Quick math questions

[LiLRiceBoi]

I'm trying to help my friend out w/ some "Related Rates" problems. I just cant get 1 of them to match the ans in the back, and i dont know how to do the second.

1. Find the rate of change of the distance between the origin and a moving point on the graph y=x^2 + 1 if dx/dt=2 centimeters per second.

2. A winch at the top of a 12-meter building pulls a pipe of the same length to a vertical position. The winch pills in rope at a rate of -.2 meters per second. Find the rate of vertical change and the rate of horizontal change at the end of the pipe when y=6.

y is the coordinate of the very end of the pole (where the winch is attatched). So its like the bottom end touching the building is (0,0) and the end of the pole is (12,0) when it isnt pulled.

One end of the pipe is at the base of the building like a fulcrum? And the other is pulled by the winch. Hope you guys can understand this.


THANKS


I need to get this to him by 10 tomarrow morning.

 

[fatalbert]

can't remember calculus right now,

free bump

 

[gwlam12]

i hate those problems. but i might remember. if my sister stops bugging me with her hw questions (annoying AIM sound keeps going off), maybe i'll devote some time to urs. not saying i can solve it though. free bump for now

 

[gwlam12]

does number 1 happen to be 4 cm/sec?

 

[LiLRiceBoi]

The ans for number 1 isnt a numeric answer. it has x in it. i dont remember it and i wrote it down upstairs. but i do know that there were a lot of Xs in it and the denominator was a square root of some function.

i know you guys know how to do this... BUMP

 

[Skyclad1uhm1]

1. If measured in a straigth line and starting point is (0,0): sqr((2t^2+1)^2+(2t)^2)
Otherwise: Good luck, I need to work too, can't spend too much time on it

 

[CSoup]

did you get 2x(1 + 2(x^2+1))/SQRT(x^4+3x^2+1) = [4x^3 + 6x]/SQRT(x^4 + 3x^2 + 1)
for the first one?

 

[CSoup]

nm I see you updated the description for the second one.

 

[LiLRiceBoi]

<< did you get 2x(1 + 2(x^2+1))/SQRT(x^4+3x^2+1) = [4x^3 + 6x]/SQRT(x^4 + 3x^2 + 1)
for the first one? >>



You are GOOD. Thats the ans in the back of the book. Care to give me a brief explination?

THANK YOU so much.

Still waiting for help on the second one...

 

[CSoup]

this is how I did the first one.

first you have the equation x^2 + y^2 = z^2 where z is the distance from the point on the function to the origin.
you do implicit differentiation to get 2z(dz/dt) = 2x(dx/dt) + 2y(dy/dt).
dx/dt is given as 2.
dy/dt = 2x(dx/dt) from differentiating the function y = x^2 + 1.
dz/dt is what you want.
you also know z = sqrt(x^2+y^2).

you do a little substitution with these values and you should be able to get what I got.

I might look at number two if I don't feel to sleepy.

 

[CSoup]

not exactly sure on the second one.

I got dy/dt = 0.2 and dx/dt = -0.115
these seem a little odd, so I'm not sure how close they are.
I'm not even sure if I modeled this problem correctly from your description.

last time I took calc was over 5 years ago, so I'm a little rusty.

 

[thEnEuRoMancER]

Here's what I got for the second one:

- the building wall, the pipe and the rope form a equilateral triangle when y=6
- the angle between the rope and the horizontal is therefore 30°
- the x projection of speed vector (the speed vector is always pointing in the direction of rope) is -0.2*cos(30°)=-0.173 m/s
- the y projection of speed vector is 0.2*sin(30°)=0.1 m/s

 

[AaronP]

1+1=2

 

[LiLRiceBoi]

<< 1+1=2 >>



I really needed help w/ that one Aaron....


Thanks CSoup and thEnEuRoMancER, you guys got the first one right. second one isnt.

 

[LiLRiceBoi]

OK i figured out the second one in class.

 

[littlezipp]

<< OK i figured out the second one in class. >>

What class is this for, I need to remember not to take it!

 

[CSoup]

<< OK i figured out the second one in class. >>



What is the answer to the second one?

 

[LiLRiceBoi]

My friends in first semester calculus at a jc. I'm taking calc in hs, and we learned this stuff a long time ago, so i kinda forgot some of it.

I dont think calc is that hard though...