Newtons method?

Limits, differentiation, related rates, integration, trig integrals, etc.

Newtons method?

Postby TheAngryMathStudent on Mon Mar 10, 2014 10:18 pm

I have no idea where to start with these problems.

1) Find an approximate value 11^(1/3) with using 3 iteration of Newtons method with startpoint of x-3/2 on an appropriate function

2) 6,5m ladder resting up against a wall. The lader gets pulled away from the wall (while still resting on it) at a speed of 0,5 m/s. At what speed does the ladder move downward when it's 6m up on the wall.

Anyone who can help me out/nudge me in the right direction?

Any help is greatly appreciated.
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Re: Newtons method?

Postby stapel_eliz on Mon Mar 10, 2014 10:44 pm

TheAngryMathStudent wrote:I have no idea where to start with these problems.

Then the first step will be to learn the basics of this topic, so that you know enough to understand what is being asked. There are loads of online lessons, such as here.

TheAngryMathStudent wrote:1) Find an approximate value 11^(1/3) with using 3 iteration of Newtons method with startpoint of x-3/2 on an appropriate function

Does the exercise really say "x-3/2", or does it maybe say "x = 3/2"? (The latter makes much more sense.)

TheAngryMathStudent wrote:2) 6,5m ladder resting up against a wall. The lader gets pulled away from the wall (while still resting on it) at a speed of 0,5 m/s. At what speed does the ladder move downward when it's 6m up on the wall.

This is just a related rates problem. (The topic of related rates would have been covered on a different day (maybe even a different week) from Newton's Method. You must have been really sick to miss so many classes. I hope you're doing better now.)

Start by drawing a picture, just like you did back in algebra. Then label the wall (between the floor and where the ladder's top hits the wall) as "y", the floor (between the wall and the base of the ladder) as "x", and the length of the ladder as "L = 6.5". Note that L is fixed, so dL/dt = 0; and you're given that dx/dt = 0.5. Then see if maybe the Pythagorean Theorem gives you anything useful. :wink:
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Re: Newtons method?

Postby TheAngryMathStudent on Tue Mar 11, 2014 12:41 pm

Does the exercise really say "x-3/2", or does it maybe say "x = 3/2"? (The latter makes much more sense.)


This is indeed correct, just typed it in wrongly. I still don't really grasp how to work it out though,
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Re: Newtons method?

Postby FWT on Tue Mar 11, 2014 1:28 pm

TheAngryMathStudent wrote:I still don't really grasp how to work it out though,

The lesson at the link shows the formula. You just plug into it. They gave you f(x)=x^(1/3) and x_0=3/2. Find f' and then start plugging in.

x_1=x_0-[f(x_0)/f'(x_0)]

where is it going wrong? Please show all your work. Thans.
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Re: Newtons method?

Postby TheAngryMathStudent on Thu Mar 13, 2014 12:48 pm

stapel_eliz wrote:Start by drawing a picture, just like you did back in algebra. Then label the wall (between the floor and where the ladder's top hits the wall) as "y", the floor (between the wall and the base of the ladder) as "x", and the length of the ladder as "L = 6.5". Note that L is fixed, so dL/dt = 0; and you're given that dx/dt = 0.5. Then see if maybe the Pythagorean Theorem gives you anything useful.


x^2 + y^2 = L^2

2x dx/dt + 2y dy/dt = 0

y dy/dt = -x dx/dt

dy/dt
----- = -x/y
dx/dt


dy/dt
-------- = 2,5/6 (6 is m up on the wall and 2,5 is how far (x) the ladder has been drawn out?)
0,5


So dy/dt = 0,2083? Is this correct?
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Re: Newtons method?

Postby FWT on Thu Mar 13, 2014 6:13 pm

TheAngryMathStudent wrote: dy/dt
----- = -x/y
dx/dt

dy/dt
-------- = 2,5/6 (6 is m up on the wall and 2,5 is how far (x) the ladder has been drawn out?)
0,5

How did "-"x/y turn into "+"2.5/6?
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