# Math Help - Joint variation

1. ## Joint variation + percent

The weight (W) that can be supported by a wooden beam varies directly as the square of its diameter (d) and inversely as its length (l).

a) What percentage increase in the diameter would be necessary for a beam twice as long to support an equivalent weight?

b)
What percentage change in the weight would be capable of being supported by a beam
three times as long with twice the diameter?

thank you.

2. ## Re: Joint variation

The setup is

(1) $W = \frac{kd^2}{l}$

where k is the constant of variation.

The percentage increase of d is $\frac{x}{100} \cdot d$ W remains the same but l becomes 2l.

$W = \frac{k(d + \frac{xd}{100})^2}{2l}$

(2) $W = \frac{kd^2(1 + \frac{x}{100})^2}{2l}$

Set (1) = (2) and solve for x

$\frac{kd^2}{l} = \frac{kd^2(1 + \frac{x}{100})^2}{2l}$

We see k and l and d^2 cancel immediately giving

$2 = (1 + \frac{x}{100})^2$

Can you continue from here?

3. ## Re: Joint variation

Originally Posted by agentmulder
The setup is

(1) $W = \frac{kd^2}{l}$

where k is the constant of variation.

The percentage increase of d is $\frac{x}{100} \cdot d$ W remains the same but l becomes 2l.

$W = \frac{k(d + \frac{xd}{100})^2}{2l}$

(2) $W = \frac{kd^2(1 + \frac{x}{100})^2}{2l}$

Set (1) = (2) and solve for x

$\frac{kd^2}{l} = \frac{kd^2(1 + \frac{x}{100})^2}{2l}$

We see k and l and d^2 cancel immediately giving

$2 = (1 + \frac{x}{100})^2$

Can you continue from here?