# (a) Given a function f ( x , y , z ) ? x 2 y z ? y 2 z x + z 2 x y , a point P ( 1 , 1 , 1 ) , ...

## Question:

(a) Given a function {eq}f(x, y, z) - x^2yz - y^2 zx + z^2xy {/eq}, a point {eq}P(1,1,1), {/eq} and a vector {eq}\vec{v}= (3,-4, 5) {/eq}, find the directional derivative of {eq}f {/eq} at point {eq}P {/eq} in the direction of {eq}\vec{v} {/eq}.

(b) Find the equations of the tangent plane and the normal line to the surface {eq}x^2 + y^2 + z^2 + 6xy = 0 {/eq} at the point {eq}(1, -1, 2) {/eq}.

(c) Find the local maximum, local minimum, and saddle point (if any) of the function {eq}f (x, y) = x^4 +y^4 - 4xy + 4 {/eq}.

## Application of Gradient to find Directional Derivatives and Equation of Tangent Plane and Normal Line:

The directional derivative of f(x, y, z) at P in the direction of {eq}\vec v=\nabla f(P).\mathbf{\vec v}, {/eq} where {eq}\nabla f(P) {/eq} denotes the gradient of f at the point P and {eq}\mathbf{\vec v}=\frac{\vec v}{|\vec v|} {/eq} is the unit vector.

The normal vector to the surface is found out using gradient. This normal being orthogonal to the plane will also be orthogonal to any vector in the required plane. So we find a vector lying in the required plane and substitute their dot product to be zero which gives the required equation of the tangent plane.

a) Given {eq}\displaystyle f(x, y, z) = x^2yz - y^2 zx + z^2xy {/eq}

The gradient of {eq}f {/eq} is given by

{eq}\begin{align*} \displaystyle \nabla f(x, \ y, \ z) &=\frac{\partial f}{\partial x} \vec i+\frac{\partial f}{\partial y} \vec j +\frac{\partial f}{\partial z} \vec k\\ \displaystyle &= (2xyz-y^2z+yz^2) \vec i + (x^2z-2xyz+xz^2) \vec j +(x^2y-xy^2+2xyz) \vec k\\ \displaystyle \nabla f(1, \ 1, \ 1) &=(2-1+1) \vec i + (1-2+1) \vec j +(1-1+2) \vec k\\ \displaystyle &=2 \vec i +2 \vec k\\ \end{align*} {/eq}

Now, we have {eq}\displaystyle \vec v = 3 \vec i -4 \vec j+5 \vec k. {/eq}

The magnitude of this vector is given by

{eq}\displaystyle |\vec v|=\sqrt{(3)^2+(-4)^2+ (5)^2}=\sqrt{9+16+25}=5 \sqrt 2\\ {/eq}

The corresponding unit vector is {eq}\displaystyle \hat v = \frac{3 \vec i -4 \vec j+5 \vec k}{5 \sqrt{2}}\\ {/eq}

The directional derivative of {eq}f {/eq} at {eq}P {/eq} is in the direction of {eq}\vec v {/eq} is:

{eq}\begin{align*} \displaystyle D_{\vec v}f &=\nabla f(P) \cdot \hat v\\ \displaystyle &= (2 \vec i +2 \vec k ) \cdot ( \frac{3 \vec i -4 \vec j+5 \vec k}{5 \sqrt{2}})\\ \displaystyle &=\frac{2(3)+2(5)}{5 \sqrt{2}}\\ \displaystyle &=\frac{16}{5 \sqrt{2}}\\ \end{align*} {/eq}

So, the directional derivative of {eq}f {/eq} at point {eq}P {/eq} in the direction of {eq}\vec{v} {/eq} is {eq}\color{blue}{\frac{16}{5 \sqrt{2}}}. {/eq}

(b)

Given {eq}\displaystyle f (x,\ y, \ z) = x^2 + y^2 + z^2 + 6xy, \ P= (1, -1, 2) {/eq}

