# Need calculus help

1. Sep 24, 2006

### Kristal

Can anyone help me with this problem,please?

xo=latitude of arctic circle

S(x)={a+barcsin(tanx/tanxo) when 0<x<xo
24 when xo[less-than equal to]x[less-than equal to]90

Find a and b so that S(x) is continuous.

My work:
xo= 66(degrees)30' which I think equals 66.30 (correct?)
arcsin=sin(to the negative 1) (I think?)

And from there I really don't know how to do it or start to do it.

Last edited: Sep 24, 2006
2. Sep 25, 2006

### HallsofIvy

Staff Emeritus
If you are asking if 66 degrees 30 minutes is 66.30 degrees, the answer is no. Since there are 60 minutes in a degree, 30 minutes is 1/2 degree. 66 degrees, 30 minutes is 66.5 degrees.

Yes, "arcsine" is the inverse sine function- often written sin-1 but there are some of use who prefer the "arcsin" notation since the other is easy to confuse with -1 power.

You should know that the arcsine function is differentiable in the domain give here and, of course, the constant function, y= 24, is differentiable so the only question is differentiability at x= x0. What is the value of a+barcsin(tanx/tanxo) at x= x0? What is the value of 24 at x= x0 (yes, that's a trivial question!). What is the derivative of a+b arcsin(tanx/tanxo)? What is the derivative of 24? What do a and b have to be so that both the values and the derivatives are the same at x= x0?

3. Sep 25, 2006

### Kristal

Thank you so much... I still do not understand how to get the values of a and b... but if need be I can guess and check values (I'm quite good at it by now.) You helped alot at giving me an idea of what to do and what I should consider.

4. Sep 25, 2006

### HallsofIvy

Staff Emeritus
First, tell us what you have for the derivative of a+barcsin(tanx/tanxo) at x= x0. (Use the chain rule differentiating.) What is the value of a+barcsin(tanx/tanxo) itself when x= x0(that's easy)?

5. Sep 25, 2006

### Kristal

Thank you. I think I got it...

12+(48/pi)arcsin(tanx/tanxo)

Last edited: Sep 25, 2006