Find the Function x(t) | Bug Climbing a 30-Foot Ladder Against a Wall

[nowiz68710]

Problem: A 30-foot ladder rests vertically against a wall. A bug starts at the bottom of the ladder and climbs up at a rate of 3.5 feet per minute. At the same time, the foot of the ladder is being pulled along the ground at a rate of 1.5 feet per minute until the top of the ladder reaches the ground. Let x be the distance of the bug from the wall at time t.



Question: Find the function x(t). This function gives the horizontal distance of the bug to the wall as a function of time, t.



My lame attempt: okay so I know the ladder will always be 30 foot long so a^2 + b^2 = 30^2 assuming a is the distance pulled away from the wall and b is the distance from the top of the ladder to the ground. Now a starts at zero and b at 30 feet with the ladder resting against the wall and the bug at the bottom of the ladder. a will constantly be growing per minute at a rate of 1.5 feet. b is where I'm stuck, it will be shrinking not only that the top of the ladder is sliding down the wall but also that the bug is moving up the ladder creating a similar triangle inside but I am stuck Please help if you can. Thank You.

 

[tiny-tim]

welcome to pf!

hi nowiz68710! welcome to pf!

(try using the X² icon just above the Reply box )

hint: the bug is a certain proportion of the way along the line AB

 

[nowiz68710]

A certain proportion? well let's see da/dt = 1.5 ft/min and dz/dt = 3.5 ft/min (z being distance from base of ladder to the bug) then a/30 = x/30-z so 1.5t/30 = x(t)/30-3.5t am I even close?

 

[tiny-tim]

i'm not really following that …

write A and B as vectors (pairs of coordinates), then work out how far along the line AB the bug is at time t …

that will give you the coordinates of the bug

 

[zgozvrm]

You don't even need to calculate b in order to solve this problem!

Try drawing a right triangle where the hypotenuse is the ladder. This picture represents the ladder at some time t. You already know how to calculate the distance of the bottom of the ladder from the wall. What about some given point along the hypotenuse? For example, 2 feet up the ladder.

Using similar triangles, I think you'll find the answer.