Problem Solving with Algebraic Manipulation and Variation

Perimeter and Area Applications

• Perimeter Definition: The perimeter $(P)$ of any shape is the sum of all its sides.

• Rectangle Perimeter: For a rectangle, with length $(l)$ and width $(w)$, the perimeter is given by the formula: $P = l + l + w + w = 2l + 2w$

• Problem Example: Expressing Area in terms of Width

  • Given: The perimeter of a rectangle is $84$. We want to express the area $(A)$ solely in terms of its width $(w)$.

  • Step 1: Use the perimeter equation to express length in terms of width.

    • The perimeter equation is: $2l + 2w = 84$

    • Divide by $2$: $l + w = 42$

    • Solve for $l$: $l = 42 - w$

    • (Alternatively, solving for $w$ would be $w = 42 - l$)

  • Step 2: Recall the area formula for a rectangle.

    • Area is given by: $A = l \times w$

  • Step 3: Substitute the expression for length into the area formula.

    • Replace $l$ with $(42 - w)$: $A = (42 - w)w$

    • Distribute $w$: $A = 42w - w^2$

  • Significance: This demonstrates how, given a constraint (perimeter), we can express a two-variable function (area) as a single-variable function, making it possible to calculate the area if only the width is known, along with the perimeter. If one knows the perimeter and one width, the exact area can be found.

Volume and Substitution: Right Circular Cone

• Volume Formula (Given): The volume $(V)$ of a right circular cone is given by: $V = \frac{1}{3}\pi r^2 h$

  • Where $r$ is the radius and $h$ is the height.

• Part A: Express Volume as a Function of the Radius (R)

  • Key Concept: "Express as a function of the radius $(r)$" means the final volume equation should only contain the variable $r$.

  • Hidden Equation/Constraint: The problem states that "the height of the certain cone is twice the radius."

    • This translates to the equation: $h = 2r$

  • Substitution: Substitute the expression for $h$ into the original volume formula.

    • $V = \frac{1}{3}\pi r^2 (2r)$

  • Simplification: Combine constants and variables.

    • Multiply coefficients: $\frac{1}{3} \times 2 = \frac{2}{3}$

    • Combine powers of $r$: $r^2 \times r = r^3$

    • The simplified volume formula in terms of $r$ is: $V = \frac{2}{3}\pi r^3$

• Part B: Find the Volume of a Paper Snow Cone with a Radius of 2 Inches

  • Given: Radius $r = 2 \text{ inches}$.

  • Substitute into the derived formula from Part A:

    • $V(2) = \frac{2}{3}\pi (2)^3$

  • Calculation for Exact Answer (in terms of $\pi$):

    • Calculate $2^3$: $2^3 = 8$

    • $V(2) = \frac{2}{3}\pi (8) = \frac{16}{3}\pi$

    • This is the exact answer. If a question asks for "exact" or "in terms of $\pi$", this format is required.

    • Reason for Exactness: Calculators approximate $\pi$, so typing it directly into a calculator will not yield an exact answer.

  • Calculation for Approximate Decimal Answer:

    • $V(2) \approx 16.755 \text{ cubic inches}$

    • This decimal approximation is generally acceptable if not specified otherwise, but AP exams often prefer exact answers or answers rounded to a specific number of decimal places.

Direct and Inverse Variation (Hooke's Law)

• Identifying Variation Problems: These problems often use specific phrases like "varies directly," "directly proportional," "varies inversely," or "indirectly proportional."

Direct Variation

• Definition: When a quantity $(D)$ varies directly with another quantity $(F)$, it means that $D$ is a constant multiple of $F$. As one quantity increases, the other increases proportionally.

• Equation Form: $D = kF$

  • Where $k$ is the constant of proportionality (also called the constant of variation).

• Example 3: Hooke's Law Application

  • Problem Statement: "How far will a force of $620 \text{ Newtons}$ stretch the spring? Distance varies directly with force."

  • Step 1: Set up the direct variation equation.

    • Distance $(D)$ varies directly with force $(F)$: $D = kF$

  • Step 2: Use given data to find the constant of proportionality $(k)$.

    • Given Data Point: A force of $275 \text{ Newtons}$ stretches the spring $0.2 \text{ meters}$.

    • Substitute: $0.2 = k(275)$

    • Solve for $k$: $k = \frac{0.2}{275} = \frac{2/10}{275} = \frac{2}{2750} = \frac{1}{1375}$

  • Step 3: Write the specific equation for this spring.

    • Now that $k$ is known: $D = \frac{1}{1375}F$

  • Part A: Calculate distance for a force of 620 Newtons.

    • Substitute $F = 620$: $D = \frac{1}{1375}(620)$

    • Result: $D \approx 0.45 \text{ meters}$

  • Part B: Calculate force needed to stretch the spring 0.1 meters.

    • Substitute $D = 0.1$ into the specific equation: $0.1 = \frac{1}{1375}F$

    • Solve for $F$: $F = 0.1 \times 1375 = 137.5 \text{ Newtons}$

  • Self-Check/Proportion: This result makes sense because $\text{0.1 m}$ is half of the initial $\text{0.2 m}$ stretch, and $\text{137.5 N}$ is half of the initial $\text{275 N}$ force. This is consistent with direct variation (a proportional relationship).

Inverse Variation

• Definition: When a quantity $(D)$ varies inversely with another quantity $(F)$, it means that $D$ is a constant divided by $F$. As one quantity increases, the other decreases proportionally.

• Equation Form: $D = \frac{k}{F}$ or $DF = k$

  • Where $k$ is the constant of proportionality.

• Contrast with Direct Variation: In direct variation, multiplication by the constant occurs. In inverse variation, division by the variable (or multiplication by its inverse) occurs. Both types of problems are solved by first finding the constant $k$ using a given data point, and then using that constant to solve for unknown values.