## Calculus 10th Edition

The solutions are: a) $f(2,\pi/4) =\sqrt{2}.$ b) $f(3,1) = 3\sin1.$ c) $f(-3,\pi/3) =-\frac{3\sqrt{3}}{2}.$ d) $f(4,\pi/2) = 4.$
We will substitute $x$ and $y$ from the first and the second coordinate from the given point into the function $f(x,y) = x\sin y$. a) $f(2,\pi/4) = 2\sin(\pi/4) = 2\times \frac{\sqrt{2}}{2} = \sqrt{2}.$ b) $f(3,1) = 3\sin1.$ c) $f(-3,\pi/3) = -3\sin(\pi/3) = -3\times \frac{\sqrt{3}}{2}=-\frac{3\sqrt{3}}{2}.$ d) $f(4,\pi/2) = 4\sin(\pi/2)=4\times 1 = 4.$