Let's dive into finding dy/dx when we're given x = a cos(log t) and y = a sin(log t). This problem involves a bit of calculus and chain rule, but don't worry, we'll break it down step by step so it's super easy to follow. We aim to express dy/dx in terms of t. This means we'll need to find dx/dt and dy/dt first, and then use the chain rule to combine these derivatives to get our final answer. So, grab your favorite beverage, and let's get started!

Step 1: Find dx/dt

Okay, our first mission is to find dx/dt. Remember, we have x = a cos(log t). To find the derivative of x with respect to t, we'll need to use the chain rule. The chain rule basically says that if you have a function inside another function, you take the derivative of the outer function, keeping the inner function the same, and then multiply by the derivative of the inner function.

So, let's break it down:

• The outer function is cos(u) where u = log t.
• The inner function is log t.

First, the derivative of cos(u) with respect to u is -sin(u). So, we have -sin(log t). Next, we need to find the derivative of log t with respect to t. If we're using the natural logarithm (base e), then the derivative of log t is 1/t. Putting it all together, and remembering that a is a constant, we get:

dx/dt = a * (-sin(log t)) * (1/t) = (-a sin(log t))/t

So, that's our dx/dt. Not too bad, right? Now, let's move on to finding dy/dt.

Step 2: Find dy/dt

Alright, next up is finding dy/dt. We know that y = a sin(log t). Just like before, we'll use the chain rule to find the derivative of y with respect to t.

• The outer function is sin(u) where u = log t.
• The inner function is log t.

The derivative of sin(u) with respect to u is cos(u). So, we have cos(log t). And, as before, the derivative of log t with respect to t is 1/t. So, putting it all together, remembering that a is a constant, we get:

dy/dt = a * (cos(log t)) * (1/t) = (a cos(log t))/t

Great! We've found dy/dt. Now we have both dx/dt and dy/dt. We're in the home stretch now. Let's use these to find dy/dx.

Step 3: Find dy/dx using the Chain Rule

Okay, now for the grand finale: finding dy/dx. The chain rule tells us that:

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

We already found dy/dt and dx/dt, so let's plug those in:

dy/dx = [(a cos(log t))/t] / [(-a sin(log t))/t]

Now, we can simplify this expression. Notice that we have t in the denominator of both the numerator and the denominator, so those cancel out. Also, a is a common factor, so that cancels out too. We're left with:

dy/dx = cos(log t) / (-sin(log t)) = -cot(log t)

And there you have it! The derivative dy/dx is equal to -cot(log t). That's our final answer.

Summary

Let's recap what we did:

1. We started with x = a cos(log t) and y = a sin(log t).
2. We found dx/dt using the chain rule: dx/dt = (-a sin(log t))/t
3. We found dy/dt using the chain rule: dy/dt = (a cos(log t))/t
4. We used the chain rule to find dy/dx: dy/dx = (dy/dt) / (dx/dt)
5. We simplified the expression to get dy/dx = -cot(log t)

So, the final answer is:

dy/dx = -cot(log t)

Great job, guys! You've successfully navigated through this calculus problem. Remember, the key is to break it down into smaller, manageable steps and apply the chain rule carefully. Keep practicing, and you'll become a pro in no time! This detailed walkthrough should not only help in solving similar problems but also provide a solid understanding of the underlying calculus principles involved.

Additional Insights

While we've successfully found dy/dx, it's worth exploring some additional insights to deepen our understanding. The relationship between x and y can give us clues about the underlying geometry or physics of the system. Let's consider some additional points.

Understanding the Relationship Between x and y

We have x = a cos(log t) and y = a sin(log t). Notice the similarity to the parametric equations of a circle, which are typically given by x = r cos(θ) and y = r sin(θ), where r is the radius and θ is the angle. In our case, instead of a simple angle θ, we have log t. This suggests that the path traced by the point (x, y) might be related to a spiral rather than a circle. As t varies, log t also varies, causing the point to move in a way that spirals either inward or outward, depending on the behavior of t.

Implicit Relationship

Let's see if we can find an implicit relationship between x and y by eliminating t. Square both equations and add them:

x^2 = a^2 cos^2(log t) y^2 = a^2 sin^2(log t)

Adding these gives:

x^2 + y^2 = a^2 cos^2(log t) + a^2 sin^2(log t) = a^2 [cos^2(log t) + sin^2(log t)]

Since cos^2(θ) + sin^2(θ) = 1 for any θ, we have:

x^2 + y^2 = a^2

This is the equation of a circle with radius a. However, it's crucial to remember that this is an implicit relationship. The actual path traced by (x, y) as t varies is not the entire circle, but rather a spiral that is constrained within this circle. The logarithmic function log t restricts the movement along this circle, creating the spiral-like behavior.

Physical Interpretation

In physics, such equations could represent damped harmonic motion or the motion of a particle in a central force field. The logarithmic term introduces a damping or spiraling effect. For example, if t represents time, the equations could describe a particle moving towards the origin in a spiral path due to some form of resistance or damping force.

Visualizing the Path

To visualize this path, you could plot x and y for different values of t. You would see that as t increases, the point (x, y) spirals around the origin, getting closer and closer to it if the damping effect is strong enough. This visualization can provide a better understanding of the behavior of the system.

Alternative Forms of dy/dx

We found that dy/dx = -cot(log t). While this is a correct answer, it might be useful to express dy/dx in terms of x and y directly. From our original equations, we have cos(log t) = x/a and sin(log t) = y/a. Therefore, cot(log t) = cos(log t) / sin(log t) = (x/a) / (y/a) = x/y. Substituting this into our expression for dy/dx gives:

dy/dx = -x/y

This form is often more useful in certain contexts, as it directly relates the rate of change of y with respect to x to the actual coordinates x and y.

Implications and Applications

Understanding derivatives like dy/dx is essential in many fields, including physics, engineering, and economics. They help us analyze rates of change, optimize systems, and model complex phenomena. In engineering, for instance, these equations could model the behavior of a damped oscillator or the trajectory of a projectile in a resistive medium. In economics, they could be used to model the growth or decay of investments over time, where the logarithmic term represents some form of diminishing returns or increasing risk.

By exploring these additional insights, we gain a deeper appreciation for the mathematical relationships and their practical implications. The combination of trigonometric functions and logarithms creates a rich set of behaviors that can model a wide range of real-world phenomena. Keep exploring, and you'll continue to uncover the beauty and power of calculus!

Practice Problems

To solidify your understanding, here are a few practice problems similar to the one we just solved.

1. If x = b cos(log t) and y = b sin(log t), find dy/dx.
2. If x = a e^t cos(t) and y = a e^t sin(t), find dy/dx.
3. If x = a cos(t^2) and y = a sin(t^2), find dy/dx.

Happy solving, guys!