Help with a trig differentiation problem again

[Asphyxiated]

Homework Statement

Ok so I found out last time that the derivative of the function they asked me to find was:


[tex] \frac {dF}{d\theta} = \frac { - \mu^{2}Wcos(\theta)+\mu W sin(\theta)} { (\mu sin(\theta) + cos (\theta))^{2}} [/tex]

and now they are asking me when is this 0, specifically they want the general formula for theta to be zero and i really don't have a clue where to start. I have looked at a bunch of trig identities to see if anything would help but I can not see anything, sure there are a couple of swaps I could make but it would not quickly make things much easier in the problem. I really don't think it is suppose to be that hard. Perhaps I am over thinking it somehow. I figure that you would just set the equation equal to zero and solve for theta but I really don't know where to start doing that... point me in the right direction!

thanks!

 

[tiny-tim]

Hi Asphyxiated!

… when is this 0, specifically they want the general formula for theta to be zero …

You mean, when is dF/dθ = 0 ?

That'll be when the top line is zero …

so just rewrite Acosθ + Bsinθ in the form Csin(θ - φ). ​

 

[Asphyxiated]

ok so then if:

[tex] asin(\theta)+bcos(\theta) = \sqrt {a^{2}+b^{2}} * sin(\theta + \phi) \;\; where \;\; \phi = arctan(\frac {b}{a}) [/tex]

then:

[tex]
- \mu^{2}Wcos(\theta)+\mu W sin(\theta) = \mu W sin(\theta) + (-\mu^{2}W)cos(\theta)
[/tex]

and

[tex] \mu W sin(\theta) + (-\mu^{2}W)cos(\theta) = \sqrt {(\mu W)^{2}+(-\mu^{2}W)^{2}} * sin(\theta + \phi) \;\;where\;\; \phi = arctan(\frac{\mu^{2}W}{\mu W}) [/tex]

which could be said as:

[tex] \mu W sin(\theta) + (-\mu^{2}W)cos(\theta) = \sqrt {(\mu W)^{2}+(-\mu^{2}W)^{2}} * sin(\theta + arctan(\frac{\mu^{2}W}{\mu W})) [/tex]

but i still don't know how to start solving for theta... I could use an angle addition identity such as:

[tex] sin(\theta + \phi) = sin(\theta)cos(\phi)+sin(\phi)cos(\theta) [/tex]

but i really don't see how that is going to help me at all.. The problem is that if you set that new equation to zero then you are still going to have to multiply or divide before you can get anything on the other side of the equation. Thanks for any additional help!

 

[tiny-tim]

… but i really don't see how that is going to help me at all.. The problem is that if you set that new equation to zero then you are still going to have to multiply or divide before you can get anything on the other side of the equation. Thanks for any additional help!

I'm not following you …

if sin(θ - φ) = 0, then θ = φ or = π + φ

 

[Asphyxiated]

Well ok, yes, fair enough, i can tell when it is going to be zero because the sin is multiplied with the radical but that's not what its asking for. The program that we use is look for:

[tex] \theta = arcsin(\sqrt{a^{2}+b^{2}} [/tex]

you know, they want the full general formula, so you are suppose to solve for theta using dF/dθ, you see?

 

[tiny-tim]

(just got up :zzz: …)

… and now they are asking me when is this 0

Well ok, yes, fair enough, i can tell when it is going to be zero because the sin is multiplied with the radical but that's not what its asking for. The program that we use is look for:

[tex] \theta = arcsin(\sqrt{a^{2}+b^{2}} [/tex]

you know, they want the full general formula, so you are suppose to solve for theta using dF/dθ, you see?

sorry, I'm not following you at all …

you said originally that the question asks you when dF/dθ is 0, isn't that all you need to know?

and what is arcsin√(a² + b²) supposed to be the formula for?

 

[Asphyxiated]

No no, I have tried entering simple answers such as

[tex] \theta = \phi [/tex]

or

[tex] \theta = \pi + \phi [/tex]

that is not what they are looking for. I just wrote out that equation to show you what they were actually looking for. That of course is not the answer but it would be a general formula and that's what they are asking for here.

You know, since it is broken down to just 1 sin term now, but with two varibles, you should be able to do some algebraic manipulation to just use:

[tex]
\sqrt {(\mu W)^{2}+(-\mu^{2}W)^{2}} * sin(\theta + arctan(\frac{\mu^{2}W}{\mu W})) = 0
[/tex]

and then solve for theta, you see?

 

[Asphyxiated]

Perhaps this would make it clearer, let's just say its this:

[tex]

\sqrt {(\mu W)^{2}+(-\mu^{2}W)^{2}} * sin(\theta + arctan(\frac{\mu^{2}W}{\mu W})) = 1

[/tex]

ok? so then what we could do is this:

[tex]

sin(\theta + arctan(\frac{\mu^{2}W}{\mu W})) = \frac {1} {\sqrt {(\mu W)^{2}+(-\mu^{2}W)^{2}}}

[/tex]


[tex]

\theta = arcsin(\frac {1} {\sqrt {(\mu W)^{2}+(-\mu^{2}W)^{2}}})-arctan(\frac{\mu^{2}W}{\mu W})

[/tex]

they want the FULL general formula for theta, that is theta = 1.

 

[tiny-tim]

(just got up :zzz: …)

I'm really confused … where did the "= 1" in your first line come from?

(and why do you keep writing µ²W/µW instead of just µ ? )

 

[Asphyxiated]

Hey, so I asked my Calc teacher today to explain what they wanted here and it is much simpler than what I was doing before. Since mu and W are constants that are multiplied by the µcos(θ)-sin(θ) it should be setup like this:

[tex] \mu cos(\theta)-sin(\theta) = 0 [/tex]

[tex] \mu cos(\theta) = sin(\theta) [/tex]

[tex] \mu = \frac {sin(\theta)}{cos(\theta)} [/tex]

which is:

[tex] \mu = tan(\theta) [/tex]

and since we are solving for theta:

[tex] \theta = tan^{-1}(\mu) [/tex]

thats what they were looking for, do you get what I was trying to show you? They did not want numerical answers or simple statements, you had to use the actual formula to solve for theta...

Thanks for all the help though anyway!

 

[tiny-tim]

Yes, you had to get the top line to be zero …

µWsinθ - µ²Wcosθ = 0,

so tanθ = µ.