solve these pls. calculus problems?

Question

1. You are given a piece of paper that is 12 inches in length and 6 inches wide. The lower right-hand corner is folded over so as to reach the leftmost edge of the paper. Find the minimum length of the resulting crease.
2. A 40ft ladder is leaning against a 40ft inclined plane with an angle of elevation of 30 degrees with respect to the horizontal. If the foot of the ladder is pulling away the inclined plane at a rate of 5ft/s when the distance of the top of the ladder is 10ft from the top of the inclined plane, how fast is the top of the ladder sliding down the inclined plane?

Answer 1

1. First I wanted to show where the minimum occurs rather than just say something. So hope you can follow the reasoning here and hope it is sound. I followed the crease as the bottom point of the crease moves along the bottom edge and then up the left hand side. It shows that the length depends on the angle the crease makes with the bottom edge or a line parallel to the bottom edge and that it decreases as this angle decreases. I then used similar triangles to find what the length of the crease is.

c = length of crease
b = length along bottom edge from from right corner to where the crease begins.
A = the angle at the point where the crease begins at the bottom edge

c = b/cos(A)

A second triangle is formed using the remaining part (6 - b) of the bottom edge. If B is the angle at the right vertex along this edge then you can write:
pi = 2A + B

cos(B) = (6 - b)/b = cos(pi - 2A) = -cos(2A) = cos^2(A) - sin^2(A)
6/b = sin^2(A) - cos^2(A) + 1
b = 6/(sin^2(A) - cos^2(A) + 1

c = [6/(sin^2(A) - cos^2(A) + 1)] /cos(A)
c = [6/(2 cos^3(A))]
c = 3/cos^3(A)

This function decreases as A increases. However it only goes so far as the point where the crease starts at the lower left corner at which point it is equal to 6SQRT(2). At this point the b in the original equation becomes fixed at 6.

When the crease left point starts moving up the left edge of the paper:

c = 6/cos(A)

At A = 45 this gives c=6SQRT(2) which agrees with the above and is correct for this angle.

Again this function decreases as A decreases so the minimum for c will be when A is the smallest and this is determined by the constraints on the size of the piece of paper. This will be when the lower right corner is brought up to the upper left corner.

If you look at the triangle at the bottom. The crease is the hypoteneuse and one side is a line parallel to the bottom edge. If the angle at the lower left vertex is D and w is the traingle side along the right edge of the paper then:

tan(D) = w/6 = (12 - 2e)/6 = (6 - e)/3

Where e is the distance from the right end of the crease up to the upper right corner of the piece of paper. A little thinking about similar triangles and such will show you that w = (12 - e) is indeed the side of the triangle along the right edge of the paper. And, if you look at the triangle formed around the upper right corner you will see that the next relation is true:

tan(2D) = 6/e = 2tan(D)/(1 - tan^2(D))

Put these together and you get:
3/e = [(6 - e)/3] / [1 - (6 - e)^2/9]
9[1 - (36 - 12e + e^2)/9] = 6e - e^2
9 - 36 + 12e - e^2 = 6e - e^2
6e = 27
e = 27/6 = 9/2 = 4.5

so that w = 12 - 2e = 3

c^2 = 6^2 + 3^2 = 45
c = 3SQRT(5) = 6.708 inches

2. Let "a" be the distance along the incline up to the point where the ladder meets it. "b" is the length along the ground from the ladder to that incline. So:

db/dt = 5 when a = 30 and we want to fine da/dt

First find b when a = 30 :
c^2 = a^2 + b^2 - 2abcos(180 - 30)
40^2 = 30^2 + b^2 + 30bSQRT(3)
700 = b^2 + 30SQRT(3)b

b = (-30SQRT(3) +/- SQRT(2700 + 2800)) / 2
b = -15SQRT(3) + 5SQRT(55)
b = 5 (SQRT(55) - 3SQRT(3)) = 11.1

Now find the relationship between db/dt and da/dt:
40^2 = a^2 + b^2 - 2abcos(180 - 30)
40^2 = a^2 + b^2 + SQRT(3)ab

0 = 2a da/dt + 2b db/dt + aSQRT(3) db/dt + bSQRT(3) da/dt
0 = da/dt(2a + bSQRT(3)) + db/dt (2b + aSQRT(3))
da/dt = db/dt (2b + aSQRT(3)) / (2a + bSQRT(3))

Substitute in the known values:
db/dt = 5
a = 30
b = 5 (SQRT(55) - 3SQRT(3)) = 11.1

(2b + aSQRT(3)) = 2(11.1) + 30SQRT(3) = 74.16
(2a + bSQRT(3)) = 2(30) + 11.1SQRT(3) = 79.226

da/dt = 5(74.16/79.226) = 4.68 ft/s

Or worked out with SQRTs and stuff:

(2a + bSQRT(3) = 5(12 + (SQRT(55) - 3SQRT(3))SQRT(3) )
= 5(3 + SQRT(55)SQRT(3)

(2b + aSQRT(3)) = 5(6SQRT(3) + 2(SQRT(55) - 3SQRT(3))
= 5(2SQRT(55)) = 10SQRT(55)

da/dt = db/dt [10SQRT(55))] / [5(3 + SQRT(55)SQRT(3)
da/dt = 5 [(2SQRT(55))] / [(3 + SQRT(55)SQRT(3)]
da/dt = 4.68

Answer 2

1. crease length will minimize when it touches the highest leftmost.
diagonal of paper=sqrt(12^2+6^2)
=sqrt(180)
=6sqrt(5)
when folded from 1 end of a diagonal to the other, we get an isoceles triangle with base of crease and height of half the diagonal. the angle btw base and edge is equal to the 1 of the btw of diagonal and width.
crease=6*6sqrt(5)/2*12
=1.5sqrt(5)

2.
top distance travelled=10(directly)

foot distance travelled
=sqrt(40^2-15^2)-sqrt(30^2-15^2)
=5[sqrt(55)-sqrt(27)]

v/10=5/5[sqrt(55)-sqrt(27)]

v=10/[sqrt(55)-3sqrt(3)]