Expert: Scotto - 5/13/2009

Question
1)A fence 3 feet tall runs parallel to a tall building at a distance of 2 feet from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building?

I cannot figure out how to set this up. please help!



2)Find the minimum distance from the parabola
to the point {(0,3)}.

x-y^2=0

minimum distance is??




(sorry I sent so many questions, everyone else was maxed out or on vacation)

Answer
1. You can draw a triangle with base A+2 and height H.  The height H would be for the building.  The A+2 would be the 2 feet to the fence plus A untis to the ladder.  There is another triangle with this triangle.  It has base A and height 3 for the fence.  The two triangles both are right triangles and have the same bottom angle, therefore the top angle is the same as well.

For the small triangle, the height is 3 and the base i A.
For the large triangle, the height is H and the base is A+2.

The ratio of 3/A = H/(2+A) => H = 3(2+A)/A = (6+3A)/A.

We know that for the big triangle, the sum of each side squared is the square of the hypoteneuse.  That is,
((6+3A)/A)² + (2+A)² = L², where L is the ladder length.
Working this out gives 36/A² + 36/A + 9 + 4 + 4A + A².

Since L is always positive, L² is an ever increasing function.
This means that we can find d(L²)/dA.

It is d(L²)dA =  -72/A^3 - 36/A² + 4 + 2A.
Setting this to 0 gives (-36/A^3)(2+A) + 2(2+A) = 0.
That is, (-36/A^3 + 2)(2+A) = 0.

The second expression, 2+A, could be 0 at A = - 2,
but that makes no sense.

The first expression would be 0 when 36/A^3 = 2,
or when A^3 = 18, so A = 18^(1/3).
This means that B = 2 + 18^(1/3).
It is known that H = (6+3A)/A.
This means that H  = (6 + 3*18^(1/3))/18^(1/3).
That can be reduced to H = 6*18^(2/3)/18 + 3.
Lets cancel the 3 in the first one and switch the terms, giving
H = 3  +  18^(2/3)/3.
That means B² = 4 + 4*18^(1/3) + 18^(2/3) and
H² = 9 + 2*18^(2/3) + 18^(4/3)/9.

It is known that 18^(4/3) = 18*18^(1/3),
so 18^(4/3)/9 = 18*18^(1/3)/9 = 2*18^(1/3).

This said, it can be worked out that
B² + H² = 13 + 6*18^1/3 + 3*18^(2/3).

We already have seen that B² + H² = L²,
so now just take the squareroot to find L.

It took me awhile, but this looks right.
I checked it several times since the answer was so weird.


2. The distance to a parabola is the squareroot of the sum of the squares.  It is also known that at the point y0 on the line and the parabola, x0 = y0^2.  The distance is then (x0-0)² + (y0-3)².
Replace x0 with y0² and differentiate with respect to y0.
The equation is (y0²)² + (y0-3)² = D² where D is the distance.
Take the derivative, set to 0, solve for y0, find x0, get the slope, and you have the line.