Question

Carry out the following steps. a. Verify that the given point lies on the curve. b. Determine an equation of the line tangent to the curve at the given point. $$x^{4}-x^{2} y+y^{4}=1 ;(-1,1)$$ (Graph cant copy)

Short answer

Answer: Yes, the point (-1, 1) lies on the curve. The equation of the tangent line to the curve at this point is \(y = \frac{2}{3}x + \frac{5}{3}\).

Step-by-step solution

Verify if the point lies on the curve

Plug in the given point's coordinates, (-1, 1), into the equation of the curve: $$(-1)^{4}-(-1)^{2}(1)+(1)^{4}=1$$ Simplify the equation: $$1 - 1 + 1 = 1$$ Since the equation is true, the point (-1, 1) lies on the curve.

Find the first derivative of the given function

First, find the implicit derivative of the equation with respect to x: $$\frac{d}{dx}(x^4-x^2y+y^4)=\frac{d}{dx}(1)$$ Find the derivatives of each term on the left side of the equation: $$4x^3-2xy-x^{2}\frac{dy}{dx}+4y^{3}\frac{dy}{dx}=0$$ Rearrange the equation to obtain \(\frac{dy}{dx}\): $$x^2\frac{dy}{dx}-4y^{3}\frac{dy}{dx}=-4x^{3}+2xy$$ Factor out \(\frac{dy}{dx}\): $$\frac{dy}{dx}(x^{2}-4y^{3})=-4x^{3}+2xy$$ Now, divide both sides by \((x^{2}-4y^{3})\) to get \(\frac{dy}{dx}\): $$\frac{dy}{dx}=\frac{-4x^{3}+2xy}{x^{2}-4y^{3}}$$

Plug in the coordinates of the given point into the first derivative

Substitute the coordinates (-1, 1) into the equation to find the slope of the tangent line at the given point: $$\frac{dy}{dx}\Big|_{(-1,1)}=\frac{-4(-1)^{3}+2(-1)(1)}{(-1)^{2}-4(1)^{3}}$$ Simplify the expression: $$\frac{dy}{dx}\Big|_{(-1,1)}=\frac{(-4+2)}{1-4}=-\frac{2}{-3}=\frac{2}{3}$$

Determine the tangent line's equation

Now that we have the slope and the given point, we can use the point-slope formula to find the equation of the tangent line: $$y - y_{1} = m(x - x_{1})$$ Plug in the coordinates and slope: $$y - 1 = \frac{2}{3}(x - (-1))$$ Distribute the slope: $$y - 1 = \frac{2}{3}x + \frac{2}{3}$$ Add 1 to both sides, and express y in terms of x: $$y = \frac{2}{3}x + \frac{2}{3} + 1$$ $$y = \frac{2}{3}x + \frac{5}{3}$$ The tangent line's equation at the given point (-1, 1) is: $$y = \frac{2}{3}x + \frac{5}{3}$$

Key concepts

Implicit Differentiation

Implicit differentiation is a powerful technique that allows us to find the derivative of functions that are not explicitly solved for one variable in terms of another. This method is essential when dealing with equations where one variable is complicated or impossible to isolate. For example, if we have an equation like \(x^4 - x^2y + y^4 = 1\), it's not straightforward to express \(y\) solely in terms of \(x\).To apply implicit differentiation:

• Differentiating each term of the equation with respect to \(x\), considering \(y\) as a function of \(x\) (i.e., \(y = f(x)\)), requires the application of the chain rule whenever there is a \(y\) inside the term.
• For example, the derivative of \(x^2y\) with respect to \(x\) is \(2xy + x^2\frac{dy}{dx}\). This step crucially uses the product rule and chain rule together.

In our specific problem, after differentiating, rearranging provides a differential equation in terms of \(x\), \(y\), and \(\frac{dy}{dx}\). Solving for \(\frac{dy}{dx}\) gives us the derivative which we can then use to find the tangent line's slope.

Point-Slope Formula

The point-slope formula is one of the foundational tools for working with linear equations. It is perfect for finding the equation of a line when you're given a point on the line and the slope of the line. This formula is expressed as:\[y - y_1 = m(x - x_1)\]Where:

• \((x_1, y_1)\) are the coordinates of the given point on the line.
• \(m\) is the slope of the line.

Using this formula is straightforward:

• Simply plug in the coordinates of the point and the derived slope to construct the line's full equation.

In our exercise, once the slope was determined as \(\frac{2}{3}\) at the point \((-1, 1)\), substituting into the point-slope form allowed us to quickly arrive at the equation for the tangent line. This formula neatly encapsulates the relationship between a slope and a point into an instantaneous equation for the full line.

Coordinate Substitution

Coordinate substitution is a simple, yet critical step in the process of finding the tangent line. After conducting implicit differentiation to find the derivative \(\frac{dy}{dx}\), we obtain a general expression for the slope. To find the specific slope at a particular point:1. **Substitute**: Simply plug in the coordinates of the point of interest into the derivative equation. This gives you the slope of the tangent line at that specific point.2. In the example provided, by substituting \((-1, 1)\) into the derivative equation, we calculate the slope as \(\frac{2}{3}\).This value represents the rate of change of the function at the point. It is critical for forming the tangent line, as it indicates how steep the line will be at that point on the curve.Through coordinate substitution, we harness specific values from a general derivative formula, thus enabling the precise formation of the tangent line equation in the original curve's context.