Problem 47
Question
In Exercises \(47-56,\) verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. $$ x^{2}+x y-y^{2}=1, \quad(2,3) $$
Step-by-Step Solution
Verified Answer
The tangent line equation is \(y = \frac{7}{4}x + \frac{3}{2}\) and the normal line equation is \(y = -\frac{4}{7}x + \frac{29}{7}\).
1Step 1: Verify the Point is on the Curve
Substitute the given point \((2,3)\) into the equation \(x^2 + xy - y^2 = 1\) to verify it lies on the curve.Substituting, \(2^2 + 2\cdot3 - 3^2 = 4 + 6 - 9 = 1\).Since the left side equals the right side, the point \((2,3)\) is on the curve.
2Step 2: Differentiate the Curve
To find the slope of the tangent line, we first differentiate the curve implicitly with respect to \(x\):\[\frac{d}{dx}(x^2 + xy - y^2) = \frac{d}{dx}(1)\]Using implicit differentiation, we have:\[2x + y + x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0\]Reorganize to solve for \(\frac{dy}{dx}:\)\[\frac{dy}{dx} = \frac{2x + y}{2y - x}\]
3Step 3: Evaluate the Derivative at the Given Point
Substitute \(x = 2\) and \(y = 3\) into the expression for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{2(2) + 3}{2(3) - 2} = \frac{4 + 3}{6 - 2} = \frac{7}{4}\]The slope of the tangent line at \((2,3)\) is \(\frac{7}{4}\).
4Step 4: Find the Equation of the Tangent Line
Using the point-slope form of the line, \(y - y_1 = m(x - x_1)\), where \(m = \frac{7}{4}\), and \((x_1, y_1) = (2, 3)\):Substitute the values:\[y - 3 = \frac{7}{4}(x - 2)\]Solve for \(y\) to find the equation of the tangent line:\[y = \frac{7}{4}x - \frac{7}{2} + 3\]\[y = \frac{7}{4}x + \frac{6}{4}\]\[y = \frac{7}{4}x + \frac{3}{2}\]
5Step 5: Find the Equation of the Normal Line
The slope of the normal line is the negative reciprocal of the tangent slope. Thus, it is \(-\frac{4}{7}\).Using the point-slope form of the line with this slope:\[y - 3 = -\frac{4}{7}(x - 2)\]Solve for \(y\) to find the equation of the normal line:\[y = -\frac{4}{7}x + \frac{8}{7} + 3\]\[y = -\frac{4}{7}x + \frac{29}{7}\]
Key Concepts
Implicit DifferentiationSlope CalculationCurve VerificationEquation of Tangent Line
Implicit Differentiation
When dealing with equations involving both variables, we often encounter situations where it's not feasible to solve for one variable explicitly in terms of the other.
That's where implicit differentiation comes in handy. Instead of finding a single variable, we differentiate every term with respect to a common variable, usually with respect to \( x \).
This method is crucial for equations like our exercise's curve, \( x^2 + xy - y^2 = 1 \), where both \( x \) and \( y \) are tangled together.Throughout the differentiation process, when differentiating a term involving \( y \), we multiply the derivative by \( \frac{dy}{dx} \).
This allows us to account for \( y \) being a function of \( x \). For example, while differentiating the term \( xy \), we apply the product rule: the derivative of \( xy \) is \( x \cdot \frac{dy}{dx} + y \cdot 1 = x \cdot \frac{dy}{dx} + y \).
This concept is the core of implicit differentiation.
That's where implicit differentiation comes in handy. Instead of finding a single variable, we differentiate every term with respect to a common variable, usually with respect to \( x \).
This method is crucial for equations like our exercise's curve, \( x^2 + xy - y^2 = 1 \), where both \( x \) and \( y \) are tangled together.Throughout the differentiation process, when differentiating a term involving \( y \), we multiply the derivative by \( \frac{dy}{dx} \).
This allows us to account for \( y \) being a function of \( x \). For example, while differentiating the term \( xy \), we apply the product rule: the derivative of \( xy \) is \( x \cdot \frac{dy}{dx} + y \cdot 1 = x \cdot \frac{dy}{dx} + y \).
This concept is the core of implicit differentiation.
Slope Calculation
In calculus, determining the slope of a curve at any given point is a fundamental skill. Once the implicit differentiation is done, you solve the derivative equation to uncover the slope of the tangent line.
For our curve, the derivative we found was \( \frac{dy}{dx} = \frac{2x + y}{2y - x} \).To get the slope at a particular point, like \( (2, 3) \) in our exercise, you substitute the coordinates into your differentiated equation.
Inserting \( x = 2 \) and \( y = 3 \) into \( \frac{dy}{dx} \), the expression boils down to \( \frac{7}{4} \).
This result represents the slope of the tangent line at the point \( (2, 3) \). Understanding this slope helps us define both tangent and normal line equations.
For our curve, the derivative we found was \( \frac{dy}{dx} = \frac{2x + y}{2y - x} \).To get the slope at a particular point, like \( (2, 3) \) in our exercise, you substitute the coordinates into your differentiated equation.
Inserting \( x = 2 \) and \( y = 3 \) into \( \frac{dy}{dx} \), the expression boils down to \( \frac{7}{4} \).
This result represents the slope of the tangent line at the point \( (2, 3) \). Understanding this slope helps us define both tangent and normal line equations.
Curve Verification
Before diving into complex operations on a curve, confirming that the given point lies on the curve is essential. This prevents unnecessary calculations and ensures accuracy in subsequent steps.
In our exercise, substituting the point \( (2,3) \) into the curve's equation \( x^2 + xy - y^2 = 1 \) verifies this.Here's how: Replace \( x \) with 2 and \( y \) with 3.
The calculation becomes \( 2^2 + 2 \cdot 3 - 3^2 = 4 + 6 - 9 \).
This simplifies to 1, which matches the equation's right side, confirming the point is indeed on the curve. Confirming points like this is a simple yet powerful step in solving calculus problems involving curves.
In our exercise, substituting the point \( (2,3) \) into the curve's equation \( x^2 + xy - y^2 = 1 \) verifies this.Here's how: Replace \( x \) with 2 and \( y \) with 3.
The calculation becomes \( 2^2 + 2 \cdot 3 - 3^2 = 4 + 6 - 9 \).
This simplifies to 1, which matches the equation's right side, confirming the point is indeed on the curve. Confirming points like this is a simple yet powerful step in solving calculus problems involving curves.
Equation of Tangent Line
Finding the tangent line at a specific curve point tells us how the curve behaves instantly at that position. This involves using the slope-intercept or point-slope form of line equations.In our example, we utilize the point-slope form: \( y - y_1 = m(x - x_1) \), which is ideal for lines.
Given the slope \( m = \frac{7}{4} \) and the point \( (2, 3) \), substituting into the form gives us \( y - 3 = \frac{7}{4}(x - 2) \).Reorganize to isolate \( y \), resulting in \( y = \frac{7}{4}x + \frac{3}{2} \).
This equation tells us the line tangent to our original curve at point \( (2, 3) \). Tangent lines are instrumental for approximations and for understanding the immediate direction of curves.
Given the slope \( m = \frac{7}{4} \) and the point \( (2, 3) \), substituting into the form gives us \( y - 3 = \frac{7}{4}(x - 2) \).Reorganize to isolate \( y \), resulting in \( y = \frac{7}{4}x + \frac{3}{2} \).
This equation tells us the line tangent to our original curve at point \( (2, 3) \). Tangent lines are instrumental for approximations and for understanding the immediate direction of curves.
Other exercises in this chapter
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