Problem 51 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(y=\sqrt{x}\) and consider the points \(M, N, O\), and \(P\) with coordinates \((1,0),(0,1),(0,0)\), and \((x, y)\) on the graph of \(y=\sqrt{x}\), respectively. Calculate (a) \(\lim _{x \rightarrow 0^{+}} \frac{\text { perimeter of } \Delta N O P}{\text { perimeter of } \Delta M O P}\) (b) \(\lim _{x \rightarrow 0^{+}} \frac{\text { area of } \Delta N O P}{\text { area of } \Delta M O P}\)

Step-by-Step Solution

Answer

(a) Limit is 1. (b) Limit is 0.

Step 1: Calculate Perimeter of ΔNOP

To find the perimeter of triangle ΔNOP as a function of x, calculate the length of each side:- NO: The distance between points N(0,1) and O(0,0) is \[ \sqrt{(0-0)^2 + (1-0)^2} = 1 \]- NP: The distance between points N(0,1) and P(x, \sqrt{x}) is \[ \sqrt{(x-0)^2 + (\sqrt{x}-1)^2} = \sqrt{x + (\sqrt{x}-1)^2} = \sqrt{x + x - 2\sqrt{x} + 1} = \sqrt{2x - 2\sqrt{x}+1} \]- OP: The distance between points O(0,0) and P(x, \sqrt{x}) is \[ \sqrt{(x-0)^2 + (\sqrt{x}-0)^2} = \sqrt{x + x} = \sqrt{2x} \]The perimeter of ΔNOP is the sum of these side lengths:\[ \text{Perimeter of } \Delta NOP = 1 + \sqrt{2x - 2\sqrt{x}+1} + \sqrt{2x} \]

Step 2: Calculate Perimeter of ΔMOP

To find the perimeter of triangle ΔMOP as a function of x, calculate the length of each side:- MO: The distance between points M(1,0) and O(0,0) is \[ \sqrt{(1-0)^2 + (0-0)^2} = 1 \]- MP: The distance between points M(1,0) and P(x, \sqrt{x}) is \[ \sqrt{(x-1)^2 + (\sqrt{x}-0)^2} = \sqrt{(x-1)^2 + x} \]- OP has already been calculated as \( \sqrt{2x} \).The perimeter of ΔMOP is:\[ \text{Perimeter of } \Delta MOP = 1 + \sqrt{(x-1)^2 + x} + \sqrt{2x} \]

Step 3: Evaluate the Limit of Perimeters

We are interested in the limit as \(x\) approaches zero from the positive side:\[ \lim_{x \to 0^+} \frac{\text{Perimeter of } \Delta NOP}{\text{Perimeter of } \Delta MOP} = \lim_{x \to 0^+} \frac{1 + \sqrt{2x - 2\sqrt{x}+1} + \sqrt{2x}}{1 + \sqrt{(x-1)^2 + x} + \sqrt{2x}} \]As \(x\) tends to 0, \(\sqrt{x}\) and other terms involving \(x\) also tend to zero. Thus, both perimeters tend towards 1 + 0 + 0 = 1,implying:\[ \lim_{x \to 0^+} \frac{\text{Perimeter of } \Delta NOP}{\text{Perimeter of } \Delta MOP} = 1 \]

Step 4: Calculate Area of ΔNOP

The area of ΔNOP can be found using the determinant method:\[ \text{Area of } \Delta NOP = \frac{1}{2} \left| 0(0 - \sqrt{x}) + 0(\sqrt{x} - 1) + x(1-0) \right| = \frac{1}{2} |x| = \frac{x}{2} \]

Step 5: Calculate Area of ΔMOP

The area of ΔMOP using the determinant method is:\[ \text{Area of } \Delta MOP = \frac{1}{2} \left| 1(0-\sqrt{x}) + 0(\sqrt{x}-0) + x(0-0) \right| = \frac{1}{2} |-\sqrt{x}| = \frac{\sqrt{x}}{2} \]

Step 6: Evaluate the Limit of Areas

Now calculate the limit:\[ \lim_{x \to 0^+} \frac{\text{Area of } \Delta NOP}{\text{Area of } \Delta MOP} = \lim_{x \to 0^+} \frac{\frac{x}{2}}{\frac{\sqrt{x}}{2}} = \lim_{x \to 0^+} \frac{x}{\sqrt{x}} = \lim_{x \to 0^+} \sqrt{x} = 0 \]

Step 7: Conclusion

Thus, the calculated limits for the given expressions as \(x\) approaches zero from the positive side are as follows:(a) The limit of the perimeter ratio is 1.(b) The limit of the area ratio is 0.

Key Concepts

Perimeter and Area

Understanding how to find the perimeter and area of geometric shapes, particularly triangles, is crucial in calculus limit problems. In our exercise, we deal with two triangles, \( \Delta NOP \) and \( \Delta MOP \), formed by specific points determined by the function \(y = \sqrt{x}\). To calculate their perimeters, you need to measure the distance between each pair of points forming the triangle's sides. This employs the distance formula:

• For a pair of points \((x_1, y_1)\) and \((x_2, y_2)\), the distance is \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).

Similarly, when calculating areas, particularly of triangles in coordinate geometry, utilizing the determinant method allows a structured approach. The areas involve the determinants of the positions of these points in the coordinate plane. In this context, the area formula simplifies to expressions involving \(x\) and \(\sqrt{x}\). Thus, grasping these calculations is essential for solving limit problems involving geometric figures.

Determinant Method

The determinant method is a practical tool used to calculate the area of a triangle when vertices are known. This method relies on creating a matrix from the coordinates and taking the determinant. The area of triangle \(\Delta ABC\) with points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:

• \(\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\)

This approach provides a way to handle area calculations without needing the triangle's height or needing points on horizontal and vertical lines. In our specific problem:

• \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) correspond to coordinates like \((0,0)\), \((x, \sqrt{x})\), etc.
• The determinant approach is particularly valuable when comparing the areas for different \(x\) values.

Mastering this method simplifies the process of finding area expressions in function-related problems.

Limit Evaluation

Limit evaluation in calculus is about understanding the behavior of functions as input values approach a certain point. In our task, we focus on limits as \(x\) approaches 0 from the positive side, i.e., \(x \rightarrow 0^+\). Limits help us predict and simplify functions' outcomes without actually computing them at a particular point. For perimeter and area ratios of triangles, evaluating limits tells us how these ratios behave as \(x\) nears zero.

• To evaluate limits of fractions \(\frac{f(x)}{g(x)}\), observe both numerator and denominator tending to a value.
• As \(x\) goes to zero, know the values \(\sqrt{x}\) and small \(x\) terms vanish, simplifying expressions.

In our example, the perimeter ratio approaches 1 and the area ratio approaches zero. This indicates that as the triangle's size gets infinitesimal, the perimeter difference diminishes, while the area contrast sharply due to differing rates of decay in \(x\). Understanding these concepts is pivotal in mastering calculus and its applications.