Wikipedia:Reference desk/Archives/Mathematics/2014 March 6

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March 6

derivative of (y=x^(2x+1))

If I want to find d/dx of y=x^(2x+1), the normal way is taking ln of both sides:

ln(y) = ln(x^(2x+1))

ln(y) = (2x+1)(ln(x))

d/dx (ln(y)) = d/dx ((2x+1)(ln(x)))

(1/y)(dy/dx) = (2x+1)(1/x) + (ln(x))(2)

dy/dx = y((2xln(x) + 2x + 1) / x)

dy/dx = (x^(2x+1))((2xln(x) + 2x + 1) / x)

But I can also write the original equation as log base x of y = 2x+1

d/dx (log base x of y) = d/dx (2x+1)

( - (lny) / (x ((lnx)^2 )) (dy/dx) = 2

dy/dx = 2x((lnx)^2) / (-(ln(x^(2x+1))))

The two answers are different. What gives? 75.80.145.53 (talk) 07:38, 6 March 2014 (UTC)

Your derivative of $\log _{x}y$ is wrong. It should be $[(dy/dx)(1/y)(\ln x)-(\ln y)/x]/(\ln ^{2}x)$.--149.148.255.161 (talk) 09:13, 6 March 2014 (UTC)

quintic polynomial

I know that the general quintic is insoluble in radicals. But I have a specific quintic polynomial, with integer coefficients which neither I nor mathematica can solve except numerically. Is it possible to prove that a specific integer coefficient quintic polynomial has no solutions in radicals? If so, how to go about it? thanks, 164.11.203.58 (talk) 08:50, 6 March 2014 (UTC)

See Quintic_function#Solvable_quintics, which gives (rather messy) criteria for a quintic to be solvable by radicals. AndrewWTaylor (talk) 09:18, 6 March 2014 (UTC)

(OP) perfect, thanks. It never occurred to me to look at knowledge specific to the quintic.

Resolved

164.11.203.58 (talk) 09:37, 6 March 2014 (UTC)

Tryes slipping upon deceleration equation

For those that did A-level maths, this is Mechanics 1, but what I have got just doesn't feel right.

(1) Force = mass * acceleration i.e. F=ma

(2) Force = coefficient of friction * mass * acceleration due to gravity i.e. F=µmg ∴ a=µg

In other words, the maximum acceleration (or deceleration) of a car before the tyres slip is only ten times (approx.) the coefficient of friction. That's what the maths seems to say, but it just doesn't feel right; that feels to low. I suspect I might be misusing equation (2). Could you please confirm if that makes sense or not. If not, could you please explain what does make sense. Thank you! 143.210.123.73 (talk) 17:01, 6 March 2014 (UTC)

Note that the coefficient of friction can be more than one. This can happen if the two surfaces aren't completely flat, so a piece of tire tread extends down into dip and pushes against a ridge. Also, if the tire partially melts you can get molten rubber acting as a adhesive. StuRat (talk) 17:20, 6 March 2014 (UTC)

Yes, I was guesstimating a dry friction coefficient of 1.5 for my Hoosier 19.5X6.5-10 R25B tyres. But still a deceleratn of 15ms^-2 before the car skids seems low, so I think maybe I can't use that second equation in that way, but I'm not sure. For a wet friction coefficient (really worst case scenario), I'm guessing a friction coefficient of 0.3. This is for Formula Student by the way 143.210.123.73 (talk) 17:53, 6 March 2014 (UTC)

For a race car there is quite a bit of down force due to aerodynamics. This can amount to a force equal to the weight of the car. It's this enhanced force on the tires that you must multiply by the coef. of friction to get the maximum horizontal force before skidding. Combine this with the fact that racing tires have increased coef. of friction compared to street tires (which last much longer) and it may account for higher acceleration before skidding that you're thinking of. Of course all I know about racing comes from Gran Turismo so take my explanation with a grain of salt. --RDBury (talk) 19:11, 6 March 2014 (UTC)

The car my team are building doesn't have any aerodynamics (it's hard to do) so it's just the weight of the car (320kg). But, how does that fit in mathematically, as the mass cancels out when equating equation (1) and (2). (I know my IP address will change now, but I'm still the OP) 81.101.120.9 (talk) 21:34, 6 March 2014 (UTC)

The calculation feels about right to me. On normal roads, I would expect my tyres to skid with a horizontal acceleration or deceleration approaching one "g", and I notice that my seat belts lock at less than a quarter of this. On wet leaves, I've noticed that tyres skid at about one tenth "g", corresponding to a coefficient of friction of about 0.1

I haven't tested out the deceleration required to trigger air bags. Does anyone have any figures? I would expect them to operate at about 2g because this cannot be achieved by normal braking. Dbfirs 09:00, 7 March 2014 (UTC)

I think you're underestimating how much acceleration you're getting. 1 ms-2 = 2.237 mph/s, so at 15 ms-2, you're doing 0 to 60 (or vice versa) in a little under 2 seconds. That seems like a decent upper bound for acceleration/deceleration to me, at least for any car with no downforce effects. (0-60 time = 60/(2.237*acceleration in ms-2) MChesterMC (talk) 09:12, 7 March 2014 (UTC)

Yes, I've seen drivers trying to achieve that on British road surfaces, and their tyres skid and smoke impressively, but they don't achieve that acceleration because many British roads have a coefficient of friction of around 0.5 see Road slipperiness#Reduction Dbfirs 10:21, 7 March 2014 (UTC)

Note that the coefficient of friction isn't solely a property of the road, but of the road/tire combo, as well as anything between them, like grains of sand and moisture, with even air temperature playing a role. There's also a static and dynamic coefficient of friction and the tires may exhibit a different coefficient when moving forward or sideways, too. StuRat (talk) 16:37, 7 March 2014 (UTC)

True, but I think you'd need special tyres to get more than 0.5 on many British roads. Dbfirs 16:57, 7 March 2014 (UTC)

Thanks for the info! Maybe I'm overestimating my friction coefficient then - I'll reduce it :) 143.210.123.106 (talk) 21:05, 7 March 2014 (UTC)