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?

Explanation:

The equation for kappa ( κ) is

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

Now, the conditions to find a maximum at are:

Taking the first derivative:

This clearly will be zero when

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

The second derivative is

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

at this point, this give us only the first term

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

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.