# 1st derivative in the fireman's problem

• JOhnJDC
In summary, the problem involves finding the length of the shortest ladder that will reach from the ground across the top of a fence to a high building. The solution involves using similar triangle ratios and Pythagorean's Theorem to determine the length of the ladder (L). The first derivative of L^2 is computed and set equal to 0 to find the minimum value of a, which is then plugged back into the equation for L^2 to find the shortest length of the ladder.
JOhnJDC
1st derivative in the "fireman's problem"

## Homework Statement

A fence q feet high is p feet from a high building. Find the length of the shortest ladder that will reach from the ground across the top of the fence to the building.

I know how to solve this problem, but I'm getting stuck trying to simplify the 1st derivative of the hypotenuse (L) of the right triangle (i.e., the length of the ladder).

## Homework Equations

Similar triangle ratios; Pythagorean's Theorem

## The Attempt at a Solution

First, I determined that the base of the right triangle that is formed from the ladder reaching from the ground over the fence to the building has a base of p + a and a height of q + pq/a.

Second, I plugged the base and the height into Pythagorean's Theorem to get L^2 = (q + pq/a)^2 + (p + a)^2.

Next, I attempted to find the minimum of L^2 by computing its first derivative, setting it equal to 0, and solving for a. The first derivative of L^2 with respect to a = 2(q + pq/a)(-pq/a^2) + 2(p + a). I get lost in the algebra when I try to solve for a. Please show me the way.

Last edited:

I finally figured this out, if anyone is interested.

dL^2/da = 2(q + pq/a)(-pq/a^2) + 2(p + a)

To simplify, create common denominator for left side:

[2(-apq^2 - p^2q^2)/a^3] +2(a+p)

Factor out -pq^2 from the left side, yielding:

[-pq^2(2(a+p))/a^3] + 2(a+p)

Factor out 2(a+p) from both sides, yielding:

2(a+p)[1 - (pq^2)/a^3]

Setting 2(a+p)[1 - (pq^2)/a^3] equal to 0 and solving for a yields:

a = (p^1/3)(q^2/3), which is a minimum.

Now, plug a back into the equation for L^2 above:

L^2 = (q + pq/(p^1/3)(q^2/3))^2 + (p + (p^1/3)(q^2/3))^2

L = sqrt[(q + pq/(p^1/3)(q^2/3))^2 + (p + (p^1/3)(q^2/3))^2], which reduces to:

L = sqrt[(q + (p^2/3)(q^1/3))^2 + (p + (p^1/3)(q^2/3))^2]; this is the shortest length of the ladder consistent with the above constraints.

## What is the "1st derivative" in the fireman's problem?

The "1st derivative" in the fireman's problem refers to the rate of change of the total number of firemen needed to extinguish a fire. It is calculated by taking the first derivative of the equation that models the problem.

## How is the "1st derivative" used in the fireman's problem?

The "1st derivative" is used to find the minimum number of firemen needed to extinguish a fire in the shortest amount of time. It helps to determine the optimal strategy for deploying firemen and allocating resources.

## What is the relationship between the "1st derivative" and the "2nd derivative" in the fireman's problem?

The "1st derivative" represents the rate of change of the total number of firemen needed, while the "2nd derivative" represents the rate of change of the "1st derivative" or the acceleration of the total number of firemen needed. The "2nd derivative" can help to determine the stability of the solution found using the "1st derivative".

## How does the "1st derivative" change as the fire spreads?

The "1st derivative" changes as the fire spreads because the rate of change of the total number of firemen needed will increase as the fire grows in size and intensity. This is due to the fact that more resources will be needed to contain and extinguish the fire.

## What are some limitations of using the "1st derivative" in the fireman's problem?

Some limitations of using the "1st derivative" in the fireman's problem include the assumption that the fire spreads in a uniform and predictable manner, which may not always be the case. Additionally, the model may not take into account external factors such as weather conditions or obstacles that may affect the deployment of firemen.

• Calculus and Beyond Homework Help
Replies
3
Views
815
• Calculus and Beyond Homework Help
Replies
4
Views
767
• Calculus and Beyond Homework Help
Replies
0
Views
146
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
20
Views
310
• Calculus and Beyond Homework Help
Replies
5
Views
989
• Calculus and Beyond Homework Help
Replies
5
Views
2K
• Calculus and Beyond Homework Help
Replies
7
Views
2K
• Calculus and Beyond Homework Help
Replies
7
Views
263