Difficult Problems

A walk is perpendicular to a long wall, and a man strolls along it, away from the wall, at the rate of 4 feet per second. There is a light 10 feet from the walk and 25 feet from the wall. How fast is his shadow moving along the wall when he is 15 feet from the wall?

x/25=s/(10+s)                                  15/25=sd/(10+s)
25s+10x+xs                                     150+15s=25s
(25)ds/dt=(10)dx/dt+dx/dt(s)+(x)ds/dt          150=10s
(25)ds/dt=(10)(4)+(4)(15)+(15)ds/dt            15=s
(10)ds/dt=10
ds/dt=10

A walk is perpendicular to a long wall, and a man strolls along it away from the wall at the rate of 3 ft per second. There is a light 8 feet from the walk and 24 feet from the wall. How fast is his shadow moving along the wall when he is 20 feet from the wall?
 
 

x/24=s/(8+s)                                   20/24=s/(8+s)
24s=8x+xs                                      24s=160+20s
(24)ds/dt=(8)dx/dt+(s)dx/dt+(x)ds/dt           4s=160
(24)ds/dt=(8)(3)+(3)(40)+20ds/dt               s=40
(4)ds/dt=144                                            
ds/dt=36ft/sec

A man, 6 feet tall, is walking away from a light, 14 feet above the ground, at the rate of 5 feet per second. How fast is his shadow changing in length?

(In general, shadow problems are set up as proportions!)

6/14=s/(x+s)                                 
14s=6x+6s
8s=6x
(8)ds/dt=(6)dx/dt
(8)dx/dt=(6)(5)
dx/dt=30/8=15/4

Solve. Support your answer!

abs(x+2)+abs(x-5)@10
a)  x+2+x-5     when     x+2<0 and x-5<0="2x-3" x<-2 and x<5 x<5 b) x+2-(x-5) when x+2<0 and x@5="x+2-x+5" x<-2 and x@5="7" 2*x@5 c) (x+2)-(x-5) when x+2@0 and x-5<0 dne x<-2 and x<5 impossible d) (x+2)-(x-5) when x+2<0 and x-5<0="-x-2-x+5" x<-2 and x<5="-2x+3" x<-2 a)2x-3<10 b)7<10 d)-2x+3<10 2x<13 r 2x<7 x<13/2 x>-7/2

Solution: -7/2
These inequalities do not govern the solution set.  They 
determine if a solution is possible or not for each 
equation.

A weight W is attached to a rope that is 50 feet long which passes
over a pulley at P, 20 feet above the ground. The other end of the
rope is attached to a truck at point A, 2 feet above the ground.  
If the truck moves off at a rate of 9 ft./sec., how fast is the
 weight rising when it is 6 feet above the ground?

•  
• As the truck lifts the weight, you no longer have a right triangle.
• This one is just plain hard.

A 10 pound monkey hangs at the end of a 20 foot chain that weighs .5 pounds per foot. How much work does it do in climbing the chain to the top? Assume that the end of the chain is attached to the monkey.

• 
• 1. You must account for the fact that not all of the chain goes up to the top - some of it is still hanging down behind the monkey.
• 2. There are many ways of looking at this problem, and we are not sure which ones are correct yet.

A point moves along the graph of y = 3x2 - 4 such that the rate of change of its x-coordinate is 8 inches per second. How fast is the y-coordinate changing when the point is at (2,8)?

givens:

1. dx/dt = 8 in./sec.
2. y = 3x^2 - 4 3) dy/dt = 6x(dx)

• dy/dt = 6(2)(8)
• Substitute for the values of x and dx.
• = 96 in./sec.

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Unsolved Mysteries

These are problems that for some reason or another, we never
quite decided what to do with.  Maybe you'll have better luck!

Plane A is flying west at the rate of 600mph and a second plane B, 300 miles south of A, is flying north at the rate of 500 mph.

a) At what rate is the distance between the two planes changing after 15 minutes?

b) When do the planes cease to approach one another?

Technically, the lengths of the sides of the triangle should be expressed as a rate times a time, so A would be 600t, and B, 300-500t. However, when you differentiate, you get dt/dt. What do you do with that?

b. The rate of change of x would be negative--or would it?

One ship A is sailing due south at 16 mph and a second ship B, 32 miles south of A, is sailing due east at 12 mph.

a) At what rate are they approaching or separating at the end of 1 hour?

b) When do they cease to approach each other and how far apart are they at that time?

This one suffers from the same problems as the previous question.
 

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Problems

A rectangle has two of its vertices as
the x-axis.  The other two vertices are on the parabola whose 
equation is y=18-x^2.  What are the dimensions of the rectangle
if its area is to be a maximum?

A=2xy A=2x(18-x2) [plug in for y from eqn of parabola] A=36x-2x3 A'=36-6x2 [take derivative] 0=36-6x2 -36=-6x2 6=x2 x=6 x=[0,18] [Set domain. x cannot be bigger than 18, or else it is outside the parabola and the vertices can't be on the graph. A(0)=0 A(6)=58.788=max A(18)=0 Rectangle = 26 by 18-(6)2 26 by 12

A wire 14 feet in length is cut into two pieces. One part is used to make a square. The other part is bent into a rectangle such that L=3w. How should the wire be cut if the sum of the areas is to be a minimum?

P=4x+8w       [x is side of square, w is short side of rectangle]
14=4x=8w
14-8w=4x
7/2 -2w=x

A=x2+3w2
A=(7/2 -2w)2+3w2
A=49/4 -14w+4w2+3w2
A=49/4 -14w+7w2
A'=-14+14w
0=-14+14w
14=14w
1=w

w=[0,7/2]      [Set domain.  w can be no bigger than 7/4, because when w=7/4,
                the perimeter of the rectangle is 14.]
A(0)=49/4      [test limits of domain and answer]
A(1)=21/4=min
A(7/4)=147/16

The piece for the rectangle is 8 ft long.  The piece for the square is 6 ft long.

The area bounded by the curve y=x2 and the line y=4 generated various solids of revolution when rotated

a) about the y-axis        b) about the line y=4       c) about the x-axis
d) about the line y=-1     e) about the line x=2

Find the volume generated in each case.

a)

[To use the disk method and partition the y-axis, the function must be solved for x.]
 
 

b)

[In this case, the graph was translated so y=4 is on y=0. There are other ways to approach it. For ease, half the graph was integrated and the result will be doubled.]
 
 

c)


d)

[In this case, the shell method was used. The average radius is y+1, since the graph was moved 1 down.]

e)

[Shell method used. Average radius is x-2 since the graph was moved 2 to the right. In this case, the function must be integrated from -2 to 2.]

Note! If any theoretical or calculation errors are found in the work of these problems, please notify us. If you are able to solve one of the “unsolved mysteries,” please, please, please let us know! (And congratulations -your one head was better than our thirty-two heads!)