Understanding an Equation Simplification: Help Needed

[vorcil]

Homework Statement

1. Homework Statement [/b
I need help understanding how this equation is simplified
$$(\frac{dx}{gt})*(\sqrt(\frac{g}{2h}))$$

to
$$(\frac{1}{2*(\sqrt{hx})})*dx$$

The Attempt at a Solution

I understand how the g cancels out,

but how exactly does the t disappear,
and square root of 2h become 2 root hx?

I lack the mathematical tools to solve this, please help

 

[vorcil]

(edited: was trying to figure out how latex works.)

Question:

how does, $$(\frac{dx}{gt})*(\sqrt(\frac{g}{2h}))$$
simplify to

$$(\frac{1}{2*(\sqrt{hx})})*dx$$

 

[vorcil]

[tex](\frac{1}{2(\sqrt{hx})}) dx[\tex]

 

[Mark44]

I don't think you've given us enough information to be able to help you. What's the context in which this appears? Did you start with some other expression and manipulate it to get the first one you show in your first post?

 

[Mark44]

$$(\frac{1}{2(\sqrt{hx})}) dx$$

The only real mistake is in your final [ /tex] tag, which needs to have a forward slash - /, not the backward slash you used.

Also, you have more parentheses than you need. Here it is cleaned up:
$$\frac{dx}{2\sqrt{hx}}$$

 

[vorcil]

I don't think you've given us enough information to be able to help you. What's the context in which this appears? Did you start with some other expression and manipulate it to get the first one you show in your first post?

Whoops sorry, i'll write out the entire question:

Suppose I drop a rock off a cliff of height h.
As is falls, I snap a million photographs, at random intervals.
On each picture I measure the distance the rock has fallen.

question: What is the average of all these distances? that is to say, what is the time average of the distance traveled?

------------------------------------
My solving:
The rock starts out at rest and picks up speed as it falls,
so the average distance is less than h/2
ignoring air resistance, the distance x at time t is

$$x(t) = \frac{1}{2}gt^2$$

the velocity is $$\frac{dx}{dt}=gt$$, from v = a*t

solving for the total flight time

$$h(t) = \frac{1}{2}gt^2$$

=$$\frac{2h}{g}=t^2$$

Total flight time, $$T=\sqrt{\frac{2h}{g}}$$

-

The probability the camera flashes in the interval dt, is $$\frac{dt}{T}$$

since $$\frac{dx}{dt}=gt$$

$$dt=\frac{dx}{gt}$$
---------------------------------------

now I have $$dt=\frac{dx}{gt}$$
and $$T=\sqrt{\frac{2h}{g}}$$

and I want $$\frac{dt}{T}$$

so $$\frac{\frac{dx}{gt}}{\sqrt{\frac{2h}{g}}}$$


= $$\frac{dx}{gt}\sqrt{\frac{g}{2h}}$$

-
This is the point where I'm stuck,

I don't understand how
$$\frac{dx}{gt}\sqrt{\frac{g}{2h}}$$

becomes
$$\frac{1}{2\sqrt{hx}}$$

and then solve the rest of the question,

p.s really getting a hang of LATEX now :P

 

[Mark44]

Whoops sorry, i'll write out the entire question:

Suppose I drop a rock off a cliff of height h.
As is falls, I snap a million photographs, at random intervals.
On each picture I measure the distance the rock has fallen.

question: What is the average of all these distances? that is to say, what is the time average of the distance traveled?

------------------------------------
My solving:
The rock starts out at rest and picks up speed as it falls,
so the average distance is less than h/2
ignoring air resistance, the distance x at time t is

$$x(t) = \frac{1}{2}gt^2$$

the velocity is $$\frac{dx}{dt}=gt$$, from v = a*t

Or x'(t) = gt, or v(t) = gt. You can get this by differentiating x(t) = (1/2)gt². Minor point - the comma at the end of your equation above makes it look like t' and threw me off for a bit.

solving for the total flight time

$$h(t) = \frac{1}{2}gt^2$$

=$$\frac{2h}{g}=t^2$$

The equations above are separate. They aren't "equal" so don't connect them with =.

Total flight time, $$T=\sqrt{\frac{2h}{g}}$$

-

The probability the camera flashes in the interval dt, is $$\frac{dt}{T}$$

A better choice is $\Delta t$ rather than dt. I used [ itex]Delta t[ /itex] to write this, but without the leading spaces inside the brackets.

You haven't used the given information that the camera takes 1,000,000 pictures during the rock's fall. This means it takes a picture every T/1,000,000 of a second.

I think your problems come from your assumption about the probability, but I have to get ready for work now, so can't nail it down exactly.

since $$\frac{dx}{dt}=gt$$

$$dt=\frac{dx}{gt}$$
---------------------------------------

now I have $$dt=\frac{dx}{gt}$$
and $$T=\sqrt{\frac{2h}{g}}$$

and I want $$\frac{dt}{T}$$

so $$\frac{\frac{dx}{gt}}{\sqrt{\frac{2h}{g}}}$$


= $$\frac{dx}{gt}\sqrt{\frac{g}{2h}}$$

-
This is the point where I'm stuck,

I don't understand how
$$\frac{dx}{gt}\sqrt{\frac{g}{2h}}$$

becomes
$$\frac{1}{2\sqrt{hx}}$$

and then solve the rest of the question,

p.s really getting a hang of LATEX now :P

 

[vorcil]

I think your problems come from your assumption about the probability, but I have to get ready for work now, so can't nail it down exactly.

