-=RX-7 KID=-

I have to figure out a whole bunch of related rates for my calc 1 class and there are 2 i am pretty much totaly stuck on, if any one could show me a step by step on these that would be great.



1. At what rate is the diagonal of a cube increasing if the sides are increasing at the rate of 2cm/sec?



2. A ladder 8 meters long leans against a wall 4 meters high. The lower end of the ladder is pulled away from the wall at a rate of 2 m/sec. How fast is the angle between the top of the ladder and the wall changing when the angle is pie/3 radians?



Any help would be great thanks guys.

jokerx7

I'm a little fuzzy on Calc 1 stuff, but I remember the fundamental steps to those problems:



1. You just need to set up an equation for the distance of the diagonal, something along the lines of the square root of (x^2+y^+z^2), and just plug in 2 cm/s, and that should be it.



2. If it were me, I'd split this up in vectors 4j m, and use sin to figure out the horizontal distance. From there, use the same ratio on the x velocity, and solve for y velocity. from there, just find that angle, and it should be the answer.



I'll see if I can get a couple answers real quick too if you wish

jokerx7

Answers (I Hope...)



1. 2*sqrt(3) ~ 3.464cm/s



2. 6.928i+4j = 2i+xj, x=1.1547

so, tan-1(2/1.1547)=

1.494 rad/s



I am by no means sure about 2, but if someone remembers a little more about calc 2, maybe they can help. I have a sneaking suspicion these would be pretty easy with simple derivation/integration

jokerx7

Also for 2, I just remembered an equation from Dynamics that would be useful:



V(b)=V(a)+w x r,



where v(b) and v(a) are the velocities of the endpoints of the ladder (in vectors), w is angular velocity, and r is the length of the ladder @ your moment in time, also in vectors, ie xi+yj, and zk for your angular.

-=RX-7 KID=-

Thanks me and some friends looked over 1 and we think your right. No clue about 2 still.

Dramon_Killer

If youre gonna solve it like a dynamics problem you would find how long the angle of the wall is when the angle of the bottom is pi/3. Now you solve for time since you have a velocity.



Plug it into v = ds/dt which when you rearrange it becomes vdt = ds which when you take the integral of both sides it yields 2t = 4 - wall distance at pi/3



Now that you have a time you can plug it into your rigid body rotation equations.



You use w = d(theta)/dt which becomes wdt = d(theta) take the integral of both sides with the limits of the first side being 0 and whatever time you found. and the other side is just theta and that is your final answer.



You can't use any of the rigid body relative motion problems becuase the angular acceleration isn't constant.

Dramon_Killer

**** just kidding, actually i'm not sure if you can use that becuase there really isn't any rotation taking place...and the velocity is constant.



Actually, maybe it would work? If you imagined the top of the ladder to be on a hinge so as it got pulled out it slid down.



I dunno maybe you would have to use cylindrical coordinates to solve it?