Pick any exercise in Chapter 12 of the calculus textbook. Post the solution or your attempt. Discuss any issues you might have had.
Pg.704 Ex.13 a function \(z\) = \(f(x,y)\) and a point \(P\) are given. Find the directional derivative of \(f\) in the indicated directions.
\(f(x,y)\) = \(-x^2y + xy^2 + xy\), \(P\) = (2,1)
gradient of f:
\(\nabla f\) = \((\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})\) = \((y(-2x+y+1),x(-x+2y+1))\)
\(\nabla f(2,1)\) = \((-2,2)\)
(a) In the direction of \(\overrightarrow{v}\) = <3,4>
unit vector for <3,4>: \(\sqrt{3^2+4^2}\) = \(5\) \(\Rightarrow\) <\(\frac{3}{5}\),\(\frac{4}{5}\)>
\(D(-x^2y + xy^2 + xy)_{\vec{u}}(2,1)\) = \((-2,2)\cdot(\frac{3}{5},\frac{4}{5})\) = \(\frac{2}{5}\)
(b) In the direction toward the point \(Q\) = (1,-1)
direction vector from \(P\)(2,1) to \(Q\)(1,-1): (2,1) + \(<a,b>\) = (1,-1) \(\Rightarrow\) <-1,-2>
unit vector for <-1,-2>: \(\sqrt{(-1)^2+(-2)^2}\) = \(\sqrt5\) \(\Rightarrow\) <\(-\frac{\sqrt{5}}{5},-\frac{2\sqrt{5}}{5}\)>
\(D(-x^2y + xy^2 + xy)_{\vec{u}}(2,1)\) = \((-2,2)\cdot(-\frac{\sqrt{5}}{5},-\frac{2\sqrt{5}}{5})\) = \(-\frac{2\sqrt{5}}{5}\)