Question

find the directional derivative of f =xy^2+yz^2+zx^2 the tangent to the curve x =t, y=t^2,z=t^3 at the point (1,1,1)​

Answer 1

Answer:

18/root 14

Step-by-step explanation:

Answer 2

Given:

$f\left(x, y, z\right)=x y^{2}+y z^{2}+z x^{2}$ the tangent to the curve $x =t, y=t^2,z=t^3$ at the point (1,1,1).

To find:

Directional derivative

Solution:

$\Rightarrow f\left(x, y, z\right)=x y^{2}+y z^{2}+z x^{2} \\point (1,1,1) \\ fx{=y^{2}+2 x z \quad \Rightarrow f x(1,1,1)=3 \\ fy=2y x+z^{2} \Rightarrow fy(1,1,1)=3 \\ fz=2 z y+x^{2} \Rightarrow fz(1,1,1)=3 \\Then\\ \bar{u}=\left(\frac{\partial x}{\partial t} \hat{i}+\frac{\partial y}{\partial t} \\\\\hat{j}+\frac{\partial z}{\partial t} \hat{k}\right) \\ =\left(1 \hat{i}+2 t \hat{j}+3 t^{2} \hat{k}\right) \\but t=1 \\ =(1,2,3) \\ D_{\bar{u}}{ }^{f}=(3,3,3) \frac{(1,2,3)}{\sqrt{14}}$

$=\frac{18}{\sqrt{14} }$

Therefore, the directional derivative is $\frac{18}{\sqrt{14} }$.