## anonymous one year ago Last differentiation problem for the night (hopefully). I'm not sure if it's the way it's worded, but this problem is just downright confusing to me.

1. anonymous

Refer to a 16-ft ladder sliding down a wall. The variable h is the height fo the ladder's top at time t, and x is the distance from the wall to the ladder's bottom. a) Assume that the bottom slides away from the wall at a rate of 3 ft/s. Find the velocity of the top of the ladder at t = 2 if the bottom is 5 ft from the wall at t = 0.

2. anonymous

|dw:1444353532318:dw|

3. anonymous

you are told $$x'=3$$

4. anonymous

when $$t=2$$ how far is the ladder away form the wall?

5. anonymous

Is there a formula to use on how to calculate that answer?

6. jim_thompson5910

hints:  the bottom slides away from the wall at a rate of 3 ft/s the bottom is 5 ft from the wall at t = 0

7. jim_thompson5910

|dw:1444353987783:dw|

8. jim_thompson5910

|dw:1444354010477:dw|

9. anonymous

not really a formula...

10. anonymous

So x would equal 11 at t=2 and that means that h would be about 11.6.

11. jim_thompson5910

|dw:1444354236271:dw|

12. anonymous

put those in last use pythagoras to find the relation between $$x,h$$ and $$16$$

13. anonymous

I ended up getting about 2.84

14. jim_thompson5910

I'm getting about the same $\Large x^2 + h^2 = 16^2$ $\Large x^2 + h^2 = 256$ $\Large \frac{d}{dt}(x^2 + h^2) = \frac{d}{dt}(256)$ $\Large 2x*\frac{dx}{dt}+ 2h*\frac{dh}{dt} = 0$ $\Large 2*11*3+ 2*11.61895004*\frac{dh}{dt} = 0$ $\Large 66+ 23.23790008\frac{dh}{dt} = 0$ $\Large 23.23790008\frac{dh}{dt} = -66$ $\Large \frac{dh}{dt} = \frac{-66}{23.23790008}$ $\Large \frac{dh}{dt} = -2.840187787$

15. anonymous

Oops! Forgot my negative sign. Thanks! I think it was just the way the problem was worded is what threw me off. Once you guys helped explain it, it actually made sense. Thank you! :D

16. jim_thompson5910

you're welcome

17. anonymous

This is the last part of this problem. Could you help me with it by chance? Show that the velocity dh/dt approaches infinity as the ladder slides down to the ground (assuming dx/dt is constant). This suggests that our mathematical description is unrealistic, at least for small values of h. What wound in fact, happen as the top of the ladder approaches the ground?

18. anonymous

@jim_thompson5910

19. jim_thompson5910

Let p = dx/dt q = dh/dt $\Large x^2 + h^2 = 16^2$ $\Large x^2 + h^2 = 256$ $\Large \frac{d}{dt}(x^2 + h^2) = \frac{d}{dt}(256)$ $\Large 2x*\frac{dx}{dt}+ 2h*\frac{dh}{dt} = 0$ $\Large 2x*p+ 2h*q = 0$ $\Large 2hq = -2xp$ $\Large q = \frac{-2xp}{2h}$ $\Large q = -\frac{xp}{h}$ $\Large \frac{dh}{dt} = -\frac{xp}{h}$ We can see that as the ladder slides to the ground (h --> 0), the fraction dh/dt approaches -infinity. If you have dx/dt be some negative constant, then the fraction dh/dt approaches +infinity.

20. jim_thompson5910

Obviously in the real world, the speed would hit some limit just before the ladder hits the ground

21. anonymous

Thank you so much!