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Assuming that the equations in Exercises \(25-28\) define \(y\) as a differentiable function of \(x,\) use Theorem 8 to find the value of \(d y / d x\) at the given point. $$ x^{2}+x y+y^{2}-7=0, \quad(1,2) $$

Short Answer

Expert verified
The value of \( \frac{dy}{dx} \) at point (1,2) is \( -\frac{4}{5} \).

Step by step solution

01

Write down the equation

We're given the equation \( x^2 + xy + y^2 - 7 = 0 \) and the point \((1,2)\). We need to find \( \frac{dy}{dx} \) at this point.
02

Find the partial derivative with respect to x

Take the derivative of both sides of the equation \( x^2 + xy + y^2 - 7 = 0 \) with respect to \(x\). This gives: \( 2x + y + x\frac{dy}{dx} = 0 \).
03

Find the partial derivative with respect to y

Now, differentiate the equation \( x^2 + xy + y^2 - 7 = 0 \) with respect to \(y\). The result is: \( x + 2y\frac{dy}{dx} = 0 \).
04

Use implicit differentiation to solve for dy/dx

From implicit differentiation, we need to combine the derivatives to find \(\frac{dy}{dx}\). Apply implicit differentiation: \[ \frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = 0 \] gives: \[ 2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \].
05

Simplify and substitute the point (1,2) into the derivative equation

After simplifying, we rearrange terms to get \( x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - y \). Substitute \( x = 1 \) and \( y = 2 \) into this equation: \[ 1 \frac{dy}{dx} + 2(2)\frac{dy}{dx} = -2(1) - 2 \].
06

Solve for dy/dx

Simplify \(1\frac{dy}{dx} + 4\frac{dy}{dx} = -4\) to \(5\frac{dy}{dx} = -4\), which gives \(\frac{dy}{dx} = -\frac{4}{5}\).

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Key Concepts

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

Partial Derivatives
When dealing with multivariable functions, we often use partial derivatives to understand how the function changes with respect to one variable while keeping others constant. In this exercise, we have an equation involving both \(x\) and \(y\), so we are interested in its behavior with each variable separately.
  • **Partial derivative with respect to \(x\):** This considers \(y\) as a constant and finds the rate of change as \(x\) varies. For the equation \(x^2 + xy + y^2 - 7 = 0\), the partial derivative with respect to \(x\) is \(2x + y + x\frac{dy}{dx}\).
  • **Partial derivative with respect to \(y\):** This considers \(x\) as a constant and observes changes with varying \(y\). For the same equation, the partial derivative with respect to \(y\) is \(x + 2y\frac{dy}{dx}\).
Partial derivatives are crucial for implicit differentiation, which we use to find \(\frac{dy}{dx}\) in equations where \(y\) is not explicitly solved in terms of \(x\). Understanding how each variable influences the function is key.
Implicit Function Theorem
The implicit function theorem is a fundamental concept in calculus, allowing us to differentiate functions even when they are not given explicitly. In our scenario, the equation \(x^2 + xy + y^2 - 7 = 0\) relates \(x\) and \(y\) implicitly.
This theorem states that if a function \(F(x, y) = 0\) is continuously differentiable near a point \((x_0, y_0)\), and the partial derivative of \(F\) with respect to \(y\) is non-zero at this point, then \(y\) can locally be a differentiable function of \(x\).
In simpler terms, at the point \((1, 2)\), because the partial derivative \(\frac{\partial F}{\partial y} = 2y\) is non-zero, the theorem assures us that \(y\) varies smoothly with \(x\). We combine both partial derivatives derived earlier to solve for \(\frac{dy}{dx}\), following the logic of the implicit function theorem. This is why we equate and manipulate them to eventually solve for the derivative.
Equation of a Curve
Equations that describe curves on a plane, especially those involving multiple variables, sometimes define relationships implicitly. The given equation \(x^2 + xy + y^2 - 7 = 0\) represents such a curve. Unlike functions that express \(y\) explicitly in terms of \(x\), implicit equations describe relationships without isolating a variable.
  • These equations describe level curves or contour lines, representing specific values on a surface or in space.
  • Analyzing these geometric shapes or paths helps understand how variables depend on one another.
In the example provided, by examining the partial derivatives and using implicit differentiation, we find the slope of the curve at a given point. This slope \(\frac{dy}{dx}\) tells us the rate of change or steepness at the point \((1, 2)\) on the curve described by the equation. Understanding the equation of a curve in this way combines the beauty of geometry with the precision of calculus.

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Most popular questions from this chapter

By considering different paths of approach, show that the functions in Exercises \(35-42\) have no limit as \((x, y) \rightarrow(0,0) .\) $$ f(x, y)=-\frac{x}{\sqrt{x^{2}+y^{2}}} $$ Graph cannot copy

In Exercises \(65-70,\) you will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level critical plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. $$ f(x, y)=x^{2}+y^{3}-3 x y, \quad-5 \leq x \leq 5, \quad-5 \leq y \leq 5 $$

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In Exercises \(31-36,\) find the linearization \(L(x, y)\) of the function \(f(x, y)\) at \(P_{0} .\) Then find an upper bound for the magnitude \(|E|\) of the error in the approximation \(f(x, y) \approx L(x, y)\) over the rectangle \(R\) . $$ \begin{array}{l}{f(x, y)=1+y+x \cos y \text { at } P_{0}(0,0)} \\ {R :|x| \leq 0.2, \quad|y| \leq 0.2} \\ {\text { (Use }|\cos y| \leq 1 \text { and }|\sin y| \leq 1 \text { in estimating } E . )}\end{array} $$

In Exercises \(65-70,\) you will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level critical plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. $$ \begin{array}{l}{f(x, y)=5 x^{6}+18 x^{5}-30 x^{4}+30 x y^{2}-120 x^{3}} \\\ {-4 \leq x \leq 3, \quad-2 \leq y \leq 2}\end{array} $$

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