1st derivative in the fireman's problem

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JOhnJDC
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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).

EDIT: I added some additional information below.

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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Answers and Replies

  • #2
JOhnJDC
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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.
 

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