Hi,
I was wondering how one would go about finding the domain of a 3-dimensional surface. For example, if the equation was f(x,y) = (sqrt(x + 2y + 1)) * (sqrt(x - 2y + 1)) * (ln(1-sqrt(tan(x^2 + 2x - 4y^2)))) what would the domain be? Thanks!
Hi,
I was wondering how one would go about finding the domain of a 3-dimensional surface. For example, if the equation was f(x,y) = (sqrt(x + 2y + 1)) * (sqrt(x - 2y + 1)) * (ln(1-sqrt(tan(x^2 + 2x - 4y^2)))) what would the domain be? Thanks!
$\displaystyle f(x,y)=\left[\sqrt{x+2y+1}\right]\left[\sqrt{x-2y+1}\right]\left[\ln(1-\sqrt{tan(x^2+2x-4y^2)}\right]$
find the domain for the first and the second and third each one alone then find the intersect of the result you have
you should know that the square root of the negative numbers is undefined and ln the natural log is the same and for the tan(s) ''s is arbitrary function'' cos(s) should not equal zero
$\displaystyle x+2y+1 \geq 0 $ the domain of the first one for the second
$\displaystyle x-2y+1 \geq 0 $ for the third
$\displaystyle 1-\sqrt{tan(x^2+2x-4y^2)}> 0 $ so
$\displaystyle 1>\sqrt{tan(x^2+2x-4y^2)} $ and
$\displaystyle tan(x^2+2x-4y^2) \geq 0 $ tan is positive in the first and the third quarter so
$\displaystyle 0\leq tan(t)<1 $
if $\displaystyle 0\leq t< \frac{\pi}{4}$
this in the first quarter in the third
$\displaystyle \pi \leq t <\frac{\pi}{4}+\pi $ in general
$\displaystyle n\pi \leq t < \frac{\pi}{4}+n\pi $ so
$\displaystyle n\pi \leq x^2+2x-4y^2 < \frac{\pi}{4} +n\pi$ n integer
lats return to the domain of the first and the second
$\displaystyle (x+2y+1)(x-2y+1)\geq 0$ since the both of them is positive
$\displaystyle x^2-4y^2+2x+1 \geq 0 $
$\displaystyle x^2-4y^2+2x \geq -1 $ forget it
the domain is
$\displaystyle \left(0 \leq x^2+2x-4y^2 < \frac{\pi}{4}\right)\bigcup \left(\pi \leq x^2+2x-4y^2 < \frac{\pi}{4} + \pi\right) \bigcup$$\displaystyle \left( 2\pi \leq x^2+2x-4y^2 < \frac{\pi}{4}+2\pi\right)\bigcup ..... $
that is the way you can find the domain