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The prompt is to 1. find the directional derivatives of the function $f$ at a point $P(1, -1,2)$ in the direction towards origin and we are given $$f(1, -1, 2) = 5$$ $$f_x(1, -1, 2) = 1$$ $$f_y(1, -1, 2) = 3$$ $$f_z(1, -1, 2) = -1$$

  1. Find the plane tangent to the surface $f(x, y, z) = 5$ at point $P = (1, -1, 2)$

  2. Suppose we are told that there is a direction $u$ in which the directional derivative of $f$ is $6$, is this possible? If yes, find such a direction $u$, else explain why not.

  3. The equation $f(x, y, z) = 5$ determines $z(x, y)$, evaluate $z_x(1, -1, 2)$

Here's what I tried doing,

For part 1, $$\nabla f = (1, 3, -1)$$ $$\vec v = (0, 0, 0) - (x, y, z) = (-x, -y, -z)$$ Unit vector along $\vec v$ $$\vec v = \frac{(-x, -y, -z)}{\sqrt{x^2 + y^2 + z^2}}$$ Directional vector at point $P(1, -1, 2)$ $$\nabla f.\vec v = (1, 3, -1).\frac{(-x, -y, -z)}{\sqrt{x^2 + y^2 + z^2}}$$ $$\nabla f.\vec v = \frac{(-x)}{\sqrt{x^2 + y^2 + z^2}} +\frac{(-3y)}{\sqrt{x^2 + y^2 + z^2}} + \frac{(z)}{\sqrt{x^2 + y^2 + z^2}}$$

For part 2,

The plane tangent is given by $$\nabla f_{(1, -1, 2)} = (x - 1) + 3(y + 1) -(z - 2) = 0$$ $$x +3y-z= -4$$

I'm not sure how to go on about part 3 and 4 also I'm not sure if part 1 and 2 are correct.

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  • $\begingroup$ For 3, is there a unit vector $\vec n$ such that $\vec n \cdot \nabla f = 6$? Hint: Cauchy–Schwarz inequality. $\endgroup$ – md2perpe Jun 23 '17 at 21:52
  • $\begingroup$ For 4, during the lectures you probably have showed some expression for the partial derivatives of an implicitely defined function. $\endgroup$ – md2perpe Jun 23 '17 at 21:58
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The unit vector at $P(1, -1,2)$ in the direction of the origin is $\mathbf{u}=\left(-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}\right)$ and the gradient is $\nabla f = (1, 3, -1)$ therefore the directional derivative at $P$ in the direction $\mathbf{u}$ is

$$ D_{\mathbf{u}}f(1,-1,2)=\nabla f\cdot\mathbf{u}=(1, 3, -1)\cdot\left(-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}\right)=\frac{4}{\sqrt{6}}=\frac{2\sqrt{3}}{3} $$

Your equation for the tangent plane is correct.

The largest possible value for the directional derivative of $f$ at $P$ is in the direction of the gradient, which would be

$$ \mathbf{v}=\left(\dfrac{1}{\sqrt{11}},\dfrac{3}{\sqrt{11}},-\dfrac{1}{\sqrt{11}}\right) $$

giving

$$ D_{v}=\nabla f\cdot\mathbf{v}=\sqrt{11}<6 $$

so this is not possible.

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