Problem 52 - Calculus Early Transcendentals

Question

Find the \(x\) -coordinate of the point \(P\) on the parabola \(y=1-x^{2} \quad(0

Step-by-Step Solution

Answer

The x-coordinate of point P is \(\frac{1}{\sqrt{3}}\).

Step 1: Understand the Problem

We need to find the point on the parabola where the tangent line forms a triangle with the coordinate axes and the area of this triangle is minimized.

Step 2: Find the Derivative

The derivative of the function will give us the slope of the tangent at any point on the parabola. The function is given by:\[ y = 1 - x^2 \]Taking the derivative with respect to \(x\), we have:\[ \frac{dy}{dx} = -2x \]

Step 3: Equation of the Tangent Line

The tangent line at any point \(x\) on the parabola is given by:\[ y - y_1 = m(x - x_1) \]where \(m = -2x\) is the slope and the point is \((x, 1-x^2)\). Substituting values:\[ y - (1-x^2) = -2x(x - x) \]Simplifying, the equation of the tangent line is:\[ y = -2x x + c \]

Step 4: Find Intersection with Axes

The tangent line intersects the x-axis when \(y = 0\), giving:\[ 0 = -2x x + c \]. Solve for \(x\) to find the x-intercept. It intersects the y-axis when \(x = 0\), the equation becomes \(y = c\) for the y-intercept.

Step 5: Calculate Triangle Area

The triangle's area can be found using the formula:\[ \text{Area} = \frac{1}{2} |xy| \]where \(x\) and \(y\) are the intercepts we found. We need to express these intercepts using a single variable and set up the area function to minimize.

Step 6: Minimize the Area Function

Express the area function using the variables and differentiate to find the critical points. By setting the derivative of the area function to zero, solve for the value of \(x\) which gives the minimum area.

Step 7: Verify the Minimum

Confirm the critical point actually gives a minimum by either the second-derivative test or analyzing the behavior of the function. Conclude with finding the \(x\)-coordinate at which this minimum occurs.

Step 8: Final Step: Conclusion

After following through the calculations and verifications, determine that the \(x\)-coordinate of point P that minimizes the enclosed triangle's area is the solution.

Key Concepts

Understanding Parabolas

A parabola is a U-shaped curve, and in this case, it is defined by the function \( y = 1 - x^2 \). This specific parabola opens downwards because of the negative sign in front of the \( x^2 \) term. It reaches its highest point, known as the vertex, at \( x = 0 \) with a value of \( y = 1 \). As the x-value increases or decreases from this vertex, the y-value decreases, forming the curve's characteristic "U" shape upside down.

In the given problem, the point \( P \) lies on this curve and is important because it will help us determine the tangent line at that specific point. Understanding the properties of this parabola, including the way it opens and its vertex, will assist in further calculation steps, especially when discussing intersections with axes.

• Vertex: Highest point (0,1)
• Orientation: Opens downwards
• Limits: Analysis is within \( 0 < x \leq 1 \)

Tangent Line's Role

A tangent line is a straight line that touches a curve at exactly one point without crossing it. In calculus, the tangent line represents the instantaneous rate of change of the curve at that point. For our parabola \( y = 1 - x^2 \), the tangent line at point \( P(x, 1-x^2) \) will help us form a triangle with the coordinate axes.

To find the equation of the tangent line, we need two things: the slope of the line, and a point that it passes through. The slope of the tangent line is derived from the derivative of the parabola, which we'll discuss next. Once we have the slope, we can use the point-slope form formula: \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the point of tangency.

• Touch point: \( P(x, 1-x^2) \)
• Defines shape and size of resulting triangle

The Derivative and Slope

A derivative represents the slope of a function, showing how the function's output changes with respect to its input. For the parabola \( y = 1 - x^2 \), the derivative \( \frac{dy}{dx} = -2x \) tells us the slope of the tangent line at any given point \( x \).

For our problem, this derivative helps us determine the steepness of the tangent line touching the parabola at \( P(x, 1-x^2) \). Proper understanding of this derivative is crucial as it directly affects the positions where the tangent line will intersect the x and y axes, thus forming a triangle.

Remember, finding this slope is a critical step because:

• It informs us about the direction and steepness of the tangent line.
• It aids in forming the equation of the tangent line accurately.

Calculating Triangle Area

The area of a triangle is crucial to solving our problem. The formula for the area of a triangle is \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). For our scenario, the tangent line intersects the x-axis and y-axis, creating a right-angled triangle.

To find the triangle's base and height, we need the x-intercept and y-intercept, which we derive from the tangent line's equation. These points correspond to where the triangle's sides meet the axes. Armed with this information, the formula becomes:

• Base = x-intercept
• Height = y-intercept

Understanding how these intercepts are calculated and applied lets us adapt the area formula to this specific context, preparing us for the minimization task.

Minimization of Triangle Area

Minimization is the process of finding the smallest possible value of a function, which, in this case, is the area of the triangle formed by the tangent line. Once we express the area as a function of \( x \) using the intercepts from the tangent line's equation, our next task is to find the x-value that minimizes this area.

To accomplish this, we take the derivative of the area function with respect to \( x \) and set it to zero to find the critical points. These points indicate potential minimums or maximums. Confirming that a point gives a minimum can involve further analysis, such as the second-derivative test, or checking the behavior of the function around this critical point.

• Set the derivative of the area function to zero
• Find critical points
• Test to ensure a minimum

After following these steps, you find the specific \( x \)-coordinate that results in the smallest possible triangle area.