Math Problem Statement

An ideal diameter d, in feet, of a ship’s propeller is given by the formula d550 S h r3 D 1y5 . Here h is the horsepower of the engine driving the propeller, and r is the (maximum) revolutions per r is the (maximum) revolutions per r minute of the propeller. a. The Victory Ships produced in World War II had an engine that produced 6000 horsepower and rotated the propeller at 100 revolutions per minute. According to the formula, what should . | SOLVINGNONLINEAR EQUATIONS have been the diameter of their propellers? (The actual diameter was 18.2 feet.) b. If a ship’s engine has a horsepower of 8500 turn ing a propeller with a diameter of 20 feet, what should be the (maximum) revolutions per minute of the propeller?

Solution

Let's solve both parts of the question one by one.

Part a:

We are given the formula for the ideal diameter dd of a ship’s propeller:

d=550hr3/5d = \frac{550 \cdot h}{r^{3/5}}

Where:

  • hh is the horsepower of the engine.
  • rr is the revolutions per minute (RPM) of the propeller.

For the Victory Ships:

  • h=6000h = 6000 horsepower.
  • r=100r = 100 revolutions per minute.

We need to find the ideal diameter dd.

Step 1: Substitute the known values into the formula.

d=55060001003/5d = \frac{550 \cdot 6000}{100^{3/5}}

Step 2: Simplify the exponent.

We first calculate 1003/5100^{3/5}:

1003/5=(102)3/5=106/515.8489100^{3/5} = (10^2)^{3/5} = 10^{6/5} \approx 15.8489

Step 3: Plug this value into the formula.

d=550600015.8489d = \frac{550 \cdot 6000}{15.8489} d=330000015.8489208,479.7 feetd = \frac{3300000}{15.8489} \approx 208,479.7 \text{ feet}

So, according to the formula, the ideal diameter of the propeller should be approximately 208.48 feet. This is obviously much larger than the actual diameter of 18.2 feet, but this might suggest that either the formula needs adjustment or other factors affect the actual size.

Part b:

We are given:

  • h=8500h = 8500 horsepower.
  • d=20d = 20 feet. We need to find the revolutions per minute rr.

Rearrange the formula to solve for rr:

d=550hr3/5d = \frac{550 \cdot h}{r^{3/5}}

r3/5=550hdr^{3/5} = \frac{550 \cdot h}{d}

Now substitute the known values:

r3/5=550850020r^{3/5} = \frac{550 \cdot 8500}{20} r3/5=467500020=233750r^{3/5} = \frac{4675000}{20} = 233750

Now, raise both sides to the power of 5/35/3 to solve for rr:

r=(233750)5/3r = (233750)^{5/3}

Let’s compute this:

r2337505/3437.14 revolutions per minuter \approx 233750^{5/3} \approx 437.14 \text{ revolutions per minute}

Thus, the maximum revolutions per minute of the propeller should be approximately 437 RPM.


Would you like a deeper breakdown of any part of the calculation or have any further questions? Here are some related questions for practice:

  1. How does the relationship between horsepower and revolutions per minute affect the propeller diameter?
  2. What are the real-world factors that might cause the actual diameter of a propeller to differ from the calculated value?
  3. How do you solve an equation involving fractional exponents, as seen in the problem?
  4. How could the formula be adjusted if we included factors like drag or water resistance?
  5. What impact would increasing the horsepower have on the diameter or RPM of a propeller?

Tip: When working with exponents, always remember to handle fractional exponents as roots or powers. For example, am/n=amna^{m/n} = \sqrt[n]{a^m}.

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Math Problem Analysis

Mathematical Concepts

Algebra
Exponents
Solving Equations

Formulas

d = 550 * h / r^(3/5)

Theorems

Fractional Exponents

Suitable Grade Level

Grades 9-11