Question

In Exercises, use implicit differentiation to find an equation of the tangent line to the graph at the given point. $$ x+y-1=\ln \left(x^{2}+y^{2}\right), \quad(1,0) $$

Short answer

The equation of the tangent line is \(y=0\).

Step-by-step solution

Applying Implicit Differentiation

Begin by taking the derivative of both sides, applying the rules of logarithmic differentiation. The derivative becomes \(1 + y' = (2x+2yy')/(x^2+y^2)\), where \(y'\) is the derivative of y.

Solving for y'

Next, isolate \(y'\) by moving all other terms to the opposite side of the equation: \(y'=(2x+2yy'-(x^2+y^2))/(x^2+y^2)\). Now we can simplify the equation to: \(y'=(2x-x^2+2yy'-y^2)/(x^2+y^2)\). Finally, after factoring out \(y'\), we get: \(y' = (2x-x^2 - y^2)/(x^2 -2y + y^2)\).

Finding the Slope at the Given Point

Substitute the given point (1,0) into the derivative to find the slope of the tangent line: \(m=y'(1,0)= 1 - 1^2 - 0^2 / 1^2 - 2*0 + 0^2 = 0\).

Writing the Equation of the Tangent Line

With a slope of 0 and passing through the point (1,0), the equation of the tangent line is simply: \(y=0\)

Key concepts

Tangent Line Equation

Finding the equation of a tangent line is like trying to draw a line that just kisses the curve at a particular point, without cutting through it. This can tell us useful things about the function, like how it's changing at that spot.
To determine the equation of the tangent line, we need two things: the slope of the tangent line (how steep it is) and a point through which the line passes.

• The slope tells us the direction of the line and can be found using the derivative of the function at the point of interest.
• Once we have the slope, we use the point-slope form of a line equation (\(y - y_1 = m(x - x_1)\)) to write the equation, where \(m\) is the slope, and \((x_1, y_1)\) is a point on the line.

In the exercise example, we identified the slope of the tangent line was 0 at the point (1,0), which means the line is perfectly horizontal. As a result, the equation of the tangent line is simply \(y = 0\).
Understanding how to find the tangent line is a foundational skill in calculus, helping you analyze and understand the characteristics of curves.

Logarithmic Differentiation

Logarithmic differentiation is a clever method used to take derivatives of more complex or intricate functions, especially those involving products, quotients, or variables in exponents.
Here's why it's useful: taking the logarithm of both sides of an equation can simplify the process of differentiation. This is particularly handy when dealing with equations that involve logarithms, as seen in the exercise.

• Apply the properties of logarithms to break down complex expressions.
• Differentiate both sides using implicit differentiation, which means treating one variable as a function of another.

In our example, the function was \(x+y-1=\ln(x^2+y^2)\). By differentiating both sides and applying the rules of logarithmic differentiation, we determined the derivative \(y'\).
This technique streamlines complex problems, making them more manageable and is highly valuable for derivatives that aren't straightforward.

Slope Calculation

In calculus, the slope of a tangent line is the same as the slope of the curve at a given point. It's like zooming in on a single moment of the curve to find out how fast or slow it's changing.
The concept of the slope helps us understand the rate of change, which is crucial for identifying dynamic behavior of functions.

• The slope at a given point is found by evaluating the derivative at that point.
• This value tells us whether the curve is going up, down, or leveling out at that particular point.

Say the slope is 0, as in our example: the tangent is perfectly horizontal, indicating no change at that point.
The slope calculation is essential not only in creating the tangent line but also aids in comprehending the nature and behavior of functions depicted by curves.

Calculus Problems

Calculus is a vast and fascinating area of mathematics that helps us solve complex problems related to change and motion. Whether it's calculating the area beneath a curve or finding a precise tangent, calculus is all about understanding the dynamics of change within mathematical models.

• When tackling calculus problems, break them down into smaller, manageable parts.
• Use differentiation to find rates of change, slopes, or tangents.
• Apply integration to calculate the areas and accumulations.

The calculus problem presented in the exercise involves finding the tangent line through implicit and logarithmic differentiation, showcasing a practical application of these techniques.
Always approach calculus problems step-by-step, applying each concept methodically. With practice, tackling calculus challenges becomes much more intuitive, allowing for a deeper understanding of complex systems and functions.