Tangent Line Equation For Y = 4 Sin X

Hey math whizzes! Today, we're diving into a classic calculus problem: finding the equation of a tangent line. Specifically, we'll be tackling the curve $y = 4 \sin x$ and pinpointing that tangent line at the sweet spot of $\left(\frac{\pi}{6}, 2\right)$. This might sound a bit intimidating, but trust me, it's totally doable and even kind of fun once you get the hang of it. We're aiming to express our final answer in the familiar form $y = mx + b$, where $m$ is our slope and $b$ is our y-intercept. So, buckle up, and let's get this done!

Understanding Tangent Lines

Alright guys, before we jump into the nitty-gritty, let's make sure we're all on the same page about what a tangent line actually is. Think of a curve, like our $y = 4 \sin x$. A tangent line is basically a straight line that just touches the curve at a single point without crossing it nearby. It's like a perfect kiss! The most important thing about a tangent line is that its slope at that specific point is exactly the same as the slope of the curve at that same point. And how do we find the slope of a curve at a point? You guessed it – differentiation! That's where calculus comes in handy. The derivative of a function gives us the instantaneous rate of change, which is precisely the slope of the tangent line at any given point.

So, the game plan is this: first, we need to find the derivative of our function $y = 4 \sin x$. This will give us a general formula for the slope of the tangent line at any point on the curve. Next, we'll plug in the x-coordinate of our specific point, $\frac{\pi}{6}$, into this derivative formula. This will give us the exact slope, our $m$ value, at that particular point $\left(\frac{\pi}{6}, 2\right)$. Once we have the slope ($m$) and a point $\left(x_1, y_1\right)$ that the line passes through (which is given to us as $\left(\frac{\pi}{6}, 2\right)$), we can use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, to find the equation of the tangent line. Finally, we'll rearrange this equation into the desired $y = mx + b$ format to find our $b$ value. It's a step-by-step process, and each step builds on the last, so as long as we're careful, we'll nail it.

Let's talk a bit more about why this is so cool. In mathematics and science, tangent lines are incredibly useful. They help us approximate the behavior of complex functions near a specific point. Imagine you have a really wiggly function, and you only care about what's happening right around one specific spot. The tangent line gives you a simple, linear approximation of that function in that small neighborhood. This concept is fundamental to many areas, including optimization problems, error analysis, and even understanding physical phenomena like velocity and acceleration. The velocity of an object at a certain time is the slope of the tangent line to its position-time graph at that time! So, mastering tangent lines isn't just about solving homework problems; it's about understanding the local behavior of functions, which is a cornerstone of calculus.

Furthermore, the process of finding a tangent line is the very essence of differential calculus. The definition of the derivative itself is based on the limit of secant lines approaching a tangent line. A secant line connects two points on a curve, and as those two points get closer and closer, the secant line pivots and eventually becomes the tangent line. The derivative is the limit of the slope of these secant lines. So, when we calculate the derivative of $y = 4 \sin x$, we are, in a sense, performing this limiting process algebraically. This connection between the geometric idea of a tangent line and the algebraic operation of differentiation is one of the most profound insights of calculus. It allows us to translate geometric intuition into precise mathematical tools.

Think about the sine function itself. It's a wave, oscillating between -1 and 1. Multiplying it by 4 stretches this wave vertically, so $y = 4 \sin x$ oscillates between -4 and 4. The shape of the sine wave is smooth and continuous, meaning it has a well-defined tangent line at every single point. Our specific point, $\left(\frac{\pi}{6}, 2\right)$, lies on this curve because $4 \sin \left(\frac{\pi}{6}\right) = 4 \times \frac{1}{2} = 2$. This confirms we're working with a valid point on the curve. The tangent line at this point will capture the instantaneous rate at which the function is changing precisely at $x = \frac{\pi}{6}$. It’s a snapshot of the curve’s steepness at that exact moment.

