91Ó°ÊÓ

We are given the expressions for u and v as: \(u = e^x \cos y\) \(v = e^x \sin y\) We need to find the values of u and v at the point (x=0, y=2π/3). For u, we substitute x=0 and y=2π/3: \(u(0, \frac{2π}{3}) = e^0 \cos (\frac{2π}{3}) = 1 \cdot (- \frac{1}{2}) = -\frac{1}{2}\) For v, we also substitute x=0 and y=2π/3: \(v(0, \frac{2π}{3}) = e^0 \sin (\frac{2π}{3}) = 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}\) So, we have u=-1/2 and v=√3/2 at the point (x=0, y=2π/3).

We need to find the partial derivatives of z with respect to x and y using the Chain Rule. Given: \(z = u^2 - v^2\) Using Chain Rule, we can find the partial derivatives as follows: \(\frac{\partial z}{\partial x} = 2u \frac{\partial u}{\partial x} - 2v \frac{\partial v}{\partial x}\) \(\frac{\partial z}{\partial y} = 2u \frac{\partial u}{\partial y} - 2v \frac{\partial v}{\partial y}\) Now, we need to find the expressions for the partial derivatives of u and v with respect to x and y. \(\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(e^x \cos y) = e^x \cos y\) \(\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(e^x \cos y) = -e^x \sin y\) \(\frac{\partial v}{\partial x} = \frac{\partial}{\partial x}(e^x \sin y) = e^x \sin y\) \(\frac{\partial v}{\partial y} = \frac{\partial}{\partial y}(e^x \sin y) = e^x \cos y\)

Now we need to evaluate the partial derivatives at the point (x=0, y=2π/3), where u=-1/2 and v=√3/2. a) For \(\left.\frac{\partial z}{\partial x}\right|_{\left(0, \frac{2 \pi}{3}\right)}\), we simply substitute the values: \(\left.\frac{\partial z}{\partial x}\right|_{\left(0, \frac{2 \pi}{3}\right)} = 2(-\frac{1}{2})\left(1\cdot \cos \frac{2 \pi}{3}\right) - 2 \frac{\sqrt{3}}{2}\left(1\cdot \sin \frac{2 \pi}{3}\right)\) \(\left.\frac{\partial z}{\partial x}\right|_{\left(0, \frac{2 \pi}{3}\right)} = (-1) (-\frac{1}{2}) - (\sqrt{3}) \frac{\sqrt{3}}{2} = \frac{1}{2} - \frac{3}{2} = -1\) b) For \(\left.\frac{\partial z}{\partial y}\right|_{\left(0, \frac{2 \pi}{3}\right)}\), we substitute the values as well: \(\left.\frac{\partial z}{\partial y}\right|_{\left(0, \frac{2 \pi}{3}\right)} = 2(-\frac{1}{2})\left(-1\cdot \sin \frac{2 \pi}{3}\right) - 2 \frac{\sqrt{3}}{2}\left(1\cdot \cos \frac{2 \pi}{3}\right)\) \(\left.\frac{\partial z}{\partial y}\right|_{\left(0, \frac{2 \pi}{3}\right)} = (1) \frac{\sqrt{3}}{2} - (\sqrt{3}) (-\frac{1}{2}) = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3} \) Hence, the evaluated partial derivatives are: \(\left.\frac{\partial z}{\partial x}\right|_{\left(0, \frac{2 \pi}{3}\right)} = -1\) \(\left.\frac{\partial z}{\partial y}\right|_{\left(0, \frac{2 \pi}{3}\right)} = \sqrt{3} \)