# Finding $\nabla f$ of given implicit fuction.

Let $z = f(x,y)$ be the differentiable function given implicitly by $x^3 +y^3 +z^3 + xyz =9$ and such that $f(0,1)=2$.

1. Find $\nabla f$ at the point $P_0 = (0, 1)$.
2. Find the rate and direction of the steepest increase of $f$ at $P_0$.
3. Find $(D_u f)(P_0)$, where $u = (\mathbf{i}+\mathbf{j})/\sqrt{2}$.
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actually i know the whole process when the given function is not implicit, however, i could not understand how should i do when it is implicit – Yigit Oct 30 '12 at 19:28

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Find $\nabla f$ at the point $P_0 = (0, 1)$.

For the implicitly defined $z = f(x,y)$ by $$x^3 +y^3 +z^3 + xyz =9,\tag{1}$$ the general procedure from a standard multivariable calculus course is taking partial derivatives w.r.t. $x$ and $y$ in (1) respectively, treating $z$ as a function of $x$ and $y$ (chain rule kicks in). Therefore we have: $$3x^2 + 3z^2 \frac{\partial z}{\partial x} + yz + xy\frac{\partial z}{\partial x} = 0, \\ 3y^2 + 3z^2 \frac{\partial z}{\partial y} + xz + xy\frac{\partial z}{\partial y} = 0.$$ Solving for $\partial z/\partial x$ and $\partial z/\partial y$ we have: $$\frac{\partial z}{\partial x} = -\frac{3x^2+yz}{3z^2+xy},\quad\text{and}\quad \frac{\partial z}{\partial y} = -\frac{3y^2+xz}{3z^2+xy}.$$ Plugging $x= 0,y=1$ and $z=2$: $$\nabla f = \left(\frac{\partial z}{\partial x} , \frac{\partial z}{\partial y} \right) = \left(-\frac{1}{6} , -\frac{1}{4} \right).$$

Find the rate and direction of the steepest increase of $f$ at $P_0$.

At a point of a surface $z = f(x,y)$, the steepest increase direction coincides with the direction of the gradient. Rate is the magnitude of the gradient: $|\nabla f |$.

Find $(D_u f)(P_0)$, where $u = (\mathbf{i}+\mathbf{j})/\sqrt{2}$.

Using the results above: $$(D_u f)(P_0) = \nabla f\cdot u\big|_{P_0} = \left(-\frac{1}{6} , -\frac{1}{4} \right) \cdot \left( \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right) = -\frac{5}{12\sqrt{2}}.$$

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Let me give you a few hints: $$\nabla f = \left\langle {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right\rangle$$ and $${D_u}f({P_0}) = \nabla f({P_0}) \cdot \hat u.$$ Also, we know that $\nabla f$ points in the direction of max increase, therefore $|\nabla f|$ gives the max rate. It does not matter if the function is defined implicitly, the procedure is the same. Even if $z = f(x,y)$, we can still write $g(x,y,z) = f(x,y) - z$.

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