Let f(x, y, z) = x^2 + 2y^2 - 3z^2. (a) Find the gradient nabla f. (b) Find the directional derivative of f in the direction of u = (1, 2, 3) at the point P_0 (1, 1, 1). Observe that u is not a unit vector. (c) Find the directions in which the function increases and decreases most rapidly at the point P_0 (1, 1, 1).

Accepted Solution

Answer:[tex]grad(f(x,y,z))=\frac{\partial f}{\partial x}\widehat{i}+\frac{\partial f}{\partial y}\widehat{j}+\frac{\partial f}{\partial z}\widehat{k}\\\\[/tex]Applying values we get[tex]grad(f(x,y,z))=2x\widehat{i}+4y\widehat{j}-6z\widehat{k}[/tex]b) The directional derivative in the direction of [tex]\overrightarrow{u}=\widehat{i}+2\widehat{j}+3\widehat{k}[/tex] is given by[tex]\overrightarrow{\triangledown f}.\frac{\overrightarrow{u}}{|u|}[/tex]Applying values we get the directional derivative equals[tex](2x\widehat{i}+4y\widehat{j}-6z\widehat{k}).(\widehat{i}+2\widehat{j}+3\widehat{k})\times \frac{1}{\sqrt{14}}\\\\=\frac{2x+8y-18z}{\sqrt{14}}[/tex]Thus value at [tex]P_{o}=(1,1,1)=-2.13[/tex]c)The direction of rate of maximum increase at [tex]P_{o}=(1,1,1)[/tex] is given by[tex]\triangledown \overrightarrow{f}(1,1,1)=2\widehat{i}+4\widehat{k}-6\widehat{k}[/tex]