Problem 70 - A Graphical Approach to Precalculus with Limits

Question

Approximate each expression to the nearest hundredth. $$2 \pi^{3}-5 \pi^{2}-3$$

Step-by-Step Solution

Answer

9.66

Step 1: Evaluate Powers of Pi

First, calculate the approximate values of the powers of π (pi). We know the decimal approximation for π is about 3.14159. Calculate \(\pi^3\): \[\pi^3 \approx 3.14159^3 \approx 31.00628\]Calculate \(\pi^2\):\[\pi^2 \approx 3.14159^2 \approx 9.8696\]

Step 2: Substitute Values into the Expression

Now substitute the approximations of \(\pi^3\) and \(\pi^2\) into the given expression:\[2 \cdot \pi^3 - 5 \cdot \pi^2 - 3\]Substitute:\[2 \cdot 31.00628 - 5 \cdot 9.8696 - 3\]

Step 3: Perform Multiplications

Calculate the terms involving multiplication:\[2 \cdot 31.00628 = 62.01256\]\[5 \cdot 9.8696 = 49.348\]

Step 4: Perform the Subtraction

Now substitute the results from the previous step into the expression and conclude by performing the subtraction:\[62.01256 - 49.348 - 3\]Calculate:\[62.01256 - 49.348 = 12.66456\]Then:\[12.66456 - 3 = 9.66456\]

Step 5: Round to the Nearest Hundredth

Finally, round the result 9.66456 to the nearest hundredth: The nearest hundredth is 9.66.

Key Concepts

Powers of Pi

In mathematics, powers of pi are used to express pi raised to different exponents. This concept primarily arises in calculations involving circles, spherical shapes, or other geometrical contexts. The Greek letter \( \pi \) represents a fundamental constant approximately equal to 3.14159, and its powers yield more complex values.When raising \( \pi \) to higher powers, like \( \pi^2 \) or \( \pi^3 \), it is often necessary to substitute \( \pi \) with its decimal approximation. For example:

• \( \pi^2 \approx 3.14159^2 \approx 9.8696 \)
• \( \pi^3 \approx 3.14159^3 \approx 31.00628 \)

It is crucial in solving mathematical expressions involving \( \pi \) to first calculate these powers before proceeding with the rest of the problem. Always remember, powers of pi are nothing more than multiplying \( \pi \) by itself a certain number of times, and using its decimal approximation helps in practical computations. Keeping track of these calculated values assists in efficiently solving the expression in subsequent steps.

Decimal approximation

Decimal approximation is the process of estimating a number by finding its closest simpler decimal representation. It is especially useful when dealing with irrational numbers like \( \pi \), which cannot be expressed as an exact decimal.Using decimal approximations simplifies complex mathematical calculations:

• For \( \pi \), we often use 3.14159 or just 3.14 for quick estimations.
• These approximations allow us to handle complex expressions without cumbersome calculations.

In practical scenarios, it is essential to determine what level of precision is sufficient. For example, when computing powers of \( \pi \), using approximations like 9.8696 for \( \pi^2 \) and 31.00628 for \( \pi^3 \) renders the expression more manageable while maintaining a reasonable degree of accuracy. Deciding on the decimal approximation involves balancing simplicity with the required precision of results.

Rounding to nearest hundredth

Rounding numbers to the nearest hundredth is crucial in producing results that are both easy to interpret and sufficiently precise for most practical purposes. To round a number like 9.66456 to the nearest hundredth, we look at the digit in the thousandths place, which in this case is 4. The steps for rounding to the nearest hundredth are simple:

• Identify the digit in the hundredths place – for 9.66456, it's 6.
• Check the digit in the thousandths place, which is just after the hundredths – here, it is 4.
• If the digit in the thousandths place is 5 or more, increase the hundredths digit by 1.
• If it is less than 5, keep the hundredths digit as is.

For 9.66456, since 4 is less than 5, we do not need to increase the hundredths place, so we round it to 9.66. This ensures our final answer is both accessible and adequately precise for many real-world calculations.