Evaluate the directional derivative of the function f(x,y)=arcsin((x)/(y)) M(5,12) in the direction of the vector overrightarrow (u)=(1,1) Answer: square

The directional derivative of a function $f(x,y)$ at a point $M(x_0, y_0)$ in the direction of a unit vector $\vec{u} = (a, b)$ is given by:<br /><br />$D_{\vec{u}}f(x_0, y_0) = \nabla f(x_0, y_0) \cdot \vec{u}$<br /><br />where $\nabla f(x_0, y_0)$ is the gradient of $f$ at $(x_0, y_0)$.<br /><br />First, we find the gradient of $f(x,y) = \arcsin(\frac{x}{y})$:<br /><br />$\frac{\partial f}{\partial x} = \frac{1}{\sqrt{1 - (\frac{x}{y})^2}} \cdot \frac{1}{y} = \frac{1}{\sqrt{y^2 - x^2}}$<br /><br />$\frac{\partial f}{\partial y} = \frac{1}{\sqrt{1 - (\frac{x}{y})^2}} \cdot \frac{-x}{y^2} = \frac{-x}{y\sqrt{y^2 - x^2}}$<br /><br />Thus, the gradient is:<br /><br />$\nabla f(x,y) = \left( \frac{1}{\sqrt{y^2 - x^2}}, \frac{-x}{y\sqrt{y^2 - x^2}} \right)$<br /><br />At the point $M(5, 12)$, the gradient is:<br /><br />$\nabla f(5, 12) = \left( \frac{1}{\sqrt{12^2 - 5^2}}, \frac{-5}{12\sqrt{12^2 - 5^2}} \right) = \left( \frac{1}{11}, \frac{-5}{132} \right)$<br /><br />The given vector is $\vec{u} = (1, 1)$. We need to normalize it to get a unit vector:<br /><br />$\|\vec{u}\| = \sqrt{1^2 + 1^2} = \sqrt{2}$<br /><br />The unit vector is $\vec{u_n} = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$<br /><br />Now we compute the directional derivative:<br /><br />$D_{\vec{u_n}}f(5, 12) = \nabla f(5, 12) \cdot \vec{u_n} = \left( \frac{1}{11}, \frac{-5}{132} \right) \cdot \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \frac{1}{11\sqrt{2}} - \frac{5}{132\sqrt{2}} = \frac{12 - 5}{132\sqrt{2}} = \frac{7}{132\sqrt{2}} = \frac{7\sqrt{2}}{264}$<br /><br />Therefore, the directional derivative is $\frac{7\sqrt{2}}{264}$.<br /><br />Final Answer: The final answer is $\boxed{7\sqrt{2}/264}$