Question

A bag is tied to the top of a 5 m ladder resting against a vertical wall. Suppose the ladder begins sliding down the wall in such a way that the foot of the ladder is moving away from the wall. How fast is the bag descending at the instant the foot of the ladder is 4 m from the wall and the foot is moving away at the rate of 2 m/s?

Answer 1

@AccessDenied this is a related rates problem but I don't know how to approach it.

Answer 2

@iPwnBunnies

Answer 3

@amistre64

Answer 4

I would first recommend a drawing of this scene. You can label the 5 meter ladder and the house/ground perpendicular to each other and creating a right triangle. The bag tied at the top of the ladder would fall with the ladder.

Answer 5

I think the bag is descending at a rate of -1.798 do you have any choices? @enjc

Answer 6

no free response

Answer 7

Okay my thoughts on this one is that you have a ladder and that makes a triangle with the wall. They have given you that the height is 5m and that the length is 4m. We can use pythagores to find the hypotenuse.

With the height and width given, we can figure out the area = bh/2

in this case if you differentiate this function, we would get:

a = b'h/2 + h'b/2

we have the base, the height and the rate at which the base is moving which is 2m/s to the right.

Solving for h' should give us the rate at which the bag is descending.

Note that it should be a negative number because the magnitude is going down.

Hope this helps.

Answer 8

thanks

Answer 9

hmmm, using the area of a triangle to determine this ... i dont see why it wouldnt work but i never thought of using it for this :)

i would have stuck with the right triangle pythag.

h^2 + w^2 = l^2, for height, width (from wall), and ladder

implicitly derived we get:

2hh' + 2ww' = 2l l'; but since l is constant, its rate of change l' is zero; and we can factor out the 2s

hh' + ww' = 0; w is given as 4, and w' as 2, and pythag gives us h=3 at this moment in time

3h' + 4(2) = 0

if everything is typed correctly, then h' would be -8/3

Answer 10

a = bh/2

a' = b'h/2 + bh'/2 would be the area implicit which means we would have to figure out what the rate of change of area is in order to use this.

Answer 11

@amistre64 I think the rate of change of the area would be the same, because bh when the base increase the height has to decrease and vice versa. What do you think?

Answer 12

there is no rate of change given for area we can only determine a' AFTER knowing h', not before knowing it

Answer 13

I know that, but my question is do you think that the rate of change for the area will be 0? Because when the base increase the height decreases so they are both inversely proportional. What do you think?

Answer 14

if your contention is that a' = 0; we can try to prove it; but i get a different result than yours so im pretty sure its not

Answer 15

@amistre64 can you type down your proof please?

Answer 16

well, we know a 3-4-5 makes a good fit, if the area of a-b-5 is different from 3-4-5 then we have an issue; agreed?

Answer 17

in other words, if the area doesnt change with respect to the leg values ... then a'=0

Answer 18

but lets say the ladder is very close to the ground, almost touching it at a height of .0000001. is the area going to be the same as 3(4)/2 = 6?

Answer 19

http://www.wolframalpha.com/input/?i=sqrt%285%5E2-%28.0000001%29%5E2%29*%28.0000001%29%2F2

1/4000000 is not equal to 6

Answer 20

Oh yeah, I get your point now. You are saying that in this case we are changing one factor of the area not both.

Answer 21

correct; and also that the area is not constant for all right triangles of hypot 5

Answer 22

@amistre64 Thanks a lot for your help! I really appreciate your time.

Answer 23

your welcome; the area application was interesting, but regretbly not doable.