Find D²y/dx² If X=sin(3t) And Y=cos(3t)

by Jhon Lennon 40 views

Hey guys! Today, we're diving into a fun 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 parametric equations and applying the chain rule multiple times. Buckle up, because it's going to be a derivative-filled ride!

Step 1: Find dy/dx

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

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

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

Finding dy/dt

We have y = cos(3t). To find dy/dt, we'll differentiate y with respect to t:

dy/dt = d/dt [cos(3t)]

Using the chain rule, we get:

dy/dt = -3sin(3t)

Finding dx/dt

Now, we have x = sin(3t). To find dx/dt, we differentiate x with respect to t:

dx/dt = d/dt [sin(3t)]

Again, using the chain rule, we get:

dx/dt = 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, dy/dx = -tan(3t). Great job, we've completed the first step!

Step 2: Find d²y/dx²

Now comes the slightly trickier part: finding the second derivative, d²y/dx². Remember, d²y/dx² is the derivative of dy/dx with respect to x, not t. So, we need to use the chain rule again:

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

We already know dy/dx = -tan(3t). Let's find d/dt [dy/dx] first.

Finding d/dt [dy/dx]

We need to differentiate -tan(3t) with respect to t:

d/dt [-tan(3t)] = -d/dt [tan(3t)]

The derivative of tan(u) is sec²(u) * du/dt, so:

-d/dt [tan(3t)] = -sec²(3t) * (3) = -3sec²(3t)

Finding dt/dx

We already found dx/dt = 3cos(3t). To find dt/dx, we simply take the reciprocal:

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

Calculating d²y/dx²

Now we have all the pieces we need to find d²y/dx²:

d²y/dx² = d/dt [dy/dx] * (dt/dx) = (-3sec²(3t)) * (1 / (3cos(3t)))

Simplify it:

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

So, d²y/dx² = -sec³(3t). You nailed it!

Summary and Key Points

Let's recap what we did:

  1. Found dy/dx: We used the chain rule to find dy/dt and dx/dt, then divided dy/dt by dx/dt to get dy/dx = -tan(3t).
  2. Found d²y/dx²: We used the chain rule again to find d/dt [dy/dx] and dt/dx, then multiplied them to get d²y/dx² = -sec³(3t).

Key points to remember:

  • The chain rule is your best friend when dealing with parametric equations.
  • Always remember to differentiate with respect to the correct variable.
  • Be careful with trigonometric derivatives and simplifications.

Alternative Forms and Trigonometric Identities

While d²y/dx² = -sec³(3t) is a perfectly valid answer, we can also express it in terms of other trigonometric functions if needed. Remember that sec(x) = 1/cos(x), so we can rewrite our answer as:

d²y/dx² = - (1 / cos³(3t))

This form might be useful depending on the context of the problem or if you need to simplify further with other expressions involving cosine. Also, since we know that y = cos(3t), we can write:

d²y/dx² = - (1 / y³)

However, expressing the answer in terms of t is generally more common when working with parametric equations.

Common Mistakes to Avoid

  • Forgetting the Chain Rule: This is the most common mistake. Always remember to multiply by the derivative of the inner function when using the chain rule.
  • Differentiating with Respect to the Wrong Variable: Make sure you're differentiating with respect to x when finding d²y/dx², not t directly.
  • Incorrect Trigonometric Derivatives: Double-check your trigonometric derivatives to avoid errors.
  • Algebraic Mistakes: Be careful with your algebra, especially when simplifying expressions.

Real-World Applications

While this might seem like a purely theoretical exercise, finding derivatives of parametric equations has many real-world applications. Here are a few examples:

  • Physics: Analyzing the motion of projectiles or objects moving along a curved path.
  • Engineering: Designing curves for roads, bridges, and other structures.
  • Computer Graphics: Creating smooth animations and realistic movements in video games and movies.

Understanding how to find derivatives of parametric equations is a valuable skill in many fields.

Practice Problems

To solidify your understanding, here are a few practice problems:

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

Work through these problems, and you'll become a pro at finding derivatives of parametric equations in no time!

Conclusion

So, there you have it! Finding d²y/dx² when x = sin(3t) and y = cos(3t) involves a few steps, but with practice, you'll be able to tackle these problems with ease. Remember the chain rule, be careful with your derivatives, and don't be afraid to ask for help if you get stuck. Keep practicing, and you'll become a calculus master in no time! Keep up the great work, and remember to always double-check your solutions. You've got this!