Rearranging Equations - Solving for v

  • Thread starter eric99gt
  • Start date
In summary, the person is trying to solve for v from an equation, and is having trouble because they are not seeing how to do partial derivatives.
  • #1
eric99gt
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0
Alright I'm going absolutely nuts here. It's a simple matter of rearranging an equation and for the life of me I can't do it. I've been working at it for like an hour now to no avail. Here's the equation.

P=[RT/(v-b)]-[a/v(v+b)T^0.5]

I'm trying to solve for v. I've manipulated this equation every way I can think of and still can't get it. I'm sure it's something simple but I'm stuck. Please help.
Thanks
 
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  • #2
you'll get a quadratic in V. You are aware of this, right?
 
  • #3
Nope. Completely over my head. How in the world would you get that in the form of ax^2+bx+c=0?
 
  • #4
Well here's the real problem. I'm trying to get the partial of (dv/dT) with constant pressure from that equation. What I was trying to do was to find v from the equation and take the partial of that side with respect to v and the partial of the other side with respect to T. Is this what I should be doing?
 
  • #5
Actually, you don't get a quadratic, you get a cubic! The least common denominator of the fractions, if I am reading this properly,
[tex]P= \frac{RT}{v-b}-\frac{a}{v(v+b)T^{0.5}}[/tex]
(Is that what you intended? Your fractions are ambiguous.)
is v(v-b)(v+b)T0.5, a cubic in v.
Cubics in general are hard to solve. Since you say you really want the derivative of y wrt T, P being constant, I would recommend "implicit differentiation". Writing the equation as
[tex]P= RT(v-b)^{-1}- av^{-1}(v-b)^{-1}T^{-1/2}[/tex]
differentiate both sides: using product and chain rules, wrt T:
[tex]0= R(v-b)^{-1}- RT(v-b)^{-2}v'+av^{-2}(v-b)^{-1}T^{-1/2}v'+ av^{-1}{v-b}^{-2}T^{-1/2}v'+ (a/2)v^{-1}(v-b)^{-1}T^{-3/2}[/tex]
and solve for v', the derivative of v with respect to T.
 
  • #6
I think my biggest problem is I'm not really seeing how to solve partials. Anyone have any good literature about the subject. Thanks.
 

1. How do you rearrange an equation to solve for v?

To rearrange an equation and solve for v, you need to isolate the variable v on one side of the equation. This can be done by using basic algebraic operations such as addition, subtraction, multiplication, and division to move the other terms to the opposite side of the equation.

2. Can you provide an example of rearranging an equation to solve for v?

Sure, let's take the equation d = vt, where d is distance, v is velocity, and t is time. To solve for v, we need to isolate it on one side of the equation. We can do this by dividing both sides by t, giving us v = d/t. Now we have rearranged the equation and solved for v.

3. What do you do if there are multiple v's in an equation?

If there are multiple v's in an equation, you can still solve for v by using the same method of isolating the variable on one side. However, you may need to use more advanced algebraic techniques such as factoring or combining like terms to simplify the equation and isolate the v variable.

4. Are there any rules or guidelines for rearranging equations to solve for v?

Yes, there are a few rules and guidelines to keep in mind when rearranging equations to solve for v. These include maintaining the equality of the equation by performing the same operation on both sides, keeping track of signs when moving terms to the other side, and simplifying the equation as much as possible before solving for v.

5. What are some common mistakes to avoid when rearranging equations to solve for v?

Some common mistakes to avoid when rearranging equations to solve for v include forgetting to perform the same operation on both sides of the equation, incorrectly distributing a negative sign when moving terms to the other side, and making arithmetic errors when simplifying the equation. It's always important to double-check your work and make sure you have followed the proper steps to solve for v.

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