Derivative max/ min problem.

[Bmanmcfly]

Im having some trouble with another situation, I swear every time I think I get the hang of somethig I get thrown another curve ball...

So, im looking for the min value of D. Where D=av^2+ b/v^2 and a and b are both positive constants.

Thats what's tripping me up, is that a and b being constants, but not defined.

So, I got so far dD/dv=a(2v)+v^2(1)(dD/dv)+((v^2dD/dv)-2bv)/(v^4)

This is where my brain shuts down and can't figure out what to do... I know that I need to evaluate dD/dv =0 and then check the second derivative to verify that It's positive to show that this is a minimum point. I don't know if I should just assume a value for a and b, or if they can be eliminated from the equation....
am I at least on the right track?

 

[tkhunny]

Im having some trouble with another situation, I swear every time I think I get the hang of somethig I get thrown another curve ball...

I think this is called "learning".

If D = f(v), and you draw stuff on a v-D set of coordinate axes, you should first notice the symmetry about the D-axis.

$\displaystyle \frac{d}{dv}\left(a\cdot v^{2} + \frac{b}{v^{2}}\right) = 2av - \frac{2b}{v^{3}}$

Where did you go after that?

 

[Bmanmcfly]

I think this is called "learning".

If D = f(v), and you draw stuff on a v-D set of coordinate axes, you should first notice the symmetry about the D-axis.

$\displaystyle \frac{d}{dv}\left(a\cdot v^{2} + \frac{b}{v^{2}}\right) = 2av - \frac{2b}{v^{3}}$

Where did you go after that?

(please send me to a FAQ on the tex code please)

Ok, that was the first problem, I did the (vu'-uv')/(v^2)... Why can we get away with this easier form?

The next step is to find the 0, which, I seem to have as v=(b/a)^-4 or v=(a/b)^4

If this is correct, then the result is assumed to be the positive.

Then I check the result with the second derivative of the function. (thanks), so D''= 2a+(6b)/v^4

I hate ambiguous answers like this, but it seems that when v=(a/b)^4 that D is at a minimum?

Hope that at least approaches something like a correct answer, it just feels wrong because the answer isn't really conclusive IMO.

 

[tkhunny]

Why are you using the Division Rule?

$\displaystyle \frac{d}{dx}\frac{1}{x^{2}} = \frac{d}{dx}x^{-2} = -2\cdot x^{-3} = \frac{-2}{x^{3}}$

You must get the correct expression for the derivative before you complain that the result is ambiguous.

 

[mmm4444bot]

please send me to a FAQ on the tex code please

Start HERE. :cool:

Oh, yeah, you may also right-click on any LaTex expression in a post, to follow sub-menus for displaying the actual code. That's how I learned a lot.

 

[Bmanmcfly]

Why are you using the Division Rule?

$\displaystyle \frac{d}{dx}\frac{1}{x^{2}} = \frac{d}{dx}x^{-2} = -2\cdot x^{-3} = \frac{-2}{x^{3}}$

You must get the correct expression for the derivative before you complain that the result is ambiguous.

I guess I was treating a and b as though they were variables, not constants.

 

[tkhunny]

That will do it. Here's a plan...don't do that.

Have you managed a better result?

 

[Bmanmcfly]

That will do it. Here's a plan...don't do that.

Have you managed a better result?

$\displaystyle \frac{dD}{dv}\left(a\cdot v^{2} + \frac{b}{v^{2}}\right) = 2av - \frac{2b}{v^{3}}$

As you had shown, which showed me the first error of my ways.

Next, because I'm looking for a Max / Min point I need
$\displaystyle \frac{dD}{dv} = 0$
$\displaystyle 0=2av-\frac{2b}{v^{3}}$

Then I go,
$\displaystyle \frac{2b}{v^{3}}=2av$
$\displaystyle \frac{2b}{2a}=v^{4}$

Which means $\displaystyle v=\left(\frac{2b}{2a}\right)^{-4}$
or simplified to

$\displaystyle v=\left(\frac{a}{b}\right)^{4}$

I don't know if I'm handling this wrong, or where I'm going wrong... and even though I don't like the answer, this seems to be the best I can get...

but then checking the $\displaystyle \frac{d^{2}D}{dv^{2}}= 2a+ \frac{6b}{v^{4}}$
It can be shown that any positive value of v will have a positive value as a result meaning that at any point where the derivative =0 will be a minimum for the function.

