Tangent Line Equation: Curve $x^2y^2 + 4xy = 12y$ At (2, 1)

Hey guys! Today, we're diving into a super interesting problem from calculus: finding the equation of a tangent line to a curve. Specifically, we're looking at the curve defined by the equation $x^2y^2 + 4xy = 12y$ at the point $(2, 1)$. This might sound intimidating at first, but trust me, we'll break it down step by step and you'll see it's totally manageable. So, buckle up, grab your pencils, and let's get started!

Understanding the Problem: What Are We Trying to Find?

Before we jump into the calculations, let's make sure we understand exactly what we're trying to find. When we talk about the equation of a tangent line at a specific point on a curve, we're talking about a straight line that just touches the curve at that point. Imagine drawing a line that skims the curve at $(2, 1)$ – that's the tangent line. To define any straight line, we need two key pieces of information:

1. A point on the line: We already have this! It's the point $(2, 1)$ where the tangent line touches the curve.
2. The slope of the line: This tells us how steep the line is. This is where the calculus comes in. We'll need to find the derivative of the curve's equation to determine the slope of the tangent line.

So, our main goal is to find the slope of the tangent line at the point $(2, 1)$. Once we have that, we can use the point-slope form of a linear equation to write the equation of the line. Sounds like a plan, right? Let’s dive into the first step: finding the derivative.

Step 1: Implicit Differentiation – Finding dy/dx

The equation $x^2y^2 + 4xy = 12y$ isn't in the typical $y = f(x)$ form. This means we'll need to use a technique called implicit differentiation to find $\frac{dy}{dx}$. Implicit differentiation allows us to find the derivative of $y$ with respect to $x$ even when $y$ isn't explicitly defined as a function of $x$.

Here's the basic idea:

We differentiate both sides of the equation with respect to $x$, remembering to apply the chain rule whenever we differentiate a term involving $y$. The chain rule is crucial because $y$ is a function of $x$, so its derivative with respect to $x$ is $\frac{dy}{dx}$. Let’s walk through each term:

• Differentiating $x^2y^2$: This term requires both the product rule and the chain rule. The product rule states that the derivative of $uv$ is $u'v + uv'$. Here, let $u = x^2$ and $v = y^2$. So, we get:

  • $u' = 2x$
  • $v' = 2y \frac{dy}{dx}$ (using the chain rule)

  Applying the product rule, the derivative of $x^2y^2$ is $2xy^2 + 2x^2y \frac{dy}{dx}$.

• Differentiating $4xy$: Again, we need the product rule. Let $u = 4x$ and $v = y$. So:

  • $u' = 4$
  • $v' = \frac{dy}{dx}$

  The derivative of $4xy$ is $4y + 4x \frac{dy}{dx}$.

• Differentiating $12y$: This is a straightforward application of the chain rule. The derivative of $12y$ is $12 \frac{dy}{dx}$.

Putting it all together, the derivative of the entire equation is:

$2xy^2 + 2x^2y \frac{dy}{dx} + 4y + 4x \frac{dy}{dx} = 12 \frac{dy}{dx}$

Next step: We need to isolate $\frac{dy}{dx}$. This means gathering all the terms with $\frac{dy}{dx}$ on one side and the remaining terms on the other side of the equation. Let's do it! It’s like solving for a regular variable, just a bit more involved because we have this $\frac{dy}{dx}$ hanging around.

Step 2: Isolating dy/dx

Now that we have the differentiated equation, our goal is to get $\frac{dy}{dx}$ by itself. This involves some algebraic manipulation. Think of it as a puzzle – we just need to rearrange the pieces!

Starting with the equation:

$2xy^2 + 2x^2y \frac{dy}{dx} + 4y + 4x \frac{dy}{dx} = 12 \frac{dy}{dx}$

Let's move all the terms containing $\frac{dy}{dx}$ to the left side and the other terms to the right side. This gives us:

$2x^2y \frac{dy}{dx} + 4x \frac{dy}{dx} - 12 \frac{dy}{dx} = -2xy^2 - 4y$

Now, we can factor out $\frac{dy}{dx}$ from the left side:

$\frac{dy}{dx}(2x^2y + 4x - 12) = -2xy^2 - 4y$

Finally, to isolate $\frac{dy}{dx}$, we divide both sides by the expression in the parentheses:

$\frac{dy}{dx} = \frac{-2xy^2 - 4y}{2x^2y + 4x - 12}$

We can simplify this a bit by dividing both the numerator and the denominator by 2:

$\frac{dy}{dx} = \frac{-xy^2 - 2y}{x^2y + 2x - 6}$

Awesome! We've found an expression for $\frac{dy}{dx}$. This formula will give us the slope of the tangent line at any point $(x, y)$ on the curve. The next step is to plug in our specific point, $(2, 1)$, to find the slope at that location.

Step 3: Evaluating dy/dx at (2, 1)

We've got the formula for $\frac{dy}{dx}$, which represents the slope of the tangent line at any point on the curve. Now, we need to find the slope specifically at the point $(2, 1)$. This is a simple matter of plugging in $x = 2$ and $y = 1$ into our expression for $\frac{dy}{dx}$.

Recall that:

$\frac{dy}{dx} = \frac{-xy^2 - 2y}{x^2y + 2x - 6}$

Substituting $x = 2$ and $y = 1$, we get:

$\frac{dy}{dx} = \frac{-(2)(1)^2 - 2(1)}{(2)^2(1) + 2(2) - 6}$

Let's simplify this:

$\frac{dy}{dx} = \frac{-2 - 2}{4 + 4 - 6} = \frac{-4}{2} = -2$

Fantastic! We've found the slope of the tangent line at the point $(2, 1)$. The slope, often denoted by $m$, is $-2$. This means that for every 1 unit we move to the right along the tangent line, we move 2 units down. Now that we have the slope and a point, we're ready to write the equation of the tangent line.

Step 4: Finding the Equation of the Tangent Line

We're in the home stretch! We know the slope of the tangent line ($m = -2$) and a point it passes through ($(2, 1)$). To write the equation of the line, we can use the point-slope form, which is a super handy formula:

$y - y_1 = m(x - x_1)$

where $(x_1, y_1)$ is the point and $m$ is the slope.

Plugging in our values, we have:

$y - 1 = -2(x - 2)$

Now, let's simplify this equation to get it into slope-intercept form ($y = mx + b$):

$y - 1 = -2x + 4$

Add 1 to both sides:

$y = -2x + 5$

Boom! We've found the equation of the tangent line to the curve $x^2y^2 + 4xy = 12y$ at the point $(2, 1)$. The equation is $y = -2x + 5$.

Conclusion: We Did It!

Guys, we tackled a challenging calculus problem and came out victorious! We successfully found the equation of the tangent line to the curve $x^2y^2 + 4xy = 12y$ at the point $(2, 1)$. We did this by:

1. Understanding the problem and what we needed to find.
2. Using implicit differentiation to find $\frac{dy}{dx}$.
3. Isolating $\frac{dy}{dx}$ through algebraic manipulation.
4. Evaluating $\frac{dy}{dx}$ at the point $(2, 1)$ to find the slope.
5. Using the point-slope form to write the equation of the tangent line.

Remember, the key to mastering calculus is practice and breaking down complex problems into smaller, manageable steps. Keep practicing, and you'll become a calculus whiz in no time! If you have any questions or want to try another problem, let me know. Keep up the great work!