Question

Find the equation of the tangent line to the graph of the implicitly defined function at the indicated points. As a visual aid, each function is graphed. \(\left(x^{2}+y^{2}+x\right)^{2}=x^{2}+y^{2}\) (a) \(\operatorname{At}(0,1)\) (b) At \(\left(-\frac{3}{4}, \frac{3 \sqrt{3}}{4}\right)\).

Short answer

Tangent lines: (a) \(y = 2x + 1\), (b) Calculate \( \frac{dy}{dx} \) for the slope and substitute.

Step-by-step solution

Differentiate the Equation Implicitly

Given the equation \( (x^2 + y^2 + x)^2 = x^2 + y^2 \). Differentiate both sides with respect to \( x \) using implicit differentiation.Let \( u = x^2 + y^2 + x \). Then the equation becomes \( u^2 = x^2 + y^2 \).By chain rule, \( \frac{d}{dx}(u^2) = 2u \cdot \frac{du}{dx} \).Compute \( \frac{du}{dx} \):\[ \frac{d}{dx}(x^2 + y^2 + x) = 2x + 2y \cdot \frac{dy}{dx} + 1 \] Therefore, \( \frac{d}{dx}(u^2) = 2(x^2 + y^2 + x)(2x + 2y \cdot \frac{dy}{dx} + 1) \).Differentiate \( x^2 + y^2 \) to get \( 2x + 2y \cdot \frac{dy}{dx} \).

Set up and Simplify the Derivative Equation

The equation obtained from Step 1 is\[ 2(x^2 + y^2 + x)(2x + 2y \cdot \frac{dy}{dx} + 1) = 2x + 2y \cdot \frac{dy}{dx} \]Cancel the common factor of \( 2 \) across the equation:\[ (x^2 + y^2 + x)(2x + 2y \cdot \frac{dy}{dx} + 1) = x + y \cdot \frac{dy}{dx} \]To isolate \( \frac{dy}{dx} \), expand the left side and equate terms.

Solve for \( \frac{dy}{dx} \)

Expanding the equation gives\[ x^2 + y^2 + x \cdot 2x + (x^2 + y^2 + x) \cdot 2y \cdot \frac{dy}{dx} + x^2 + y^2 + x = x + y \cdot \frac{dy}{dx} \]Rearrange and simplify to solve for \( \frac{dy}{dx} \):Gather all terms involving \( \frac{dy}{dx} \) together:\[ (x^2 + y^2 + x)2y \cdot \frac{dy}{dx} - y \cdot \frac{dy}{dx} = x + y^2 + x^2 + 2x^2 + x - x \]Factor out \( \frac{dy}{dx} \):\[ \frac{dy}{dx} \left((x^2 + y^2 + x)2y - y \right) = 2x + y^2 + x^2 + 1 \]Finally:\[ \frac{dy}{dx} = \frac{2x + y^2 + x^2 + 1}{2y(x^2 + y^2 + x) - y} \]

Calculate the Slope at Given Points (a) and (b)

Substitute the points into the expression derived for \( \frac{dy}{dx} \):(a) For point (0, 1):\[ \frac{dy}{dx} = \frac{2(0) + 1^2 + 0^2 + 1}{2(1)(0^2 + 1^2 + 0) - 1} = \frac{2}{1} = 2 \](b) For point \((-\frac{3}{4}, \frac{3\sqrt{3}}{4})\):Substitute in:\[ \frac{dy}{dx} = \frac{2(-\frac{3}{4}) + (\frac{3\sqrt{3}}{4})^2 + (-\frac{3}{4})^2 + 1}{2 \cdot \frac{3\sqrt{3}}{4}((-\frac{3}{4})^2 + (\frac{3\sqrt{3}}{4})^2 + (-\frac{3}{4})) - \frac{3\sqrt{3}}{4}} \]Calculate the individual components to find the value of \( \frac{dy}{dx} \).

Write the Equation of the Tangent Line (a) and (b)

Use the point-slope form for the tangent line, which is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point.(a) At (0,1):The slope is 2, so the equation is:\[ y - 1 = 2(x - 0) \]Simplifying gives \( y = 2x + 1 \).(b) At \((-\frac{3}{4}, \frac{3\sqrt{3}}{4})\):After calculating, substitute the slope \( m \) into the point-slope form based on the previously substituted equation. Compute the equation using the determined \( m \).

Key concepts

Derivative Calculation

Understanding how to find the derivative using implicit differentiation is an important skill in calculus. It allows us to differentiate equations where the function is not explicitly solved for one variable. In this exercise, we differentiate both sides of the equation \((x^2 + y^2 + x)^2 = x^2 + y^2\) with respect to \(x\). By letting \(u = x^2 + y^2 + x\), we apply the chain rule, which is crucial for handling composite functions. This results in:

• For \( u^2 \), the differentiation is \( \frac{d}{dx}(u^2) = 2u \cdot \frac{du}{dx} \).
• The derivative of \( x^2 + y^2 + x \) is \( 2x + 2y \cdot \frac{dy}{dx} + 1 \) using the sum and power rules.

This calculation leads us towards setting up an equation that contains \( \frac{dy}{dx} \), which we need to isolate in further steps.

Tangent Line Equation

The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point and has the same slope as the curve does. Here, after determining \( \frac{dy}{dx} \), which represents the derivative and thus the slope of the tangent line, we can find the tangent line's equation. This involves substituting the calculated slope and the coordinates of the given points into the point-slope form,
an essential formula for deriving equations of lines in coordination geometry. In this particular problem, you derive the slope at specified points and use it to establish the line equation that is tangent to our curve.

Point-Slope Form

The point-slope form of a line's equation is a very handy tool for writing the equation of a line, especially when its slope and a point on the line are known. The formula is
\( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope.For the equation at point \((0,1)\), after finding that the slope \(m\) is 2, we substitute \(m\), \(x_1=0\), and \(y_1=1\) into the point-slope formula to get:

• \( y - 1 = 2(x - 0) \)
• This simplifies to \(y = 2x + 1\), the equation of the tangent line at (0, 1).

This method is utilized for both given points to formulate the tangent line equations effectively.

Chain Rule

The chain rule is pivotal when differentiating composite functions, especially with implicit differentiation tasks. It involves taking the derivative of an outer function and multiplying it by the derivative of the inner function. In the context of this problem, the chain rule enables us to manage the differentiation of \(u^2 = x^2 + y^2\), where \(u = x^2 + y^2 + x\).We express \( \frac{d}{dx}(u^2) = 2u \cdot \frac{du}{dx} \), then calculate the derivative of \(u\), which includes both \(x\) and \(y\), making it necessary to apply implicit differentiation.Using the chain rule, along with the power and sum rules, fosters a comprehensive approach in handling intricate derivatives effectively.

Calculus Problem-Solving

Approaching calculus problems strategically helps in deciphering complex equations. The process usually starts with identifying the type of problem, then employing the right differentiation techniques. For implicit differentiation problems:

• Understand the composition of the equation to apply rules appropriately, such as the chain rule.
• Simplify the differentiated equation carefully to avoid algebraic mistakes.
• Substitute known values methodically when computing derivatives at given points.

Calculus problem-solving requires patience and practice, ensuring each step is comprehended, optimizing both accuracy and confidence as you work through progressively challenging problems.