# 1st derivative in the fireman's problem

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.

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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.