your pencils, and let's get started!

Step 1: Find dy/dx

First, we need to find dy/dx. Since we have x and y defined in terms of t, we'll use the chain rule. Recall that:

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

Let's find dy/dt and dx/dt separately.

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. The derivative of cos(u) is -sin(u), and the derivative of 3t is 3. Combining these, we get -3sin(3t).

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. Thus, we get 3cos(3t).

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). This is a crucial intermediate result that we'll use in the next step.

Step 2: Find d²y/dx²

Next, we need to find the second derivative d²y/dx². Remember that:

d²y/dx² = d/dx (dy/dx)

Since dy/dx is a function of t, we need to use the chain rule again. We can write:

d²y/dx² = d/dt (dy/dx) * (dt/dx)

We already know dy/dx = -tan(3t), so let's find d/dt (dy/dx).

Finding d/dt (dy/dx)

We have dy/dx = -tan(3t). Differentiating with respect to t, we get:

d/dt (dy/dx) = d/dt (-tan(3t)) = -3sec²(3t)

Here, the derivative of tan(u) is sec²(u), and the derivative of 3t is 3. So, we have -3sec²(3t).

Finding dt/dx

We need to find dt/dx. We know dx/dt = 3cos(3t), so:

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). Woohoo, we found it!

Detailed Explanation and Concepts

Let's break down the concepts and steps in more detail to ensure we understand everything thoroughly. This problem beautifully illustrates the application of parametric differentiation and the chain rule. It’s not just about crunching numbers; it's about understanding how rates of change relate to each other.

Parametric Equations

In this problem, x and y are defined in terms of a third variable, t. This is known as a parametric representation. Instead of y being directly expressed as a function of x, both x and y are functions of t. This is particularly useful when dealing with curves that are not easily expressed in the form y = f(x).

Chain Rule

The chain rule is a fundamental concept in calculus that allows us to differentiate composite functions. In simple terms, if y is a function of u, and u is a function of x, then the derivative of y with respect to x is given by:

dy/dx = (dy/du) * (du/dx)

We used this rule extensively in our problem.

First Derivative dy/dx

To find dy/dx, we used the formula:

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

This formula is derived from the chain rule. By finding dy/dt and dx/dt separately, we can easily compute dy/dx. In our case:

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

Therefore, dy/dx = (-3sin(3t)) / (3cos(3t)) = -tan(3t). This tells us how y changes with respect to x at any given value of t.

Second Derivative d²y/dx²

Finding the second derivative d²y/dx² is a bit more involved. We need to differentiate dy/dx with respect to x. However, since dy/dx is a function of t, we again use the chain rule:

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

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

• d/dt (-tan(3t)) = -3sec²(3t)
• dt/dx = 1 / (dx/dt) = 1 / (3cos(3t)) = (1/3)sec(3t)

Multiplying these together:

d²y/dx² = (-3sec²(3t)) * ((1/3)sec(3t)) = -sec³(3t)

This gives us the rate of change of the slope of the curve with respect to x.

Common Mistakes to Avoid

1. Forgetting the Chain Rule: One of the most common mistakes is forgetting to apply the chain rule when differentiating. Always remember to multiply by the derivative of the inner function.
2. Incorrect Differentiation: Make sure you know the derivatives of basic trigonometric functions. For example, the derivative of sin(u) is cos(u), and the derivative of cos(u) is -sin(u).
3. Not Simplifying: Simplify your expressions as much as possible. This not only makes the problem easier to solve but also reduces the chance of making errors.
4. Incorrectly Finding dt/dx: Remember that dt/dx is the reciprocal of dx/dt. Make sure you take the reciprocal correctly.

Real-World Applications

Understanding parametric equations and derivatives has numerous real-world applications. Here are a few examples:

1. Physics: In physics, parametric equations are used to describe the motion of objects. For example, the trajectory of a projectile can be described using parametric equations, where t represents time. Derivatives can then be used to find the velocity and acceleration of the object.
2. Engineering: Engineers use parametric equations to design curves and surfaces. For example, in computer-aided design (CAD), parametric equations are used to create complex shapes.
3. Computer Graphics: In computer graphics, parametric equations are used to draw curves and surfaces on the screen. Bezier curves and B-splines, which are commonly used in computer graphics, are based on parametric equations.
4. Economics: Economists use derivatives to analyze rates of change. For example, they might use derivatives to study how the price of a product changes over time.

Conclusion

So, to wrap things up, given x = sin(3t) and y = cos(3t), we found that d²y/dx² = -sec³(3t). This problem highlights the importance of understanding parametric equations, the chain rule, and trigonometric derivatives. Keep practicing, and you'll become a calculus pro in no time! Keep an eye out for more fun calculus problems coming your way!

Hope this helps, and happy calculating, guys!