Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find the equation of the line tangent to the curve \(x^{3}+y^{3}-3 x^{2} y^{2}+1=0\) at the point \((1,1) .\)

Step-by-Step Solution

Answer

The equation of the line tangent to the curve \(x^{3}+y^{3}-3 x^{2} y^{2}+1=0\) at the point (1,1) is described in the solution steps.

Step 1: Differentiate The Equation

Begin by finding the derivative of the given equation. To do this, apply the Chain Rule and Power Rule in differentiating each term with respect to x. Leave y's derivative as is, denoted by \(y'\) because y is a function of x.

Step 2: Substitute The Given Point Into The Derivative

Now you have obtained an equation in terms of x, y, and \(y'\). Next, substitute the given point (1,1) into this equation to solve for \(y'\). The solution will be the slope of the tangent line at the point (1,1).

Step 3: Form The Equation Of The Tangent Line

With the slope from Step 2 and the given point (1,1), use the slope-intercept form \(y = mx + c\) to find the equation of the tangent line. Substitute \(m = slope\) from the previous step and the point (1,1) into the equation to find c. Finally, write down the equation of the tangent line.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when differentiating equations that involve two variables, say x and y, where y is defined implicitly rather than explicitly as a function of x. In the equation \(x^{3} + y^{3} - 3x^{2}y^{2} + 1 = 0\), we treat y as an implicit function of x. This means that instead of expressing y explicitly in terms of x, we differentiate each term of the equation with respect to x while considering y as a dependent variable.

To apply implicit differentiation, follow these steps:

• Differentiate each term of the equation individually concerning x.
• When differentiating terms containing y, apply the derivative rules you would normally use, but multiply them by \(y'\), the derivative of y, as we're implicitly treating y as a function of x.
• Rearrange the differentiated equation to solve for \(y'\).

Implicit differentiation allows us to find the derivative even if we cannot isolate y explicitly.

Chain Rule

The Chain Rule is a fundamental concept in calculus used for differentiating composite functions. In the context of differentiating functions like \(y^{3}\) or \(-3x^{2}y^{2}\), the Chain Rule becomes essential. In essence, it helps us tackle functions that are nested within others.

Here's how you might apply the Chain Rule in implicit differentiation:

• Identify the outer function and the inner function in the term you're differentiating. For \(y^{3}\), the outer function is \(u^{3}\) with \(u = y\), so we apply the Power Rule to find the derivative, which is \(3u^{2}\).
• Differentiating \(u = y\) with respect to x, we end up with \(3y^{2} \cdot y'\). This is because the derivative of y concerning x is not just y, but \(y'\).
• Repeat this process for each term where the variable is embedded within another expression or function.

Understanding and applying the Chain Rule is crucial for tackling implicit differentiation problems.

Power Rule

The Power Rule is one of the simplest yet most powerful rules in calculus. It allows us to find the derivative of expressions of the form \(x^n\), where n is any real number. In our exercise, it is vital when differentiating terms like \(x^{3}\) or \(y^{3}\).

To apply the Power Rule:

• Consider the term \(x^n\). The derivative with respect to x is \(nx^{n-1}\).
• Apply this principle whenever you encounter a term which is a power of a variable, whether it’s x or y. So, the derivative of \(x^3\) is \(3x^2\).
• For terms like \(y^3\), apply the Power Rule to get \(3y^2\), and remember to multiply by \(y'\) using the Chain Rule since y is a function of x.

The Power Rule, in combination with other rules such as the Chain Rule, helps simplify the process of differentiation in functions involving powers.

Slope-Intercept Form

The slope-intercept form is a common way of writing the equation of a line. It is expressed as \(y = mx + c\), where m is the slope of the line, and c is the y-intercept. This form is highly valuable when forming the equation of a tangent line.

Here's how you can use the slope-intercept form in our exercise:

• Determine the slope of the tangent line, which has been calculated as \(y'\) at the given point, which in this case is (1,1).
• Insert this slope (m) into the equation \(y = mx + c\).
• Use the coordinates of the given point (1,1) to solve for c, ensuring the line goes through this point.

Once you're done, you'll have a complete equation in the slope-intercept form that represents the tangent line to the curve at that specific point. This form makes it easy to graph and understand the line's behavior, providing deeper insights into the function's characteristics.