Thank you very much for you help, I will continue to work on it, and yes I assumed the probability was dt/T because I did some previous questions in my book and it had the same sense of result, if you catch my drift

 

[Mark44]

Since h = (1/2)gt², then solving for t gives t = sqrt(2h/g). This is the total time the rock falls, which I'll give a new name: T.

The camera fires 1,000,000 times in T sec., so each time interval is $\Delta T$ = T/1,000,000 sec or T/10⁶ sec.

question: What is the average of all these distances? that is to say, what is the time average of the distance traveled?

I think these are asking for two different things. The average of the distances would be the sum of the incremental distances divided by how many there are (1,000,000), and would be in units of feet or meters, or whatever h is measured in. The time average of the distance seems to me to be the average of the velocities of each separate time interval, and would be in units of ft/sec or m/sec.

Assuming for the time being that we're after the average of the distances, you need to get an expression for how far the rock travels in each time subinterval, then add them all up, and divide by the number of subintervals (1 million). Obviously, you're not going to want to actually add up a million numbers, and at this point should be thinking about an integral that represents the sum.

The key to this is to get an expression that represents how far the rock has fallen in a typical time interval; i.e., from t = t_i to t_i + T/10⁶.

 

[vorcil]

Since h = (1/2)gt², then solving for t gives t = sqrt(2h/g). This is the total time the rock falls, which I'll give a new name: T.

The camera fires 1,000,000 times in T sec., so each time interval is $\Delta T$ = T/1,000,000 sec or T/10⁶ sec.
I think these are asking for two different things. The average of the distances would be the sum of the incremental distances divided by how many there are (1,000,000), and would be in units of feet or meters, or whatever h is measured in. The time average of the distance seems to me to be the average of the velocities of each separate time interval, and would be in units of ft/sec or m/sec.

Assuming for the time being that we're after the average of the distances, you need to get an expression for how far the rock travels in each time subinterval, then add them all up, and divide by the number of subintervals (1 million). Obviously, you're not going to want to actually add up a million numbers, and at this point should be thinking about an integral that represents the sum.

The key to this is to get an expression that represents how far the rock has fallen in a typical time interval; i.e., from t = t_i to t_i + T/10⁶.

I was shown that, the probability density

$$p(x)=\frac{1}{2\sqrt{hx}}$$ as long as (0 <= x <= h)

and this means that the probability that the rock is found outside of that range is 0, IE it won't be found underground, or above the point h, where it was dropped

I was told if I wanted to check it, then I integrate

$$\int_{0}^{h} \frac{1}{2\sqrt{hx}} dx = something = 1$$ (also how do i integrate this?)

I'm just not sure how, $$\frac{dx}{gt}\sqrt{\frac{g}{2h}}dx becomes \frac{1}{2\sqrt{hx}}dx$$

 

[vorcil]

$$\frac{dx}{gt}\sqrt{\frac{g}{2h}}dx, becomes \frac{1}{2\sqrt{hx}}dx$$

I can see how the g's cancel out, do they?

$$\frac{1}{g} * {\sqrt{\frac{g}{1}} = 1?$$

Can I just ignore the square root?

but I am not too sure how the 2 moves outside the square root,

nor how an x, joins $$\sqrt{hx}$$

 

[Mark44]

$$\frac{dx}{gt}\sqrt{\frac{g}{2h}}dx, becomes \frac{1}{2\sqrt{hx}}dx$$

I can see how the g's cancel out, do they?

No, they don't. The first expression makes no sense to me. If you simplify it, you get this:
$$\frac{dx}{gt}\sqrt{\frac{g}{2h}}dx = \frac{(dx)^2}{t\sqrt{2gh}} \neq \frac{1}{2\sqrt{hx}}dx$$

$$\frac{1}{g} * {\sqrt{\frac{g}{1}} = 1?$$

Can I just ignore the square root?

but I am not too sure how the 2 moves outside the square root,

nor how an x, joins $$\sqrt{hx}$$

 

[Mark44]

I was shown that, the probability density

$$p(x)=\frac{1}{2\sqrt{hx}}$$ as long as (0 <= x <= h)

and this means that the probability that the rock is found outside of that range is 0, IE it won't be found underground, or above the point h, where it was dropped

I was told if I wanted to check it, then I integrate

$$\int_{0}^{h} \frac{1}{2\sqrt{hx}} dx = something = 1$$ (also how do i integrate this?)

Mathematically this makes sense, but I don't understand how it ties into the physical problem. It's easy enough to integrate, and comes out to 1.

$$\int_{0}^{h} \frac{1}{2\sqrt{hx}} dx = \frac{1}{2\sqrt{h}} \int_0^h \frac{dx}{\sqrt{x}} = \frac{1}{2\sqrt{h}} \int_0^h x^{-1/2} dx$$

You should be able to finish this integration.

Again, I don't understand the probability slant. I've told you how I would approach this problem, so if you want to pursue another avenue, have at it.

I'm just not sure how, $$\frac{dx}{gt}\sqrt{\frac{g}{2h}}dx becomes \frac{1}{2\sqrt{hx}}dx$$