Problem 88 - Precalculus with Limits

Question

GEOMETRY The area of the shaded region in the figure is \(A=\frac{8}{3}p^{1/2}b^{3/2}\). (a) Find the area when \(p=2\) and \(b=4\). (b) Give a geometric explanation of why the area approaches 0 as \(p\) approaches 0.

Step-by-Step Solution

Answer

The area of the shaded region when \(p=2\) and \(b=4\) is approximately 30 square units. The area approaches 0 as \(p\) approaches 0 because \(p\) might represent a dimension of the figure and as this dimension shrinks, the area of the shaded region also shrinks.

Step 1: Substitution

Substitute \(p=2\) and \(b=4\) into the formula \(A=\frac{8}{3}p^{1/2}b^{3/2}\) in order to find the area when \(p=2\) and \(b=4\). The result will provide the solution to question (a).

Step 2: Simplification

Simplify the expression in Step 1 to get the area: \(A=\frac{8}{3}(2)^{1/2}(4)^{3/2}\). The square root of 2 is 1.41 and the cubic square root of 4 equals to 8, then multiply these with 8/3.

Step 3: Calculation

After performing the operations described in Step 2, the area is calculated as \(A = \frac{8}{3}*1.41*8\), which simplifies to approximately \(A = 30\) square units.

Step 4: Geometric Explanation

To answer question (b), examine the formula \(A=\frac{8}{3}p^{1/2}b^{3/2}\). According to this formula, as \(p\) approaches 0, the term \(p^{1/2}\) overall approaches 0. Because one term becomes smaller and smaller, resulting in \(A\) to approach 0. In geometric terms, \(p\) might represent a dimension of the figure such as a radius or length. As \(p\) goes to 0, this dimension shrinks, therefore the area of the shaded region also shrinks, approaching 0.

Key Concepts

Substitution in Algebra

Substitution is an essential concept in algebra, where it involves replacing a variable with a specific numerical value. It's like giving a letter in a mysterious code a concrete meaning, which then helps to unravel the secret message—in this case, the value of an algebraic expression or equation.

For example, looking at the area formula from our exercise, \(A=\frac{8}{3}p^{1/2}b^{3/2}\), to find the area when \(p=2\) and \(b=4\), we replace the variables with these numbers. This turns an abstract formula into a concrete calculation step that leads us to the exact area of the shaded region in the geometric figure. By substituting, we've essentially made a direct connection between abstract algebraic expressions and real numbers, which allows for practical problem-solving.

Remember, the precise execution of the substitution step is crucial. If we replace \(p\) with 2 and \(b\) with 4 carefully, we avoid mistakes that could derail our entire solution process. Imagine each alphabet letter as a placeholder for a unique number—it's your task to put the right number in the right place.

Simplifying Algebraic Expressions

Simplifying algebraic expressions is like tidying up a room—getting rid of the clutter to see the room’s true layout. Here, 'clutter' refers to the math operations that we can perform to reduce an expression to its simplest form. Working with algebra often involves dealing with powers, roots, and multiplication, which requires a certain order of operations.

In our exercise, after substituting \(p\) and \(b\) with 2 and 4 respectively, we follow the 'BEDMAS' or 'PEMDAS' order: brackets or parentheses first, exponents (including roots) next, and then multiplication and division. Simplification is like peeling back layers: we first deal with the exponent, the square root of 2 which is approximately 1.41, and the cube of 4 (which is actually the cube root of 4 raised to the power of 2), which is 8. All that's left after this is a simpler multiplication problem, which is easily manageable. This process not only makes calculations manageable but also lays the groundwork for understanding more complicated algebraic manipulations.

Geometric Interpretation of Limits

In geometry, a limit is like a horizon—it tells us how an area or a length behaves as we stretch or shrink its dimensions. The concept of limits reaches far beyond simple measurements and touches the essence of what shapes become under certain conditions.

The geometric interpretation of limits connects algebra with the visual world. For instance, by observing the formula for the shaded area \(A=\frac{8}{3}p^{1/2}b^{3/2}\), the geometric interpretation of the limit as \(p\) approaches 0 makes us think of one side of the figure getting smaller and smaller. If you think of \(p\) as possibly being the radius of a circle or the length of a rectangle bordering the shaded region, then as this dimension heads towards 0, the shaded area itself would also dwindle, eventually becoming a line or a point—which has no area at all.

In essence, this concept teaches us that the values we approach in algebra have tangible real-world repercussions in geometry; understanding these abstract concepts allows us to predict and explain the behavior of shapes and space in a comprehensive and meaningful way.