Step 1: Find the Derivative

Okay team, the first crucial step is to find the derivative of our function $y = 4 \sin x$. Remember, the derivative of a function tells us the slope of the tangent line at any point on that function. So, we need to whip out our differentiation rules. The function is $y = 4 \sin x$. We can treat the constant '4' as a multiplier. The derivative of $\sin x$ is $\cos x$. Therefore, using the constant multiple rule, the derivative of $y = 4 \sin x$ with respect to $x$ is $y' = \frac{dy}{dx} = 4 \cos x$. This formula, $y' = 4 \cos x$, is our golden ticket to finding the slope of the tangent line at any x-value on the curve $y = 4 \sin x$. It tells us how steep the curve is at any given horizontal position.

Let's break down why the derivative of $\sin x$ is $\cos x$. This is a fundamental result in calculus, often derived using the limit definition of the derivative. Recall that the derivative of a function $f(x)$ is given by $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$. For $f(x) = \sin x$, this becomes $f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}$. Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$, we get $f'(x) = \lim_{h \to 0} \frac{(\sin x \cos h + \cos x \sin h) - \sin x}{h}$. Rearranging terms, we have $f'(x) = \lim_{h \to 0} \frac{\sin x(\cos h - 1) + \cos x \sin h}{h} = \lim_{h \to 0} \left( \sin x \frac{\cos h - 1}{h} + \cos x \frac{\sin h}{h} \right)$. As $h$ approaches 0, we know two important limits: $\lim_{h \to 0} \frac{\sin h}{h} = 1$ and $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$. Substituting these values, we find $f'(x) = \sin x \cdot 0 + \cos x \cdot 1 = \cos x$. This rigorous proof confirms our basic rule. Applying the constant multiple rule, which states that (c (x))' = c '(x), to our function $y = 4 \sin x$, we get $y' = 4 \cdot (\sin x)' = 4 \cos x$. So, the derivative $y' = 4 \cos x$ is indeed the correct expression for the slope of the tangent line to the curve $y = 4 \sin x$ at any point $x$.

This derivative, $y' = 4 \cos x$, is a powerful tool. It doesn't just give us the slope at one point; it gives us a formula that works for all points on the curve. For instance, if we wanted to know the slope at $x=0$, we'd calculate $4 \cos(0) = 4 \times 1 = 4$. At $x=\frac{\pi}{2}$, the slope would be $4 \cos\left(\frac{\pi}{2}\right) = 4 \times 0 = 0$, indicating a horizontal tangent line at that peak of the sine wave. This understanding of the derivative as a function itself, providing slopes across the entire domain, is fundamental to calculus. It allows us to analyze the rate of change dynamically, not just at a single instance. The cosine function, as the derivative of sine, has a phase shift and amplitude that perfectly mirrors the behavior of the slope of the stretched sine wave.

It's also worth noting the domain and range of our derivative. The cosine function has a range of $[-1, 1]$. Therefore, the derivative $y' = 4 \cos x$ will have a range of $[-4, 4]$. This means the steepest the tangent line can ever be on the curve $y = 4 \sin x$ is a slope of 4, and the least steep (in the negative direction) is -4. These occur at the points where $\cos x = \pm 1$, which corresponds to $x = n\pi$ for integer $n$. This gives us valuable insight into the 'steepness' characteristics of the original function $y = 4 \sin x$ across its entire graph.

Step 2: Calculate the Slope at the Given Point

Now that we have our derivative, $y' = 4 \cos x$, we need to find the specific slope ($m$) at our given point $\left(\frac{\pi}{6}, 2\right)$. To do this, we simply substitute the x-coordinate of our point, which is $x = \frac{\pi}{6}$, into the derivative formula. So, $m = y' \left(\frac{\pi}{6}\right) = 4 \cos \left(\frac{\pi}{6}\right)$. We all know (or should quickly look up!) that $\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$. Plugging this in, we get $m = 4 \times \frac{\sqrt{3}}{2}$. Simplifying this, we find our slope $m = 2\sqrt{3}$. This is the value for the slope of our tangent line.

