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Chapter 3: Problem 31

Find an equation of the line \(l\) tangent to the graph of the equation at the given point. $$ x y^{2}=18 ;(2,-3) $$

Short Answer

Expert verified
The tangent line equation is \(y = \frac{3}{4}x - \frac{9}{2}\).

Step by step solution

01

Identify the Function and the Point

The function given is an implicit equation: \( xy^2 = 18 \). You are asked to find the tangent line at the point \((2, -3)\).
02

Differentiate Implicitly

Since the given equation is implicit, use implicit differentiation to find \(\frac{dy}{dx}\). Start by differentiating both sides of the equation \(xy^2 = 18\) with respect to \(x\). The left side becomes \(y^2 + 2xy \cdot \frac{dy}{dx} = 0\). The right side is constant so its derivative is 0.
03

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

From the differentiated equation \(y^2 + 2xy \cdot \frac{dy}{dx} = 0\), solve for \(\frac{dy}{dx}\):\[2x y \frac{dy}{dx} = -y^2 \]Then, \[\frac{dy}{dx} = \frac{-y^2}{2xy} = \frac{-y}{2x}\]
04

Evaluate the Derivative at the Given Point

Substitute \(x = 2\) and \(y = -3\) into \(\frac{dy}{dx} = \frac{-y}{2x}\) to find the slope of the tangent line:\[\frac{dy}{dx}\bigg|_{(2, -3)} = \frac{-(-3)}{2 \cdot 2} = \frac{3}{4}\]
05

Use the Point-Slope Form to Find the Tangent Line Equation

The point-slope form of the line equation is \(y - y_1 = m(x - x_1)\) where \((x_1, y_1)\) is the point \((2, -3)\) and \(m = \frac{3}{4}\). Substitute these values in:\[y + 3 = \frac{3}{4}(x - 2)\]
06

Simplify the Equation

Simplify the equation \(y + 3 = \frac{3}{4}x - \frac{3}{2}\) to obtain the tangent line equation:\[y = \frac{3}{4}x - \frac{3}{2} - 3\]\[y = \frac{3}{4}x - \frac{9}{2}\]

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Key Concepts

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

Tangent Line
A tangent line to a curve at a specific point is a straight line that just "touches" the curve at that point. It doesn't cross the curve at that point and matches the curve's slope. By understanding tangent lines, you get a glimpse into how the function behaves at a specific point.
  • Consider the curve given by the equation \( xy^2 = 18 \), and the point \((2, -3)\).
  • The tangent line will have the same slope as the curve at this point.
  • This slope is crucial because it shows the rate at which \(y\) changes with respect to \(x\), right at that point on the curve.
Understanding the concept of tangent lines helps in analyzing the instantaneous rate of change of a function.
Point-Slope Form
The point-slope form of a line is a convenient way to write the equation of a line when you know one point on the line and the slope. The formula is \(y - y_1 = m(x - x_1)\) where \((x_1, y_1)\) is a point on the line, and \(m\) is the slope.
  • For this exercise, the point \((2, -3)\) is given.
  • The slope, found using the derivative at that point, is \(\frac{3}{4}\).
  • By substituting these into the point-slope form, you derive the equation for the tangent line.
This is often a starting point for finding the equation of a line, facilitating a deeper understanding of linear equations.
Derivative
A derivative indicates how a function changes as its input changes. It is essentially the slope of the tangent line at a particular point on a curve. Finding the derivative when dealing with implicit functions, like \( xy^2 = 18 \), involves implicit differentiation.
  • Implicit differentiation allows us to find the derivative without explicitly solving for \(y\) in terms of \(x\).
  • In this scenario, differentiating implicitly gives \(\frac{dy}{dx} = \frac{-y}{2x}\).
  • This result helps in calculating the slope of the tangent line precisely where needed, here at the specific point \((2, -3)\).
Understanding derivatives is critical for comprehending rates of change and for when explicit functions can鈥檛 be directly solved.

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