Problem 33 - University Calculus: Early Transcendentals

Question

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. $$x^{2} y^{2}=9, \quad(-1,3)$$

Step-by-Step Solution

Answer

The tangent line is \(y = 3x + 6\), and the normal line is \(y = -\frac{1}{3}x + \frac{8}{3}\).

Step 1: Verify the Point is on the Curve

To confirm the point \((-1, 3)\) is on the curve defined by the equation \(x^2 y^2 = 9\), substitute \(x = -1\) and \(y = 3\) into the equation:\[(-1)^2 (3)^2 = 1 imes 9 = 9\]Since the left side of the equation equals the right side, the point \((-1, 3)\) is indeed on the curve.

Step 2: Find the Derivative

We need to differentiate the equation \(x^2 y^2 = 9\) with respect to \(x\). Using implicit differentiation, rewrite the equation:\[ \frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(9)\]Apply the product rule to \(x^2 y^2\):\[2x y^2 + x^2(2y \frac{dy}{dx}) = 0\]Solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = - \frac{2x y^2}{2x^2 y} = -\frac{y}{x}\]

Step 3: Calculate the Slope of the Tangent Line

Substitute \((-1, 3)\) into the derivative \(\frac{dy}{dx} = -\frac{y}{x}\) to find the slope of the tangent line:\[\frac{dy}{dx}\bigg|_{(-1,3)} = -\frac{3}{-1} = 3\]The slope of the tangent line at \((-1, 3)\) is \(3\).

Step 4: Write the Equation of the Tangent Line

Using the point-slope form of the line equation, \(y - y_1 = m (x - x_1)\), where \((x_1, y_1) = (-1, 3)\) and \(m = 3\):\[y - 3 = 3(x + 1)\]Simplify:\[y = 3x + 6\]This is the equation of the tangent line.

Step 5: Calculate the Slope of the Normal Line

The slope of the normal line is the negative reciprocal of the tangent line's slope. Since the slope of the tangent line is \(3\):\[m_{\text{normal}} = -\frac{1}{3}\]

Step 6: Write the Equation of the Normal Line

Using the point-slope form again for the normal line:\[y - 3 = -\frac{1}{3}(x + 1)\]Simplify:\[y = -\frac{1}{3}x + \frac{8}{3}\]This is the equation of the normal line.

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at exactly one point, without crossing it. At that point, the tangent line has the same slope as the curve. This means that if you drew the tangent line, it would look like it gently meets the curve and then runs parallel to it, at least for a tiny stretch.
To find the tangent line to a curve at a given point, you first need to calculate the slope of the curve at that point. This involves finding the derivative of the curve's equation. Implicit differentiation is often used when the equation of the curve is given in an implicit form, like our current equation, where both variables, x and y, are mixed together in one equation.
Once you have the derivative, evaluate it at the particular point you're interested in to get the slope. You then use this slope in the point-slope form of the line equation to determine the equation of the tangent line. That form is written as:

• Point-slope form: \[ y - y_1 = m(x - x_1) \]

Plugging in our point and slope gives us the actual equation of the tangent line.

Normal Line

The normal line is a line perpendicular to the tangent line at the point where the curve and the tangent meet. It's like a line standing at a right angle to the curve at a particular point.
To find the normal line's slope, you take the slope of the tangent line and find its negative reciprocal. This is because perpendicular lines have slopes that are negative inverses of each other. If the tangent line's slope is 3, the normal line's slope is \(-\frac{1}{3}\).
Just like with tangent lines, you can find the normal line’s equation by using the point-slope form:

• Point-slope form:\[ y - y_1 = m(x - x_1) \]

Substitute the point and the slope you found into this formula to find the normal line equation. This helps you understand how the curve's angle changes at that single point.

Curve Verification

To ensure that a given point lies on a specific curve, you substitute the point's coordinates into the curve’s equation. If both sides of the equation balance, the point is indeed on the curve.
In our given exercise, the equation is \(x^2 y^2 = 9\). By substituting \(x = -1\) and \(y = 3\) into the equation, we verified that \((-1)^2 (3)^2\) simplifies to \(9\), which matches the original equation. This calculation confirms that the point is on the curve.
Curve verification is crucial because it confirms that the subsequent calculations for the tangent and normal lines are meaningful. Without the point being on the curve, these lines wouldn't intersect the curve at that point, making our results irrelevant for understanding the curve's behavior at the supposed intersection.