I will like it if someone helps me out and also checks my work for me for this equation:
$f(x,y)=\arcsin(6y-5x)$
Domain
$-1\leq6y-5x\leq1$
$y\geq\frac{5x}{6}-\frac{1}{6}$ and $y\leq\frac{5x}{6}+\frac{1}{6}$
I think the shape of the domain is a region bounded by two lines.
Range
I have no clue how to find the range. I know the range of arcsin is [$\frac{−π} {2}$ , $\frac{π}{2}$].
Level Curves
Not very good at this either haha. This is what I did:
$\arcsin(6y-5k)=k$
$6y-5x=\sin k$
I did some test points:
$k=0$: $y=\frac{5}{6}x + 0$
$k=1$: $y=\frac{5}{6}x+\frac{\sin1}{6}$
$k=2:$ $y=\frac{5}{6}x+\frac{\sin2}{6}$
I think that these are unequal parallel lines because $\frac{\sin1}{6}-0 = 0.14$ and $\frac{\sin2}{6}-\frac{\sin1}{6} = 0.011$.
Boundary of Domain
I think the boundary is two lines
Open set, closed set, neither, or both for Domain
I think the domain is a closed set.
Is domain bounded or unbounded
I think the domain is bounded.
I will really appreciate if you helped me on this problem with the range and check my answers and tell me what I did wrong. I feel like I am not very good at these (I take so long to do them).
Thank you, and you might see similar questions like these in the future.
Best Answer
The domain of $f$ consists of lines of the form $y_k = \frac{5x}{6} + k$, where $k \in [-\frac{1}{6},\frac{1}{6}]$, so since $f(x,\frac{5x}{6}+k) = \sin^{-1}(6(\frac{5x}{6}+k) - 5x)) = \sin^{-1}(5x + 6k - 5x) = \sin^{-1}(6k)$. As $k$ ranges from $\frac{-1}{6}$ to $\frac{1}{6}$, the value $6k$ ranges from $-1$ to $1$. You should verify this. So the range of $f$ is $[-\frac{\pi}{2},\frac{\pi}{2}]$.