Problem 66 - University Calculus: Early Transcendentals

Question

a. Find an equation for the line that is tangent to the curve \(y=x^{3}-6 x^{2}+5 x\) at the origin. b. Graph the curve and tangent together. The tangent intersects the curve at another point. Use Zoom and Trace to estimate the point's coordinates. c. Confirm your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).

Step-by-Step Solution

Answer

The tangent at the origin is \( y = 5x \), and it intersects the curve at (6, 0).

Step 1: Find the Derivative of the Curve

Start by finding the derivative of the curve given by the function \( y = x^3 - 6x^2 + 5x \). The derivative, denoted as \( y' \), represents the slope of the tangent line at any point \( x \). Differentiate term by term: \[ y' = 3x^2 - 12x + 5. \]

Step 2: Evaluate the Slope at the Origin

Evaluate the derivative at the origin \((0,0)\) to find the slope of the tangent line at this point. Substitute \( x = 0 \) into the derivative:\[ y'(0) = 3(0)^2 - 12(0) + 5 = 5. \] Thus, the slope of the tangent line at the origin is 5.

Step 3: Write the Equation of the Tangent Line

With the slope \( m = 5 \) and the point \((0,0)\), use the point-slope form of a line to write the equation of the tangent:\[ y - 0 = 5(x - 0) \] Simplify to get the equation of the tangent line:\[ y = 5x. \]

Step 4: Graph the Curve and the Tangent Line

Plot the curve \( y = x^3 - 6x^2 + 5x \) and the tangent line \( y = 5x \) on a graph. Look for where the tangent line intersects the curve at another point besides the origin.

Step 5: Estimate the Second Intersection Point

Use a graphing calculator or software to zoom in and trace along the curve and tangent line to estimate another intersection point other than the origin. You may need to iteratively adjust your view window for clarity. This point will likely be visible where the outputs of both equations start to coincide beyond the origin point.

Step 6: Solve Algebraically to Confirm Intersecting Point

To confirm the intersection point found graphically, solve the equation \( x^3 - 6x^2 + 5x = 5x \) for \( x \). Simplify the equation by subtracting \( 5x \) from both sides:\[ x^3 - 6x^2 = 0. \]Factor out the greatest common factor, \( x^2 \):\[ x^2(x - 6) = 0. \]This equation gives us two solutions: \( x = 0 \) and \( x = 6 \). Substitute \( x = 6 \) back into the original equation:\[ y = 6^3 - 6(6)^2 + 5(6) = 0. \] Thus, the point \((6,0)\) is the second intersection point.

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing through it. This touching point is known as the "point of tangency" and it is where the curve and the line share a common slope.

• The slope of the tangent line is equal to the derivative of the curve at the point of tangency.
• In algebra, the point-slope form of a linear equation is used to write the equation of a tangent line.

For example, in the exercise provided, we calculated the tangent line to the curve \( y=x^3-6x^2+5x \) at the origin \((0,0)\). We first determined the slope by evaluating the derivative of the curve at \( x = 0 \), giving us a slope of 5. Then, we applied the point-slope form formula \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, leading to the equation \( y = 5x \). This simple equation describes how the tangent line behaves at the point of contact.

Curve Intersections

Curve intersections occur at points where two different curves meet or cross each other on a graph. In calculus, it's important to identify these points because they can reveal key insights about the behavior of the functions involved.

• An intersection point satisfies the equation of both curves simultaneously.
• This can be found both graphically and algebraically.

In the given exercise, after finding the tangent line at the origin, we looked for another point where the tangent line \( y=5x \) intersected the curve \( y=x^3 - 6x^2 + 5x \). By solving the equation \( x^3 - 6x^2 + 5x = 5x \) algebraically, we discovered that the curves also intersect at \((6, 0)\) apart from the origin. This highlights that understanding intersection points can unveil unique solutions or behaviors within functions.

Derivatives

Derivatives are fundamental in calculus as they represent the rate of change or slope of a function at any given point.

• The derivative gives us a precise way to find the slope of a tangent line to a curve.
• It describes how a function is changing at every instant.

In the problem, we began by finding the derivative of the curve \( y = x^3 - 6x^2 + 5x \). This was done through differentiation, giving us \( y' = 3x^2 - 12x + 5 \). Evaluating this derivative at \( x = 0 \), we found that the slope of the tangent line is 5. Derivatives thus allowed us to determine how steep the tangent line is at the origin point of the curve, leading us to understand the behavior of the line with respect to the curve it touches.