Problem 28
Question
Find the slope of the tangent to each curve at the given point. $$x^{2}+x y+y^{2}-3=0 \text { at }(1,1)$$
Step-by-Step Solution
Verified Answer
The slope of the tangent to the curve at the point (1,1) is 0.
1Step 1: Differentiate the equation implicitly
To find the slope of the tangent at a specific point on the curve defined by an implicit function, we first need to differentiate both sides of the equation with respect to x. Remember to use the product rule where necessary, and that the derivative of y with respect to x is dy/dx.
2Step 2: Simplify the differentiated equation
After taking the derivative of both sides of the equation implicitly, we must collect like terms and solve for dy/dx, which represents the slope of the tangent line.
3Step 3: Substitute the point into the equation
Once we have solved for dy/dx, substitute the x and y values of the given point into the differentiated equation to find the numerical value of the slope at that point.
Key Concepts
Tangent Line SlopeProduct RuleDifferential Calculus
Tangent Line Slope
Understanding the slope of a tangent line to a curve at a given point is an essential concept in differential calculus. It provides the instantaneous rate of change, effectively describing the steepness or inclination of the line in question.
To find this slope for a curve described by an explicit function (where y is given as a function of x), one simply calculates the derivative at the point of interest. However, when dealing with an implicit function (where x and y are mixed together), such as the one presented in the exercise \( x^2 + xy + y^2 - 3 = 0 \) at the point \( (1,1) \), implicit differentiation becomes necessary.
After differentiating implicitly and simplifying, you get a derivative that includes both \( x \) and \( y \) terms. Substituting the x and y coordinates of the given point into the derived equation will yield the slope of the tangent at that precise location. The result is crucial because it encapsulates the behavior of the curve exactly at the point \( (1,1) \) without requiring the entire equation to be explicitly solved for y.
To find this slope for a curve described by an explicit function (where y is given as a function of x), one simply calculates the derivative at the point of interest. However, when dealing with an implicit function (where x and y are mixed together), such as the one presented in the exercise \( x^2 + xy + y^2 - 3 = 0 \) at the point \( (1,1) \), implicit differentiation becomes necessary.
After differentiating implicitly and simplifying, you get a derivative that includes both \( x \) and \( y \) terms. Substituting the x and y coordinates of the given point into the derived equation will yield the slope of the tangent at that precise location. The result is crucial because it encapsulates the behavior of the curve exactly at the point \( (1,1) \) without requiring the entire equation to be explicitly solved for y.
Product Rule
The product rule is a cornerstone operation in differential calculus, offering a prescribed method for differentiating products of functions. It states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. Symbolically, this is expressed as \( \frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v' \).
In the context of our problem, where we are asked to differentiate \( x^2 + xy + y^2 - 3 = 0 \), the product rule must be employed to differentiate the \( xy \) term properly. You apply the rule by treating \( x \) as one function and \( y \) as the other, remembering that the derivative of \( y \) with respect to \( x \) is notated as \( \frac{dy}{dx} \).
In the context of our problem, where we are asked to differentiate \( x^2 + xy + y^2 - 3 = 0 \), the product rule must be employed to differentiate the \( xy \) term properly. You apply the rule by treating \( x \) as one function and \( y \) as the other, remembering that the derivative of \( y \) with respect to \( x \) is notated as \( \frac{dy}{dx} \).
Application of the Product Rule
Following the product rule, the derivative of \( xy \) with respect to \( x \) would be \( x \cdot \frac{dy}{dx} + y \cdot 1 \). This additional step in the differentiation process is fundamental when working with implicit functions and is a recurring theme in many calculus problems.Differential Calculus
Differential calculus is the branch of mathematics that is concerned with how things change. It is the study of the rates at which quantities change and the slopes of curves. The main tool of differential calculus is the derivative, which can be thought of as a measure of how a function value changes as its input changes.
The concept of the derivative can be applied to practically any function, whether it is expressed explicitly as \( y = f(x) \) or implicitly. Solving our exercise requires applying principles of differential calculus to an implicit function, which involves taking the derivative of both sides of the equation with respect to \( x \) and following rules such as the product rule.
Differential calculus is also integral in finding optimal solutions to problems and understanding motion—concepts that pervade all fields of science and engineering. Implicit differentiation, and by extension differential calculus, is thus not just an academic exercise but a critical tool for analyzing and understanding real-world phenomena.
The concept of the derivative can be applied to practically any function, whether it is expressed explicitly as \( y = f(x) \) or implicitly. Solving our exercise requires applying principles of differential calculus to an implicit function, which involves taking the derivative of both sides of the equation with respect to \( x \) and following rules such as the product rule.
Differential calculus is also integral in finding optimal solutions to problems and understanding motion—concepts that pervade all fields of science and engineering. Implicit differentiation, and by extension differential calculus, is thus not just an academic exercise but a critical tool for analyzing and understanding real-world phenomena.
Other exercises in this chapter
Problem 27
Limits Depending on Direction of Approach $$\lim _{x \rightarrow 2^{-}} \frac{5+x}{x-2}$$
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Find the derivative with respect to the independent variable. $$s=\left(8.3 t^{3}-3.8 t\right)^{-2}$$
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Find the derivative of each function. Verify some of your results by calculator. As usual, the letters \(a, b, c, \ldots\) represent constants. Derivative of a
View solution Problem 28
$$y=x(x+2)(x-9)^{2}$$
View solution