# Video: Solving World Problems Involving Direct Variation of a Quantity with Another Two Quantities

The horsepower (hp) that a shaft can safely transmit varies jointly with its speed in revolutions per minute (rpm) and the cube of the diameter. A shaft with a diameter of 3 inches can transmit 45 hp at 100 rpm. What is the diameter of a shaft that can transmit 60 hp at 150 rpm?

04:23

### Video Transcript

The horsepower, hp, that a shaft can safely transmit varies jointly with its speed in revolutions per minute, rpm, and the cube of the diameter. A shaft with a diameter of three inches can transmit 45 horsepower at 100 rpm. What is the diameter of a shaft that can transmit 60 horsepower at 150 rpm?

A variable in our case, horsepower, varies jointly with two other variables. And that means that the horsepower is directly proportional to both the rpm and the cube of the diameter.

Proportional values vary by a constant of 𝑘. We can take the values we’re given for the first shaft, use that to calculate our constant variation. We’ll plug the 𝑘 in to our equation a second time to find the diameter of the new shaft that can transmit 60 horsepower.

But before that, we need to solve for 𝑘. Our first shaft transmits a horsepower of 45. We don’t know its constant. We do know that it operates at 100 rpms and that its diameter is three, three cubed. Three cubed equals 27. Bring everything down. 100 times 27 equals 2700.

Remember, we’re trying to solve for 𝑘, so we divide both sides of our equation by 2700. On the right, it cancels out, and we have 𝑘 equals 45 over 2700. If we simplify this fraction, if we divide the numerator by 45 and the denominator by 45, we find that our constant value is one over 60. Our constant of variation is one over 60.

We take our 𝑘-value and we plug it back into the horsepower formula. We can simplify the formula by saying that the rpms times the diameter cubed over 60 equals the horsepower.

Remember, this time, we’re trying to solve for the diameter. That’s the value we don’t know. We want a shaft that can transmit 60 horsepower, so we plug in 60 at the horsepower. And we want its speed to be 150 rotations per minute, so we plug that in as well.

Now we’ll just use some algebra to solve for 𝑑. That means we need to isolate 𝑑, get 𝑑 by itself. I’m gonna get 60 out of the denominator by multiplying the right side by 60 and the left side. On the right side, they cancel out. And we’re left with 150 times the diameter cubed. The left side equals 3600. Next, we’ll divide both sides by 150. On the right, that cancels out, leaving us with the diameter cubed. On the left side, 3600 divided by 150 equals 24. But 24 equals the diameter cubed. And to get rid of that cube, we need to take the cube root of both sides of the equation.

However, when we’re working with higher powers, I usually think it’s easier to write fractions of powers. The cube root of the diameter cubed is the same thing as saying the diameter cubed to the one-third power. So what we’re gonna do is take both sides of this equation to the one-third power. On the right, you’ll just be left with the diameter, and on the left 24 to the one-third power equals 2.884. The diameter of a shaft that could transmit 60 horsepower at 150 rpms is equal to 2.88 inches.