Answer to Question #138339 in Calculus for Promise Omiponle

Question

Find the directional derivative of the function f(x, y, z) = 2xy-yz at the point(1,1,1) in the direction of u=<1,2,3>. Is there a direction (^v) in which f(x, y, z) has a directional derivative D^vf=-3 at the point (1,-1,1)?

Answer 1

You

16:46

$f_x(x,y,z)= 2y, f_y(x,y,z)= 2x-z , f_z(x,y,z)=-y.$. Since the partial derivatives are continuous we can easily find directional derivatives through dot product. point= $(1,1,1).$ Hence $f_x=2, f_y=1, f_z=-1$ at the given point. Unit vector= $\frac{1}{\sqrt{1^2+2^2+3^2 }} (1,2,3)=(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} )$ Hence directional derivative = $2.\frac{1}{\sqrt{14}}+1.\frac{2}{\sqrt{14}}-1.\frac{3}{\sqrt{14}}=\frac{1}{\sqrt{14}}.$

In the next case let the unit vector along the direction be $(a,b,c).$ Hence $a^2+b^2+c^2=1$ and $-2a+b+c=-3.$ Hence $(b+c)^2=(2a-3)^2.$ Hence $2bc= (2a-3)^2-(1-a^2)$ . Hence $(b-c)^2=1-a^2-(2a-3)^2+1-a^2=12a-7-6a^2.$

R.H.S.= $-[1+6(a-1)^2] <0 \Rightarrow\Leftarrow .$ Hence no such direction exists.