is there a function wich images the half of a circle in the time domain?

Question

Is there any function f(t) which images the half of a circle in the time domain more or less like this?:
(I don't mean the sinus function here)

f(t)
l
l......._---_
l..../............\ ------->this should be the half of a circle
l..l............. ..l....(the periods are only to replace the space key)
l---------------------------> t

So what the the function should do, i think, is to evaluate the distance between the diameter line (which would be t-domain line in this case) and the upper arc of the half of the circle.
Is this posible?

Thanks!!

Answer 1

Equation of a circle (x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center of the circle.
Let x correspond to time t and y to f(t). As you have surmised, we can only get a semi-circle because we want f(t) to have a single value. Let the diameter of the circle be a value of time, e.g. T. for values of t greater than T this function is undefined. The center of the circle is then at (T/2, 0) -- 0 because any other value would not produce a semi-circle of only positive values. (If you allow negative values, set the center wherever you want.)
So, manipulating the circle formula a bit, you get y^2 = (T^2)/4 - (t - T/2)^2 where 0 ≤ t ≤ T
expand and combine terms, y^2 = t(T - t)
The positive square root of this function should give the curve you show.
y(t) = |√[t(T-t)] where 0 ≤ t ≤ T

I ran it on Excel and it seems to work.

If you wanted to continue the function in time, define another segment from T ≤ t ≤ 2T with the center at 3T/2. etc, etc, etc.

Answer 2

(sin(x))^x = e^(x * ln(sin(x))) This function will needless to say be defined each time sin(x) is postive, that's whilst 2n? < x < (2n + a million)?, the place n is any integer. Now, all it is left to make sure is the places the place the function would be defined whilst sin(x) is destructive. whilst sin(x) is destructive, enable sin(x) = -a, the place a = |sin(x)|. So, the function will become: e^(x * ln(-a)) = e^(x * (i? + ln(a))) = e^(xi? + xln(a)) = e^(xi?) * e^(xln(a)) This function is defined each time e^(xi?) is a real quantity, which basically happens whilst x is an integer. So, the area of the function would be {(2?n, (2? + a million)n); n ?Z} ?{ok; ok?Z}. wish this helps. Edit : Yeah, its genuine that this fuction would not have a non-quit area the two for advantageous or destructive x. As for the rational powers, i'm having my doubts. Is the cube root of -a million, for occasion, real or no longer? This has 3 values, of which one is real and the different 2 are no longer. nicely, it quite relies upon on the way you define exponents. So, now we could desire to look for a definition for a^b, the place a and b could be any complicated numbers. we are able to define it as 'a prolonged b cases' yet this definition works basically whilst b is a superb integer. we are able to enhance it for destructive integers and ultimately all rational numbers, besides the fact that it stops there. we will not use this definition to evaluate some thing like 2^?. it is the place the exponential function is provided in. If we write 2^? as e^(?ln(2)), we are able to extremely basically evaluate its vaue utilising the potential series enhance for e^x or some thing like that. So we've got definted the exponential function for all complicated numbers a and b. Now, we are able to evaluate (-a million)^(a million/3) utilising this defintion. Write (-a million)^(a million/3) as e^(a million/3 * ln(-a million)) = e^(i?/3) = cos(?/3) + i sin(?/3) = a million/2 + ?3i/2, which isn't real. So, in accordance to this definition for exponents, the huge-unfold cube root of -a million isn't real. So, (sin(x))^x would not have a internet site for all rational x with weird and wonderful denominators and is defined basically for indispensable values of x.

Answer 3

The eq. of circle is (y-a)^2+(t-b)^2 = r^2, where (a,b) is the center of your circle, and r=radius

If you constrain: y >=a, b-r < t< b+r then there is a unique correspondence between t and y (or f(t)). Just solve the quadratic and pick the solution for which y>=a