 

[JeffM]

(please send me to a FAQ on the tex code please)

Ok, that was the first problem, I did the (vu'-uv')/(v^2)... Why can we get away with this easier form?

The next step is to find the 0, which, I seem to have as v=(b/a)^-4 or v=(a/b)^4 This is NOT correct.

$\displaystyle D = av^2 + \dfrac{b}{v^2} = av^2 + bv^{-2} \implies 0 = \dfrac{dD}{dv} = 2av - 2bv^{-3}\implies$

$\displaystyle 2av = 2bv^{-3} \implies 2av = \dfrac{2b}{v^3} \implies v^4 = \dfrac{2b}{2a} = \dfrac{b}{a} \implies v = \pm \left(\dfrac{b}{a}\right)^{(1/4)}.$

If this is correct, then the result is assumed to be the positive. It is given that a and b are positive, but what leads you to believe that v must be positive? The derivative is zero at TWO values of v.

Then I check the result with the second derivative of the function. (thanks), so D''= 2a+(6b)/v^4 This is correct, but where do you go from here.

$\displaystyle D' = 2av - 2bv^{-3} \implies D'' = 2a + 6bv^{-4} = 2a + \dfrac{6b}{v^4}> 0\ if\ v \ne 0.$

So, how many local minima are there? Where are they? What are the values of D at those values of v, which is the question asked?

I hate ambiguous answers like this, but it seems that when v=(a/b)^4 that D is at a minimum?

Hope that at least approaches something like a correct answer, it just feels wrong because the answer isn't really conclusive IMO.

I doubt you would have had any difficulty with $\displaystyle D = 2v^2 + \dfrac{4}{v^2}.$ The logic is identical.

 

[JeffM]

$\displaystyle \frac{dD}{dv}\left(a\cdot v^{2} + \frac{b}{v^{2}}\right) = 2av - \frac{2b}{v^{3}}$

As you had shown, which showed me the first error of my ways.

Next, because I'm looking for a Max / Min point I need
$\displaystyle \frac{dD}{dv} = 0$
$\displaystyle 0=2av-\frac{2b}{v^{3}}$

Then I go,
$\displaystyle \frac{2b}{v^{3}}=2av$
$\displaystyle \frac{2b}{2a}=v^{4}$

Which means $\displaystyle v=\left(\frac{2b}{2a}\right)^{-4}$ You are making an error here.
or simplified to

$\displaystyle v=\left(\frac{a}{b}\right)^{4}$

I don't know if I'm handling this wrong, or where I'm going wrong... and even though I don't like the answer, this seems to be the best I can get...

It is not an ambiguous answer. It is a general answer in terms of a and b. Given any a and b, you could plug them in to get numeric answers.

but then checking the $\displaystyle \frac{d^{2}D}{dv^{2}}= 2a+ \frac{6b}{v^{4}}$
It can be shown that any positive value of v will have a positive value as a result meaning that at any point where the derivative =0 will be a minimum for the function.

We cross-posted.

 

[tkhunny]

Not bad, but you forgot one thing I told you. You would have found it on your own if you had FACTORED rather than what you have shown.

In other words, you've only half the answer. JeffM gave it away.

BTW - Good work getting up to speed on the LaTeX!

 

[Bmanmcfly]

I doubt you would have had any difficulty with $\displaystyle D = 2v^2 + \dfrac{4}{v^2}.$ The logic is identical.

No, those ones don't cause me grief... and if it was an actual value, like Pi where I could substitute I wouldn't find this so confusing...

So, the potential d minimums are going to be :

1 - d= av^2+b/v^2... replacing V

$\displaystyle d=a((b/a)^{1/4})^2+ b/((b/a)^{1/4})^{2}$
$\displaystyle d= a\sqrt{\frac{b}{a}}+ \frac{b}{\sqrt{\frac{b}{a}}}$
$\displaystyle d= \frac{a\sqrt{b}}{\sqrt{a}} + \frac{b\sqrt{a}}{\sqrt{b}}$
$\displaystyle d= \frac{ab+ba}{\sqrt{ab}}$

This one feels useless since I'm trying to find values FOR V, not eliminating v from the equation.