Let's reflect on this value, $m = 2\sqrt{3}$. Since $\sqrt{3}$ is approximately 1.732, our slope is about $2 imes 1.732 = 3.464$. This is a positive and fairly steep slope, which makes sense visually if you picture the graph of $y = 4 \sin x$ at $x = \frac{\pi}{6}$. The sine wave is increasing rapidly at this point, just before it reaches its peak. The value $2\sqrt{3}$ is the precise instantaneous rate of change of the function at that specific x-value. It's the exact steepness of the curve right at $\frac{\pi}{6}$.

Understanding the trigonometric values for common angles like $\frac{\pi}{6}$ (or 30 degrees) is super important in these types of problems. If you're ever unsure, a quick sketch of the unit circle or a reference triangle can save you. For $\frac{\pi}{6}$, we have a 30-60-90 triangle. The side opposite 30 degrees (or $\frac{\pi}{6}$) is 1, the side opposite 60 degrees (or $\frac{\pi}{3}$) is $\sqrt{3}$, and the hypotenuse is 2. Cosine is adjacent/hypotenuse. For the angle $\frac{\pi}{6}$ on the unit circle, the x-coordinate (which represents the cosine) is $\frac{\sqrt{3}}{2}$.

It's also good to consider the context of the point $\left(\frac{\pi}{6}, 2\right)$. We already verified that $y = 4 \sin\left(\frac{\pi}{6}\right) = 4 \times \frac{1}{2} = 2$, so the point is indeed on the curve. The value $x = \frac{\pi}{6}$ is in the first quadrant (between 0 and $\frac{\pi}{2}$), where both sine and cosine are positive. Our slope $m = 2\sqrt{3}$ is also positive, confirming that the function is increasing at this point, as expected for the sine function in this interval.

This calculated slope, $m = 2\sqrt{3}$, is the first piece of our final equation. It dictates the steepness and direction of our tangent line. Without this value, we couldn't proceed to find the full equation of the line. So, take a moment to appreciate this number – it's the heart of our tangent line's orientation.

Step 3: Use the Point-Slope Form

Now that we have our slope $m = 2\sqrt{3}$ and our point $(x_1, y_1) = \left(\frac{\pi}{6}, 2\right)$, we can use the point-slope form of a linear equation. This form is super handy because it directly uses a point and the slope to define a line. The formula is $y - y_1 = m(x - x_1)$. Let's plug in our values: $y - 2 = 2\sqrt{3}\left(x - \frac{\pi}{6}\right)$. This is the equation of our tangent line, but it's not yet in the $y = mx + b$ format we need.

Let's break down the point-slope form. It's derived directly from the definition of slope: $m = \frac{y_2 - y_1}{x_2 - x_1}$. If we let $(x, y)$ be any other point on the line, then the slope between $(x_1, y_1)$ and $(x, y)$ must also be $m$. So, $m = \frac{y - y_1}{x - x_1}$. Multiplying both sides by $(x - x_1)$ gives us the point-slope form: $y - y_1 = m(x - x_1)$. This equation fundamentally states that the change in $y$ relative to the change in $x$ is constant and equal to the slope $m$, starting from the reference point $(x_1, y_1)$.

In our case, $y_1 = 2$, $x_1 = \frac{\pi}{6}$, and $m = 2\sqrt{3}$. So we substitute these values directly into the formula: $y - 2 = 2\sqrt{3} \left( x - \frac{\pi}{6} \right)$. This equation represents the infinite set of points $(x, y)$ that lie on the tangent line. Any point $(x,y)$ that satisfies this equation is on the line. The equation visually describes a line passing through the specific point $(\frac{\pi}{6}, 2)$ with a constant slope of $2\sqrt{3}$.

It's crucial to ensure that $x_1$ and $y_1$ are correctly identified from the given point and that $m$ is correctly calculated from the derivative. A small error in any of these values will propagate through the rest of the calculation. For instance, if we accidentally used $x = \frac{\pi}{3}$ instead of $x = \frac{\pi}{6}$, our slope calculation would be different, and consequently, our final line equation would be incorrect.

Also, remember that $\pi$ is just a number (approximately 3.14159). When we have terms like $2\sqrt{3} \left( - \frac{\pi}{6} \right)$, we treat $\pi$ like any other constant. This will become important in the next step when we rearrange the equation.

