Physics, 01.10.2019 02:00 amanquen35

The curvature of the helix r(t)equals(a cosine t )iplus(a sine t )jplusbt k (a, bgreater than or equals0) is kappaequalsstartfraction a over a squared plus b squared endfraction . what is the largest value kappa can have for a given value of b?

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Answer from: masonorourke

$\kappa = \frac{1}{2 b}$

Explanation:

The equation for kappa ( κ) is

$\kappa = \frac{a}{a^2 + b^2}$

we can find the maximum of kappa for a given value of b using derivation.

As b is fixed, we can use kappa as a function of a

$\kappa (a) = \frac{a}{a^2 + b^2}$

Now, the conditions to find a maximum at a_0 are:

$\frac{d \kappa(a)}{da} \left | _{a=a_0} = 0$

$\frac{d^2\kappa(a)}{da^2} \left | _{a=a_0} < 0$

Taking the first derivative:

$\frac{d}{da} \kappa = \frac{d}{da} (\frac{a}{a^2 + b^2})$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} \frac{d}{da}(a)+ a * \frac{d}{da} (\frac{1}{a^2 + b^2} )$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 + a * (-1) (\frac{1}{(a^2 + b^2)^2} ) \frac{d}{da} (a^2+b^2)$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - a (\frac{1}{(a^2 + b^2)^2} ) (2* a)$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - 2 a^2 (\frac{1}{(a^2 + b^2)^2} )$

$\frac{d}{da} \kappa = \frac{a^2+b^2}{(a^2 + b^2)^2} - 2 a^2 (\frac{1}{(a^2 + b^2)^2} )$

$\frac{d}{da} \kappa = \frac{1}{(a^2 + b^2)^2} (a^2+b^2 - 2 a^2)$

$\frac{d}{da} \kappa = \frac{b^2 - a^2}{(a^2 + b^2)^2}$

This clearly will be zero when

a^2 = b^2

as both are greater (or equal) than zero, this implies

a=b

The second derivative is

$\frac{d^2}{da^2} \kappa = \frac{d}{da} (\frac{b^2 - a^2}{(a^2 + b^2)^2} )$

$\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} \frac{d}{da} ( b^2 - a^2 ) + (b^2 - a^2) \frac{d}{da} ( \frac{1}{(a^2 + b^2)^2} )$

$\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} ( -2 a ) + (b^2 - a^2) (-2) ( \frac{1}{(a^2 + b^2)^3} ) (2a)$

$\frac{d^2}{da^2} \kappa = \frac{-2 a}{(a^2 + b^2)^2} + (b^2 - a^2) (-2) ( \frac{1}{(a^2 + b^2)^3} ) (2a)$

We dcan skip solving the equation noting that, if a=b, then

b^2 - a^2 = 0

at this point, this give us only the first term

$\frac{d^2}{da^2} \kappa = \frac{- 2 a}{(a^2 + a^2)^2}$

if a is greater than zero, this means that the second derivative is negative, and the point is a minimum

the value of kappa is

$\kappa = \frac{b}{b^2 + b^2}$

$\kappa = \frac{b}{2* b^2}$

$\kappa = \frac{1}{2 b}$

Answer from: Quest

Energy transformations are processes that convert energy from one type (e.g., kinetic, gravitational potential, chemical energy) into another. any type of energy use must involve some sort of energy transformation.energy must obey the laws of thermodynamics.thermodynamics is the study of how energy changes from one type to another. the laws of thermodynamics apply to energy and energy conversions.the first law of thermodynamics: main articleenergy cannot be created or destroyed (which is called the conservation of energy); however, it can be transformed from one type into another. in fact, every useful process transforms energy from one form to another. there are many different forms or types of energy. some examples of everyday energy transformations are: scenario energy conversions involvedrubbing hands together to make them warm kinetic energy to thermal energyusing a battery-powered flashlight chemical energy to electrical energy (in the battery)electrical energy to radiant energy (in the bulb)an object speeding up as it falls gravitational potential energy to kinetic energythe second law of thermodynamics: main articlesome forms of energy are more useful than others. using energy always makes it less useful, even though no energy is actually destroyed.kinetic energy and electricity are the most useful forms. these are "high-quality" because they can be transformed almost completely into any other type of energy. electricity, for example, can be easily used to generate heat (thermal energy) or light (radiant energy), break chemical bonds (chemical energy), move objects (kinetic energy), or lift objects (gravitational potential energy).the least useful form of energy is low-temperature thermal energy. it can still be converted back to a higher-quality form, but useful energy is always lost in this process. converting energy to a less-useful form and then trying to work backwards never gets 100% of the useful energy back.for example, when a car runs, the engine will become hot (thermal energy). the warmth of the engine does nothing to the car move or go faster. this wasted energy is an unavoidable by-product of converting the car's fuel (chemical energy) into movement ("high-quality" kinetic energy), however it can potentially be used for heating the cabin of the car to slightly increase overall energy efficiency. it is a challenge for all forms of power generation to minimize wasted energy and be as efficient as possible.

Answer from: Quest

answer; (d) spontaneous processes are characterized by the overall conversion of order to disorder;

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