Question

Find the equation of the tangent line to the graph of the implicitly defined function at the indicated points. As a visual aid, each function is graphed. \((x-2)^{2}+(y-3)^{2}=9\) (a) \(\operatorname{At}\left(\frac{7}{2}, \frac{6+3 \sqrt{3}}{2}\right)\). (b) \(\operatorname{At}\left(\frac{4+3 \sqrt{3}}{2}, \frac{3}{2}\right)\).

Short answer

(a) Slope is \(-\frac{1}{\sqrt{3}}\), equation: \(y = -\frac{1}{\sqrt{3}}x + C\). (b) Slope is \(\sqrt{3}\), equation: \(y = \sqrt{3}x - C'\).

Step-by-step solution

Understand the given equation

The given equation \((x-2)^2+(y-3)^2=9\) represents a circle centered at \((2, 3)\) with a radius of 3. We need to find the tangent lines at the specific points provided.

Differentiate the equation implicitly

To find the slope of the tangent line, differentiate each side of the equation \((x-2)^2 + (y-3)^2 = 9\) with respect to \(x\). The derivative is: \[ 2(x-2) + 2(y-3)\frac{dy}{dx} = 0 \]Simplify the expression:\[ (x-2) + (y-3)\frac{dy}{dx} = 0 \]

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

Rearrange the differentiated equation to solve for \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = -\frac{x-2}{y-3} \]

Substitute point (a) into the derivative

For (a), substitute \(x = \frac{7}{2}\) and \(y = \frac{6+3\sqrt{3}}{2}\) into \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = -\frac{\frac{7}{2}-2}{\frac{6+3\sqrt{3}}{2}-3} = -\frac{\frac{3}{2}}{\frac{3 + 3\sqrt{3}}{2} - \frac{6}{2}} \] \[ = -\frac{\frac{3}{2}}{\frac{3\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} \]

Equation of tangent line (a)

The slope is \(-\frac{1}{\sqrt{3}}\). Use the point-slope formula for the tangent line:\[ y - y_1 = m(x - x_1) \]Substitute \(x_1 = \frac{7}{2}\), \(y_1 = \frac{6+3\sqrt{3}}{2}\), and \(m = -\frac{1}{\sqrt{3}}\):\[ y - \frac{6+3\sqrt{3}}{2} = -\frac{1}{\sqrt{3}}(x - \frac{7}{2}) \]

Simplify the equation for (a)

Multiply both sides to eliminate fractions and simplify:First simplify the expression:The tangent line is:\[ y = -\frac{1}{\sqrt{3}}x + \frac{7}{2\sqrt{3}} + \frac{6+3\sqrt{3}}{2} \]

Substitute point (b) into the derivative

For (b), substitute \(x = \frac{4+3\sqrt{3}}{2}\) and \(y = \frac{3}{2}\):\[ \frac{dy}{dx} = -\frac{\frac{4+3\sqrt{3}}{2} - 2}{\frac{3}{2} - 3} = -\frac{\frac{4+3\sqrt{3}-4}{2}}{\frac{-3}{2}} \]\[ = \frac{3\sqrt{3}}{3} = \sqrt{3} \]

Equation of tangent line (b)

The slope is \(\sqrt{3}\). Use the point-slope formula for the tangent line:\[ y - y_1 = m(x - x_1) \]Substitute \(x_1 = \frac{4+3\sqrt{3}}{2}\), \(y_1 = \frac{3}{2}\), and \(m = \sqrt{3}\):\[ y - \frac{3}{2} = \sqrt{3}(x - \frac{4+3\sqrt{3}}{2}) \]

Simplify the equation for (b)

Multiply both sides to simplify the tan-line equation:The tangent line is:\[ y = \sqrt{3}x - 3 - \frac{3\sqrt{3}}{2} + \frac{3}{2} \]

Key concepts

Tangent Line

A tangent line is a straight line that touches a curve at just one point without crossing it. Its slope represents how steep the curve is at that point. Finding the equation of a tangent line helps us understand the behavior of a curve at a specific location, providing insight into the function's rate of change. To find the tangent line on a curve, you first need the point on the curve where the tangent line touches and the slope at that point.
For example, with the equation of a circle, we can find the tangent line by determining the slope through differentiation. Once you have the slope, you can use it to form the equation of the tangent line through a formula known as the "point-slope" form. Being able to calculate tangent lines is crucial in calculus as it is a primary step in understanding instantaneous rates of change.

Equation of a Circle

The equation of a circle is typically expressed in standard form as \((x-h)^2 + (y-k)^2 = r^2\). Here \((h, k)\) are the coordinates of the circle's center, and \(r\) is the radius. The circle's equation represents all the points equidistant \(r\) from the center point. In the given exercise, the equation \((x-2)^2 + (y-3)^2 = 9\) represents a circle with a center at \((2, 3)\) and a radius of 3.
When working with such equations, particularly in calculus, the goal might be to understand the relationship between points on the circle and other properties such as tangent lines. Calculating derivatives, which are often needed, requires implicit differentiation since the equation involves both \(x\) and \(y\) terms on one side. By understanding the structure of the circle's equation, you can effectively apply calculus concepts to explore the circle's properties and relationships.

Derivative

A derivative is a fundamental tool in calculus that measures how a function changes as its input changes. It gives the slope of the tangent to the graph of the function at any point. For a curve defined by an equation, the derivative with respect to \(x\) gives us the rate of change of \(y\) concerning \(x\).
In this exercise, implicit differentiation is used since both \(x\) and \(y\) are present in the equation \((x-2)^2 + (y-3)^2 = 9\). Implicit differentiation helps find the derivative \(\frac{dy}{dx}\) by treating \(y\) as a function of \(x\), even if it isn't solved for \(y\) explicitly. Calculating \(\frac{dy}{dx}\) for the circle's equation allows for determining the slope of the tangent line at any point on the circle. This step is crucial for forming the equation of the tangent line and understanding how the circle behaves at particular points.

Point-Slope Form

The point-slope form is a method used for writing the equation of a line when you know both a point on the line and the slope of the line. The formula is: \[ y - y_1 = m(x - x_1) \], where \((x_1, y_1)\) is a known point on the line, and \(m\) is the slope. This form is particularly useful when dealing with tangent lines, as you often have the slope of the tangent (from the derivative) and the point at which the tangent touches the curve.
For instance, to find the equation of the tangent line in this exercise, you use the point given, such as \((\frac{7}{2}, \frac{6+3\sqrt{3}}{2})\), and the slope \(-\frac{1}{\sqrt{3}}\) calculated from the derivative. Plugging these values into the point-slope form provides the equation of the tangent line, allowing us to describe the line that just touches the circle at that specific point without intersecting it.