2 - Replacing A gives me
$\displaystyle d=\frac{b}{v^{4}}v^{2}+\frac{b}{v^{2}}= \frac{b}{v^{2}}+\frac{b}{v^{2}} = \frac{2b}{v^{2}}$

3 - Replacing B gives
$\displaystyle d= av^{2}+ \frac{av^{4}}{v^{2}} = 2av^{2}$

Those three points are the equivalencies for d where the rate of change of d =0, and since the graph is curving UP, this means at these values the function would be at a minimum.

Only positive values are considered, because the drag would be a different equation forward vs backwards.

Does this seem at least on the right track??

 

[JeffM]

No, those ones don't cause me grief... and if it was an actual value, like Pi where I could substitute I wouldn't find this so confusing...

So, the potential d minimums are going to be :

1 - d= av^2+b/v^2... replacing V This confuses me. We started with $\displaystyle D = av^2 + \dfrac{b}{v^2}$ Where did (1 - d) come from?

But yes, you found by looking at the first and second derivatives of D which values of v minimize D. So you substitute those values v back into the expression for D. Technically, you should substitute both values of v although, in this case, it is clear by inspection that D will have the same positive value whichever value of v you use because they have a plus/minus relationship and D is a function of v squared.

$\displaystyle d=a((b/a)^{1/4})^2+ b/((b/a)^{1/4})^{2}$
$\displaystyle d= a\sqrt{\frac{b}{a}}+ \frac{b}{\sqrt{\frac{b}{a}}}$
$\displaystyle d= \frac{a\sqrt{b}}{\sqrt{a}} + \frac{b\sqrt{a}}{\sqrt{b}}$
$\displaystyle d= \frac{ab+ba}{\sqrt{ab}}$

This one feels useless since I'm trying to find values FOR V, not eliminating v from the equation.

If you look at your very first post, you will see that you initially said that you were to find the minimum value of D. You already found what values of v generate that minimum. So you are done with v.

2 - Replacing A gives me
$\displaystyle d=\frac{b}{v^{4}}v^{2}+\frac{b}{v^{2}}= \frac{b}{v^{2}}+\frac{b}{v^{2}} = \frac{2b}{v^{2}}$

3 - Replacing B gives
$\displaystyle d= av^{2}+ \frac{av^{4}}{v^{2}} = 2av^{2}$

Those three points are the equivalencies for d where the rate of change of d =0, I have no idea what you are doing here. The problem asked you to find the minimum value of D in terms of a and b, which you already did. and since the graph is curving UP, this means at these values the function would be at a minimum.

Only positive values are considered, because the drag would be a different equation forward vs backwards. You had not, I think, previously told us that this problem was a physical problem. I am now guessing that v is a velocity. Actually, we now see that drag would have the same minimum value whether the object is moving forward at a velocity of (b / a) ^ (1 / 4) or
backward at a velocity of - (b / a)^ (1 / 4).

Does this seem at least on the right track??

.
I think that, until you get used to this sort of problem, you should ask yourself, How would I proceed if knew specifically what the constants were. Then just do exactly the same thing with the pronumerals that stand for the constants. It is the old difference between arithmetic and algebra in a new guise. You come up with a generalized answer that will work for any set of constants.

 

[Bmanmcfly]

.
1 - d= av^2+b/v^2... replacing V This confuses me. We started with [FONT=MathJax_Math]D[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]v[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Math]v[/FONT][FONT=MathJax_Main]2[/FONT] Where did (1 - d) come from?

That should have been 1. instead... my bad. there is no 1-d.

But yes, you found by looking at the first and second derivatives of D which values of v minimize D. So you substitute those values v back into the expression for D. Technically, you should substitute both values of v although, in this case, it is clear by inspection that D will have the same positive value whichever value of v you use because they have a plus/minus relationship and D is a function of v squared.

[FONT=MathJax_Math]d[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]2[/FONT]
[FONT=MathJax_Math]d[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Size2]√[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Size2]√[/FONT]
[FONT=MathJax_Math]d[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main]√[/FONT]
[FONT=MathJax_Math]d[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main]√[/FONT]

This one feels useless since I'm trying to find values FOR V, not eliminating v from the equation.

If you look at your very first post, you will see that you initially said that you were to find the minimum value of D. You already found what values of v generate that minimum. So you are done with v.

Ok... Thanks alot for the help, this gave me the answer I was looking for, I hope at least...