Solving Dimension Analysis Homework: Q1 & Q2

In summary, the conversation discusses how to calculate the derivative of u with respect to t, given that q2 is equal to u* and q1 is a function of x and t. Different approaches are suggested, including using the chain rule and dimensional analysis. Ultimately, the correct approach involves substituting the expression for q1 in terms of t into the equation for u.
  • #1
dirk_mec1
761
13

Homework Statement



http://img410.imageshack.us/img410/8495/88748860vk8.png

Homework Equations


The Attempt at a Solution


I don't know how to deal with the q1 part. q2 = u* =u x sqrt(... so that is clear.

Normally I just substitute the expression for u in terms of u* (= f* )and t in terms of t* in [tex] \frac{ du }{dt } [/tex] but how should I deal with the differentiation of the q1 -term?
 
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  • #2
You have to use the chain rule for that.
 
  • #3
Tom Mattson said:
You have to use the chain rule for that.

To be honest I don't how to begin. I have this:

[tex] u= \frac{u_0 \sqrt{q}f^*}{x} [/tex]

[tex] t = \frac{x^2}{qk} [/tex]


so [tex] \frac{du}{dt} = ? [/tex]
 
  • #4
OK, follow these steps.

1.) As directed, assume that [itex]q_2=f^*(q_1)[/itex]. So [itex]f^*[/itex] is a function of [itex]q_1=x^2/kt[/itex].

2.) Solve [itex]q_2=u\sqrt{kt}/u_0[/itex] for [itex]u[/itex]. Identify [itex]q_2[/itex] with [itex]f^*[/itex] at this point.

By now you should have the following result.

[tex]u(x,t)=\frac{u_0}{\sqrt{kt}}f^*\left(\frac{x^2}{kt}\right)[/tex]

Now you can take partial derivatives with respect to [itex]x[/itex] and [itex]t[/itex] until your heart's content. And as I said before, you'll need the chain rule to do it.

Try that and let us know if you get stuck again.
 
  • #5
Tom your approach does work I found out after 2 full pages of writing. But what I don't understand is where my approach is going wrong. You see I just used the usual method in dimensional analysis. Writing u in terms of u* and t in terms of t* and calculating the derative.

If you calculate it you'll see that u* = f* but then things get nasty because there is no t*. So what I did was rewrite the expression for q1 in terms of t. Why is that approach wrong?
 
  • #6
The problem there is that q_1 is a function of x and t. If you're going to plug u(x,t) into the heat equation then you have to see all of those hidden x's and t's, because you are taking partial derivatives of u.

dirk_mec1 said:
[tex] u= \frac{u_0 \sqrt{q}f^*}{x} [/tex]

That q under the square root sign should be q_1. Substitute the expression for q_1 in there and you'll get what I got.

[tex]u= \frac{u_0 \sqrt{q_1}f^*\left(q_1\right)}{x}[/tex]

[tex]u= \frac{u_0 \sqrt{\frac{x^2}{kt}}f^*\left(\frac{x^2}{kt}\right)}{x}[/tex]

[tex]u= \frac{u_0}{\sqrt{kt}}f^*\left(\frac{x^2}{kt}\right)}[/tex]
 

Related to Solving Dimension Analysis Homework: Q1 & Q2

What is dimension analysis?

Dimension analysis is a problem-solving technique used in science to convert units of measurement from one system to another. It involves breaking down a given quantity into its constituent units and using conversion factors to manipulate and cancel units until the desired unit is obtained.

Why is dimension analysis useful?

Dimension analysis is useful because it allows scientists to easily convert between different units of measurement, which is necessary for making accurate and meaningful comparisons in experiments and data analysis. It also helps to identify and correct errors in calculations.

How do I approach dimension analysis problems?

The first step in solving dimension analysis problems is to clearly identify the given quantity and the desired unit. Then, list out all the conversion factors needed to move from the given unit to the desired unit. Finally, use the conversion factors to manipulate and cancel units until only the desired unit remains.

What are some common mistakes to avoid when using dimension analysis?

Some common mistakes when using dimension analysis include using incorrect conversion factors, not canceling units properly, and not paying attention to the direction of the conversion. It is also important to be aware of significant figures and to carry them through the calculations.

How can I practice and improve my dimension analysis skills?

The best way to practice and improve dimension analysis skills is to solve a variety of problems using different units and conversion factors. You can find practice problems online or in textbooks, and you can also create your own problems to solve. It is also helpful to review and understand the steps and concepts involved in dimension analysis.

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