Step 4: Rearrange into $y = mx + b$ Form

Our final mission, should we choose to accept it, is to convert the equation $y - 2 = 2\sqrt{3}\left(x - \frac{\pi}{6}\right)$ into the slope-intercept form, $y = mx + b$. This means we need to isolate $y$ on one side of the equation. First, let's distribute the $2\sqrt{3}$ on the right side:

$y - 2 = 2\sqrt{3}x - 2\sqrt{3}\left(\frac{\pi}{6}\right)$

Now, simplify the term involving $\pi$: $2\sqrt{3}\left(\frac{\pi}{6}\right) = \frac{2\sqrt{3}\pi}{6} = \frac{\sqrt{3}\pi}{3}$.

So, the equation becomes:

$y - 2 = 2\sqrt{3}x - \frac{\sqrt{3}\pi}{3}$

To isolate $y$, we add 2 to both sides of the equation:

$y = 2\sqrt{3}x - \frac{\sqrt{3}\pi}{3} + 2$

Now, this equation is in the form $y = mx + b$. We can clearly see that our slope is $m = 2\sqrt{3}$ (which we already found in Step 2) and our y-intercept is $b = 2 - \frac{\sqrt{3}\pi}{3}$.

Let's take a moment to really appreciate what this $b$ value represents. The y-intercept, $b$, is the point where the tangent line crosses the y-axis (where $x=0$). In our case, $b = 2 - \frac{\sqrt{3}\pi}{3}$. Since $\frac{\sqrt{3}\pi}{3}$ is approximately $\frac{1.732 imes 3.14159}{3} \approx \frac{5.44}{3} \approx 1.81$, our y-intercept $b$ is approximately $2 - 1.81 = 0.19$. This means the tangent line crosses the y-axis just a little bit above the origin. This value is derived directly from the initial point and the slope, demonstrating how these components determine the entire position and orientation of the line.

It's important to keep the expression for $b$ exact, using $\pi$ and $\sqrt{3}$, unless the problem specifically asks for an approximation. So, $b = 2 - \frac{\sqrt{3}\pi}{3}$ is the precise value.

We have successfully transformed the point-slope equation into the slope-intercept form. The structure $y = mx + b$ is incredibly useful because it immediately tells us the slope ($m$) and the y-intercept ($b$), giving us a complete picture of the line's characteristics. The original point $\left(\frac{\pi}{6}, 2\right)$ lies on this line, and the slope $2\sqrt{3}$ ensures it's tangent to $y = 4 \sin x$ at that point.

This final form $y = 2\sqrt{3}x + \left(2 - \frac{\sqrt{3}\pi}{3}\right)$ is the equation of the tangent line. It perfectly encapsulates the geometric relationship between the curve and the line at the specified point. It's a beautiful piece of mathematical engineering!

Conclusion

So there you have it, folks! We started with the curve $y = 4 \sin x$ and the point $\left(\frac{\pi}{6}, 2\right)$, and through the magic of calculus, we found the equation of the tangent line. We first found the derivative, $y' = 4 \cos x$, to get the general slope formula. Then, we plugged in $x = \frac{\pi}{6}$ to find our specific slope, $m = 2\sqrt{3}$. Using the point-slope form, $y - 2 = 2\sqrt{3}\left(x - \frac{\pi}{6}\right)$, we rearranged everything into the slope-intercept form $y = mx + b$.

Our final equation is $y = 2\sqrt{3}x + \left(2 - \frac{\sqrt{3}\pi}{3}\right)$. This means that for the equation $y = mx + b$, we have $m = 2\sqrt{3}$ and $b = 2 - \frac{\sqrt{3}\pi}{3}$. Finding tangent lines is a fundamental skill in calculus that helps us understand the local behavior of functions. It’s all about using the derivative to find the slope at a point and then using that slope along with the point itself to construct the equation of a line.

Keep practicing these kinds of problems, guys! The more you do, the more comfortable you'll become with differentiation rules, trigonometric values, and algebraic manipulation. Calculus is a journey, and every solved problem is a step forward. You've got this!