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Chapter 13: Problem 31

Use differentials to approximate the indicated number. The \(y\) -coordinate of the point \(P\) near \((1,2)\) on the curve \(2 x^{3}+2 y^{3}=9 x y\), if the \(x\) -coordinate of \(P\) is \(1.1\)

Short Answer

Expert verified
The approximate \(y\)-coordinate is 2.08.

Step by step solution

01

Understand the Problem

We have a curve defined by the equation \( 2x^3 + 2y^3 = 9xy \). We know a point near \((1, 2)\) on the curve, and we need to approximate the \(y\)-coordinate of the point when \(x = 1.1\).
02

Differentiate the Implicit Function

Differentiate both sides of the equation \( 2x^3 + 2y^3 = 9xy \) with respect to \(x\). Apply implicit differentiation: \(6x^2 + 6y^2 \frac{dy}{dx} = 9y + 9x \frac{dy}{dx}\).
03

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

Rearrange the differential equation to solve for \( \frac{dy}{dx} \):\[6y^2 \frac{dy}{dx} - 9x \frac{dy}{dx} = 9y - 6x^2\]Thus, \( \frac{dy}{dx} = \frac{9y - 6x^2}{6y^2 - 9x} \).
04

Evaluate \( \frac{dy}{dx} \) at the Initial Point

Substitute \( x = 1 \) and \( y = 2 \) into \( \frac{dy}{dx} = \frac{9y - 6x^2}{6y^2 - 9x} \):\[ \frac{dy}{dx} = \frac{9(2) - 6(1)^2}{6(2)^2 - 9(1)} = \frac{18 - 6}{24 - 9} = \frac{12}{15} = \frac{4}{5}\]The slope at the point \((1, 2)\) is \( \frac{4}{5} \).
05

Use Differentials to Approximate \(dy\)

Use the derivative and the change in \(x\) to estimate the change in \(y\):\[ dy \approx \frac{dy}{dx} \cdot dx = \frac{4}{5} \cdot (1.1 - 1) = \frac{4}{5} \cdot 0.1 = 0.08 \].
06

Calculate the Approximate \(y\)-Coordinate

Starting from \(y = 2\) and adding the differential change \(dy\), we find the approximate \(y\)-coordinate:\[ y = 2 + 0.08 = 2.08 \].

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Key Concepts

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

Implicit Differentiation
Implicit differentiation is a technique to find derivatives when the equations are not solved explicitly for one variable. In simpler terms, it's a way to differentiate equations like our given curve, where both x and y are mixed together in the relationship.
  • We start by taking the derivative of each term with respect to x, treating y as a function of x, even though y isn't written that way.
  • This involves using the chain rule. When differentiating y, we multiply by \( \frac{dy}{dx} \) because y is implicitly a function of x.
In our problem, we took the derivative of both sides of the curve equation \( 2x^3 + 2y^3 = 9xy \) with respect to x. This helped us express the rate of change of y in terms of x, which is crucial for subsequent calculations.
Curve Approximation
Curve approximation using differentials is a method to estimate the new position of a point on a curve when its position changes slightly in one direction, like an increase in the x value. It's a powerful tool for quickly estimating values without solving the entire equation again.
  • Using the derivative \( \frac{dy}{dx} \), we can find an approximate change in y, represented by \( dy \).
  • This relies on the assumption that for small changes, the curve can be approximated as a straight line (the tangent at the point).
In this exercise, once we calculated the derivative, we used it to find \( dy \) by multiplying \( \frac{dy}{dx} \) with the small change in x (from 1 to 1.1), which provided an approximation of the curve's new y-coordinate.
Slope Calculation
Calculating the slope of a curve at a specific point involves finding the derivative at that point. This slope gives insight into how steep the curve is and the direction it moves.
  • The slope is the value of the derivative \( \frac{dy}{dx} \) at a particular point. In our case, the point was (1, 2).
  • To find the slope, substitute the x and y values of the point into the expression for the derivative.
For this exercise, after differentiating the curve equation and rearranging, we substituted the values of the initial point into the derivative formula. Calculating this gave us \( \frac{4}{5} \) as the slope at that point, a crucial factor in evaluating the change in y along the curve as x changes. This slope formed the foundation for approximating the y-coordinate.

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Most popular questions from this chapter

Find and classify the critical points of the functions . If a computer algebra system is available. check your results by means of contour plots like those in Figs. 13.10.3-13.10.5. $$ f(x, y)=10+12 x-12 y-3 x^{2}-2 y^{2} $$

Use the normal gradient vector to write an equation of the line (or plane) tangent to the given curve (or surface) at the given point \(P\). $$ 2 x^{2}+3 y^{2}=35 ; \quad P(2,3) $$

Use graphical or numerical methods to find the critical points of \(f\) to four- place accuracy. Then classify them. $$ f(x, y)=x^{4}+2 y^{4}-12 x y^{2}-20 y^{2} $$

Find and classify the critical points of the functions . If a computer algebra system is available. check your results by means of contour plots like those in Figs. 13.10.3-13.10.5. $$ f(x, y)=x^{3}+6 x y+3 y^{2} $$

Use graphical or numerical methods to find the critical points of \(f\) to four- place accuracy. Then classify them. $$ f(x, y)=2 x^{4}-12 x^{2}+y^{2}+8 x $$
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