Problem 62 - Calculus of a Single Variable

Question

Using a tangent Line In Exercises \(61-64,\) set up and evaluate the definite integral that gives the area of the region bounded by the graph of the function and the tangent line to the graph at the given point. $$ y=x^{3}-2 x, \quad(-1,1) $$

Step-by-Step Solution

Answer

12.25

Step 1: Find the Derivative

Firstly, find the derivative of the function \(y = x^3 -2x\). The derivative \(y'\) is given by :, \[y' = 3x^2 - 2\]

Step 2: Find the Tangent Slope

Now, find the slope of the tangent line at the point (-1,1). We already know that the slope of the tangent line to the curve at a specific point is given by the value of the derivative at that point. So substitute \(x = -1\) into \(y'\) to get: \[y'(-1) = 3*(-1)^2 - 2 = 3 - 2 = 1\] So the slope of the tangent line at the point (-1,1) is 1.

Step 3: Find the Equation of the Tangent Line

We can find the equation of the tangent line by substituting into the point slope form of a line equation: \(y - y1 = m(x - x1)\), where m is the slope and (x1, y1) is the point of tangency. Thus, the equation of the tangent line is: \[y - 1 = 1*(x + 1)\] which simplifies to \[y = x + 2\]

Step 4: Set Up the Integral

Now set up the definite integral that gives the area between the function and the tangent line. This area is given by: \[\int (x^3 - 2x- (x+2)) dx\] from the intersection points of the tangent line and the function.

Step 5: Find the Borders

We have to set the function equal to the tangent line to calculate the intersection points which represent our borders for definite integral: \[x^3 - 2x = x + 2\] Solving this, we get three roots, but we need the smallest and the largest, which are -2 and 1.

Step 6: Evaluate the Integral

Evaluate the definite integral from -2 to 1. Integration of \(x^3 - 2x - (x+2)\) done by elementary integration rules gives us the integrand being \(x^3 - 3x - 2\). Then \[\int_{-2}^{1}(x^3 - 3x - 2) dx = [0.25*x^4 - 1.5*x^2 -2x]_{-2}^{1}= -1.75 - (-14) = 12.25\]

Step 7: Final Answer

So, the area of the region bounded by the graph of the function and the tangent line to the graph at the given point is 12.25.

Key Concepts

Tangent Line

Imagine a graph with a smooth curve. Now picture drawing a straight line that just touches the curve at one point and follows its direction as closely as possible. This is called the tangent line.

• A tangent line to a curve represents an instantaneous rate of change of the function at that specific point.
• It shows the direction and angle of the curve at that point.

In our exercise, the tangent line touches the curve described by the function \( y = x^3 - 2x \) at the point (-1, 1). To find the tangent line, we first need the derivative of the function, as it gives us the slope of the tangent line.

Area Under Curve

When we talk about "area under the curve," we mean the total space between the graph of a function and the x-axis over a specific interval.
This area can be thought of as the sum of a series of rectangles of infinitesimally small width under the curve and is calculated using an integral.

• In this exercise, however, we are interested in the area between the curve \( y = x^3 - 2x \) and its tangent line.
• This means we calculate not just under the curve itself, but also include the space under the tangent line, effectively finding the area between them.

The definite integral helps with the calculation by evaluating the net area between these functions over specified boundaries.

Derivative

The derivative of a function is the mathematical tool that tells us how fast something is changing at any given point.

• It describes the rate of change or the slope of the function's graph at any point on that graph.
• Finding the derivative is crucial for determining the slope of the tangent line to a function's graph.

In the given exercise, the derivative of \( y = x^3 - 2x \) is calculated to be \( y' = 3x^2 - 2 \). Evaluating this derivative at the point of interest, \( x = -1 \), gives the slope of the tangent line. This slope is crucial for constructing the equation of the tangent line, which we need to find the area in question.

Point of Tangency

The concept of the point of tangency is straightforward but important. It is the exact point where the tangent line touches the curve without cutting across it.

• This point provides us with a specific location on the graph to find our tangent line's slope.
• In this exercise, the point of tangency is (-1, 1).

Using this point, along with the slope from the derivative, allows us to use the point-slope form to write the equation of our tangent line. Establishing this line is essential for accurately finding the area bounded by the curve and the tangent line.