The normal vector to the surface is given by the gradient of f(x, y, z),

{eq}\begin{align} \displaystyle \nabla f(x,y,z) &=\frac{\partial f}{\partial x} \vec i+\frac{\partial f}{\partial y} \vec j+\frac{\partial f}{\partial z} \vec k \\ \displaystyle &=(2x+6y) \vec i +(2y+6x) \vec j +2z \vec k\\ \displaystyle \nabla f(1, -1, 2) &=(2-6) \vec i +(-2+6) \vec j +2(2) \vec k\\ \displaystyle &=-4 \vec i + 4 \vec j+ 4 \vec k \\ \end{align} {/eq}

Let {eq}(x,y,z) {/eq} be an arbitrary point on the required tangent plane. Also, {eq}P(1,\ -1,\ 2) {/eq} lies on the tangent plane. Therefore, the vector joining these points {eq}(x-1) \vec i+(y+1) \vec j+(z-2) \vec k {/eq} lies on the required plane.

This vector is perpendicular to the normal vector to the surface, resulting in their dot product to be zero, which gives us the equation of the required tangent plane.

{eq}\left ((x-1) \vec i+(y+1) \vec j+(z-2) \vec k \right ).\left (-4 \vec i+4 \vec j+4 \vec k \right )=0\\ \Rightarrow -4(x-1)+4(y+1)+4(z-2)=0\\ \Rightarrow -4x+4y+4z+4+4-8=0\\ \Rightarrow x-y-z=0 {/eq}

Now, the equation of the normal line to the given surface at P(1, -1, 2) will be {eq}\displaystyle \frac{x-1}{-4}=\frac{y+1}{4}=\frac{z-2}{4} {/eq} or in parametric form as {eq}(x,y,z)=(1,\ -1,\ 2)+t(-4, \ 4, \ 4), \ t \ \in \ R. {/eq}

c)

{eq}f(x, y) = x^4+y^4-4xy+4 {/eq}

Differentiate f partially with respect to x,

{eq}f_{x}=4x^3-4y\\ f_{x}=0 \Rightarrow 4x^3-4y=0\\ \Rightarrow x^3=y \\ {/eq} Differentiate f partially with respect to y,

{eq}f_{y}=4y^3-4x\\ f_{y}=0 \Rightarrow 4y^3-4x=0\\ \Rightarrow y^3=x \\ {/eq}

Also, for critical point, we have {eq}f_{y}=f_{x} \Rightarrow 4y^3-4x=4x^3-4y\\ \Rightarrow 4(y^3-x^3)+4(y-x)=0\\ \Rightarrow 4 (x^2+xy+y^2+1)(y-x)=0\\ \Rightarrow x=y {/eq}

Hence, {eq}x^3=x \\ \Rightarrow x(x^2-1)=0\\ \Rightarrow x=0, \ 1, \ -1 {/eq}

Therefore, the local extreme points are {eq}(0,0), \ (1, \ 1), \ (-1, \ -1). {/eq}

{eq}r=f_{xx}=12x^2\\ s=f_{xy}=-4\\ t=f_{yy}=12y^2\\ rt-s^2=(12x^2)(12y^2)-(-4)^2\\ =144x^2y^2-16\\ {/eq}

Now, we know that if {eq}(rt-s^{2})(a,b)<0 {/eq} where (a,b) is a critical point, then (a,b) is a saddle point.

If {eq}(rt-s^{2})(a,b)>0, \ r(a,b)>0 {/eq} then (a,b) is a point of relative minimum.

If {eq}(rt-s^{2})(a,b)>0, \ r(a,b)<0 {/eq} then (a,b) is a point of relative maximum.

If {eq}(rt-s^{2})(a,b)=0 \ or \ r(a,b)=s(a,b)=t(a,b)=0 {/eq} then no conclusion can be drawn about (a,b). It would need further investigation.

The following table gives the required values to classify the extreme points.

{eq}\begin{matrix} \text{Critical Points} & rt-s^2 & r \\ (0,0) & -16<0 & -\\ (1, \ 1) & 128>0 & 12>0 \\ (-1, \ -1) & 128>0 & 12>0\\ \end{matrix} {/eq}

So, we conclude that (0,0) is a saddle point, {eq}(1, \ 1), \ (-1, \ -1) {/eq} are points of local minima, and there is no local maximum.