Problem 9 Find an equation for the tangent... [FREE SOLUTION]

Chapter 4: Problem 9

Find an equation for the tangent line to \(\left(x^{2}+y^{2}\right)^{2}=x^{2}-y^{2}\) at a point \(\left(x_{1}, y_{1}\right)\) on the curve, with \(x_{1} \neq 0,-1,1\). (This curve is a lemniscate.)

Short Answer

Expert verified

Differentiate implicitly, solve for \(\frac{dy}{dx}\), then evaluate at \((x_1, y_1)\), and use point-slope form.

Step by step solution

Understand the Problem

We need to find the equation of the tangent line to the given curve \((x^2 + y^2)^2 = x^2 - y^2\) at a specific point \((x_1, y_1)\) on the curve. The tangent line at this point will represent the instantaneous direction of the curve at \((x_1, y_1)\).

Differentiate Implicitly

Differentiate the entire equation \((x^2 + y^2)^2 = x^2 - y^2\) with respect to \(x\) using implicit differentiation. We'll use the chain rule for the left side and differentiate directly on the right side. The derivative is:\[2(x^2 + y^2)(2x + 2y \frac{dy}{dx}) = 2x - 2y \frac{dy}{dx}\]

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

Rearrange the differentiated equation to solve for \(\frac{dy}{dx}\) which gives the slope of the tangent line:\[2(x^2 + y^2)(2x + 2y \frac{dy}{dx}) - (2x - 2y \frac{dy}{dx}) = 0\]Simplifying this results in a formula for \(\frac{dy}{dx}\).

Evaluate the Slope at \((x_1, y_1)\)

Substitute \(x = x_1\) and \(y = y_1\) into the expression derived for \(\frac{dy}{dx}\) to find the slope \(m\) of the tangent line at the point of interest. Ensure \(x_1\) meets the condition \(x_1 eq 0, -1, 1\).

Use Point-Slope Form of a Line

With the slope \(m\) found in the previous step, use the point-slope form of a line to write the equation of the tangent line:\[y - y_1 = m(x - x_1)\]

Simplify Equation of the Tangent Line

Simplify the equation from the previous step, if necessary, to present a tidy equation for the tangent line.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangent Line

The equation of a tangent line is a fundamental concept in calculus. It represents the line that just "touches" a curve at a specific point, reflecting the slope or direction of the curve at that point. For any curve, finding the tangent line at a point involves knowing the slope of the curve there. In a mathematical sense, the tangent line is the best linear approximation of the curve at that point.

To find the equation of a tangent line, we need two key pieces of information: a point on the curve (given as \((x_1, y_1)\)), and the slope of the curve at that point.
The slope is obtained by differentiating the curve's equation, finding \(\frac{dy}{dx}\).
Using the point-slope form, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point, gives us the tangent line's equation.
Once calculated, this equation can help us understand the curve's behavior at and near \((x_1, y_1)\).

Understanding tangent lines is crucial for problems involving rates of change and local linear approximations.

Lemniscate

A lemniscate is a figure-eight-shaped curve, a distinctive type of algebraic shape. The word originates from the Latin "lemniscus" meaning "ribbon." This particular curve can be described by various equations, one of which is \([(x^2 + y^2)^2 = x^2 - y^2]\).

This curve is symmetric with respect to both axes and the origin, which often makes the calculations interesting but manageable.
Understanding its shape and properties helps in visualizing where and how tangent lines can touch the curve.
It serves as a fascinating example in calculus because, despite its complexity, the application of differentiation techniques such as the chain rule can unravel the slope and curve properties at any given point.
When dealing with lemniscates, particular attention is given to the distinct points where tangent lines possess unique behavior.

This figure provides rich opportunities for exploration in both pure and applied mathematics.

Chain Rule

The chain rule is a vital differentiation technique in calculus, crucial for implicitly differentiating complex equations like that of the lemniscate. It allows us to differentiate composite functions.

When differentiating \((x^2 + y^2)^2 = x^2 - y^2\) with respect to \(x\), the chain rule helps find the derivative of \( (x^2 + y^2)^2 \), which involves an outer function \((u^2)\) and an inner function \((x^2 + y^2)\).
In applying the chain rule, the derivative of the outer function is \(2u\), evaluated at the inner function. You then multiply it by the derivative of the inner function \((2x + 2y \frac{dy}{dx})\).
Combining these steps yields a derivative that incorporates \( \frac{dy}{dx} \), essential for solving for the slope of the tangent line.

The chain rule is especially powerful when faced with intricate expressions, making it one of the cornerstones of differential calculus techniques.

91影视

Find an equation for the tangent line to \(\left(x^{2}+y^{2}\right)^{2}=x^{2}-y^{2}\) at a point \(\left(x_{1}, y_{1}\right)\) on the curve, with \(x_{1} \neq 0,-1,1\). (This curve is a lemniscate.)

Short Answer

Step by step solution

Understand the Problem

Differentiate Implicitly

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

Evaluate the Slope at \((x_1, y_1)\)

Use Point-Slope Form of a Line

Simplify Equation of the Tangent Line

Key Concepts

Tangent Line

Lemniscate

Chain Rule

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