Problem 57 - Calculus Early Transcendentals: Pearson New International Edition

Question

57\. There are two tangent lines to the curve \(y=4 x-x^{2}\) that go through \((2,5)\). Find the equations of both of them.

Step-by-Step Solution

Answer

The equations of the tangent lines are \(y = -2x + 9\) and \(y = \frac{10}{3}x + \frac{1}{9}\).

Step 1: Determine Slope of Tangent

First, we need to determine the slope of the tangent line to the curve \(y=4x-x^2\). This is found by taking the derivative of the given function.The derivative of \(y=4x-x^2\) with respect to \(x\) is:\[\frac{dy}{dx} = 4 - 2x\] This expression \(4 - 2x\) gives us the slope of the tangent line at any point \((x_0, y_0)\) on the curve.

Step 2: Set Up Equation of Tangent Line

The equation of the tangent line at any point \((x_0, y_0)\) is given by the point-slope form:\[y - y_0 = m(x - x_0)\]where \(m\) is the slope of the tangent line. Substitute \(m = 4 - 2x_0\) and \(y_0 = 4x_0 - x_0^2\) into the equation:\[y - (4x_0 - x_0^2) = (4 - 2x_0)(x-x_0)\]

Step 3: Use Given Point (2,5)

Since the tangent lines also pass through \((2,5)\), substitute \(x = 2\) and \(y = 5\) into the tangent line equation:\[5 - (4x_0 - x_0^2) = (4 - 2x_0)(2-x_0)\]

Step 4: Simplify and Solve Equation

Simplify the equation obtained in Step 3:Substituting the values, we have:\[5 = 4x_0 - x_0^2 + (4 - 2x_0)(2-x_0)\]This simplifies to:\[5 = 8 - 2x_0 + x_0^2 - 8x_0 + 2x_0^2\]Combine like terms:\[3x_0^2 - 10x_0 + 3 = 0\]

Step 5: Solve Quadratic for x_0

Now solve the quadratic equation \(3x_0^2 - 10x_0 + 3 = 0\) for \(x_0\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 3\), \(b = -10\), and \(c = 3\):\[x_0 = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3}\]\[x_0 = \frac{10 \pm \sqrt{100 - 36}}{6}\]\[x_0 = \frac{10 \pm \sqrt{64}}{6}\]\[x_0 = \frac{10 \pm 8}{6}\]Thus, \(x_0 = 3\) or \(x_0 = \frac{1}{3}\).

Step 6: Find y_0 Values

Calculate the corresponding \(y_0\) values using \(y_0 = 4x_0 - x_0^2\). For \(x_0 = 3\):\[y_0 = 4(3) - 3^2 = 12 - 9 = 3\]For \(x_0 = \frac{1}{3}\):\[y_0 = 4(\frac{1}{3}) - (\frac{1}{3})^2 = \frac{4}{3} - \frac{1}{9} = \frac{12}{9} - \frac{1}{9} = \frac{11}{9}\]

Step 7: Write Equations of Tangent Lines

Use the point-slope form to write the equations of the tangent lines.For \((3, 3)\) with slope \(m = 4 - 2(3) = -2\):\[y - 3 = -2(x - 3)\]Simplify: \[y = -2x + 9\]For \((\frac{1}{3}, \frac{11}{9})\) with slope \(m = 4 - 2(\frac{1}{3}) = \frac{10}{3}\):\[y - \frac{11}{9} = \frac{10}{3}(x - \frac{1}{3})\]Multiply through by 9 to clear fractions:\[9y - 11 = 30x - 10\]Simplify: \[9y = 30x + 1\] or \[y = \frac{30}{9}x + \frac{1}{9}\]or simplified further, \[y = \frac{10}{3}x + \frac{1}{9}\].

Key Concepts

Tangent Line

A tangent line to a curve at a given point is like the flat path touching the curve at that solitary point. It is the line that just grazes the curve without cutting through it. To imagine it, think of gently touching the surface of a sphere. The flat point where your finger meets the sphere is the tangent point. Tangent lines are crucial in calculus because they give the best linear approximation to the curve at that point.

• A tangent line just "touches" the curve, aligning perfectly with the curve's direction at that point.
• The slope of the tangent line is the derivative value at the point of tangency.
• Understanding tangent lines is essential for solving problems involving curves and their behaviors.

Derivative

The derivative of a function is a mathematical concept that measures how the function's output changes as you tweak the input. In simple words, it gives us the slope of the tangent line to the function's graph at any point. Calculating derivatives is fundamental to studying calculus because it helps us understand rates of change.

• Derivatives signify the steepness of the function at any given point.
• The process of finding a derivative is known as differentiation.
• The derivative is represented as \( \frac{dy}{dx} \) for the function \( y \).

In the original problem, we differentiate the quadratic function \( y = 4x - x^2 \) to find \( \frac{dy}{dx} = 4 - 2x \), which indicates the slope of our tangent line at any point \( x \). This helps us set up the tangent line equation.

Quadratic Equation

A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \). Recognizable by its characteristic \( x^2 \) term, these equations portray parabolic graphs, or curves that arch upwards or downwards based on the sign of \( a \). Solving quadratic equations is a staple in algebra, and they often come up in calculus as well.

• A quadratic equation lets us find values of \( x \) that make the equation true, also known as the equation's roots.
• Quadratic formulas or factoring can help solve these equations.
• Quadratics often help model real-world situations like projectile motion.

In our exercise, once the tangent line equation is set, it simplifies down to a quadratic equation \( 3x_0^2 - 10x_0 + 3 = 0 \), which we solve using the quadratic formula for \( x_0 \).

Point-Slope Form

The point-slope form is a handy way of writing the equation of a line when you know the slope and a point on the line. It is expressed as \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is a known point on the line.

• This form is excellent for quickly creating equations without needing to backtrack to find \( y \)-intercepts first.
• Transforms easily to other forms, such as slope-intercept.
• Helpful in solving problems involving tangent lines to curves.

In this task, the point-slope form allows us to easily compose the tangent line's equation. By inserting the coordinates of the point of tangency and the derived slope, we develop both tangent line equations simplistically, illustrating how tangent lines can gracefully connect an outside point with a curve.