Tangent Line Equation: X=3sin(t), Y=cos(t) At T=π/3

by Alex Johnson 52 views

Hey there, math enthusiasts! Today, we're diving into a classic calculus problem: finding the equation of a tangent line to a curve defined by parametric equations. Specifically, we'll be tackling the equations x = 3sin(t) and y = cos(t), and our goal is to determine the tangent line at the point where t = π/3. This might sound intimidating at first, but don't worry, we'll break it down step by step in a friendly and easy-to-understand way. So, grab your pencils, and let's get started!

Step 1: Determining the Slope (dy/dx)

Let's start with finding the slope of the tangent line, often represented as dy/dx. But here's the catch: our curve isn't defined in terms of y as a function of x. Instead, both x and y are defined in terms of a third variable, the parameter t. This is where the magic of parametric equations comes in! To find dy/dx, we'll use a nifty formula derived from the chain rule: dy/dx = (dy/dt) / (dx/dt). This formula allows us to find the slope by considering the rates of change of y and x with respect to t. This is crucial because it allows us to work with parametric equations effectively. So, let’s calculate dy/dt and dx/dt separately.

First, we have y = cos(t). The derivative of cos(t) with respect to t is simply -sin(t). So, we have dy/dt = -sin(t). Remember, the derivative represents the instantaneous rate of change, and in this case, it tells us how y is changing as t changes. Now, let's move on to finding dx/dt. We are given x = 3sin(t). The derivative of sin(t) with respect to t is cos(t). Therefore, the derivative of 3sin(t) with respect to t is 3cos(t). This gives us dx/dt = 3cos(t). It's vital to remember these basic trigonometric derivatives, as they frequently appear in calculus problems. With both dy/dt and dx/dt in hand, we can now find dy/dx using our formula.

Now, we can plug our results into the formula dy/dx = (dy/dt) / (dx/dt). Substituting the derivatives we found, we get: dy/dx = (-sin(t)) / (3cos(t)). We can simplify this expression further by recognizing that sin(t) / cos(t) is equal to tan(t). Thus, our expression for the slope becomes: dy/dx = -1/3 * tan(t). This is a significant result! It gives us a general formula for the slope of the tangent line at any point t on the curve. The negative sign and the constant factor of 1/3 are important to keep track of, as they influence the direction and steepness of the tangent line. Now that we have the general slope formula, we can move on to finding the specific slope at t = π/3.

To find the slope at t = π/3, we simply substitute π/3 into our expression for dy/dx. We have dy/dx = -1/3 * tan(t), so when t = π/3, we get: dy/dx = -1/3 * tan(π/3). Recall that tan(π/3) is equal to √3. Therefore, the slope at t = π/3 is: dy/dx = -1/3 * √3 = -√3 / 3. This is the slope of the tangent line at the specific point on the curve where t = π/3. It's a negative slope, which indicates that the tangent line is decreasing as we move from left to right. We now have a crucial piece of information needed to determine the tangent line equation. With the slope calculated, we next need to find the coordinates of the point on the curve where t = π/3.

Step 2: Finding the Point (x, y) at t = π/3

Now that we've nailed down the slope, our next mission is to pinpoint the exact coordinates (x, y) on the curve where t = π/3. Remember, our parametric equations are x = 3sin(t) and y = cos(t). To find the x-coordinate, we'll substitute t = π/3 into the equation for x: x = 3sin(π/3). Recall that sin(π/3) is equal to √3 / 2. Therefore, x = 3 * (√3 / 2) = (3√3) / 2. This gives us the x-coordinate of the point of tangency. Remember, the sine function gives us the ratio of the opposite side to the hypotenuse in a right triangle, and at π/3, this ratio is √3 / 2.

Next, let's find the y-coordinate by plugging t = π/3 into the equation for y: y = cos(π/3). The cosine of π/3 is 1/2. So, we have y = 1/2. The cosine function gives us the ratio of the adjacent side to the hypotenuse in a right triangle, and at π/3, this ratio is 1/2. Putting it all together, the point on the curve where t = π/3 is ((3√3) / 2, 1/2). This point is where our tangent line will touch the curve. We now have both the slope of the tangent line and a point that it passes through, which are the key ingredients for finding the equation of the line. With these two pieces of information, we're well on our way to finding the tangent line equation.

Step 3: Determining the Equation of the Tangent Line

Alright, we're in the home stretch! We've successfully found the slope of the tangent line at t = π/3 and the coordinates of the point where the tangent line touches the curve. Now, it's time to put it all together and find the equation of the tangent line. To do this, we'll use the point-slope form of a linear equation, which is a super handy tool for this kind of problem. The point-slope form is given by: y - y₁ = m(x - x₁), where m is the slope of the line, and (x₁, y₁) is a point on the line. We have all the values we need!

We already determined that the slope, m, at t = π/3 is -√3 / 3. We also found the point (x₁, y₁) on the curve where t = π/3 to be ((3√3) / 2, 1/2). Now, we simply plug these values into the point-slope form: y - 1/2 = (-√3 / 3)(x - (3√3) / 2). This equation represents the tangent line to the curve at the point where t = π/3. It might look a bit messy right now, but we can simplify it to get a more standard form.

Let's simplify the equation. First, distribute the slope (-√3 / 3) across the terms inside the parentheses: y - 1/2 = (-√3 / 3)x + (-√3 / 3)(- (3√3) / 2). This simplifies to: y - 1/2 = (-√3 / 3)x + (3/2). Notice how the negative signs cancel out in the second term, and √3 * √3 becomes 3. Now, to get the equation in slope-intercept form (y = mx + b), we'll add 1/2 to both sides of the equation: y = (-√3 / 3)x + (3/2) + 1/2. This simplifies further to: y = (-√3 / 3)x + 2. And there you have it! The equation of the tangent line to the curve at t = π/3 is y = (-√3 / 3)x + 2.

Conclusion

And that's a wrap! We've successfully navigated the world of parametric equations and found the equation of the tangent line at a specific point. Remember, the key steps were: first, finding the slope dy/dx using the formula dy/dx = (dy/dt) / (dx/dt); second, determining the coordinates of the point on the curve at the given value of t; and third, using the point-slope form to construct the equation of the tangent line. This process can be applied to many different parametric equations, so practice makes perfect! Understanding these steps and the underlying calculus principles will help you tackle similar problems with confidence.

Calculus can be challenging, but breaking it down into manageable steps makes it much easier to grasp. Don't hesitate to revisit the concepts and practice more examples. Keep exploring, keep learning, and remember, math is an adventure! For further learning and to deepen your understanding of calculus, explore resources like Khan Academy's Calculus Course.