Undetermined Coefficients Problems
Directional Derivative Sample Problems

Suppose z =f\left (x ,y\right ) is a given function. We have seen that the first partial derivatives are

(1)   \begin{equation*}f_{x}\left (x ,y\right ) =\lim _{h \rightarrow 0}\frac{f(x +h ,y) -f\left (x ,y\right )}{h} \end{equation*}

(2)   \begin{equation*}f_{y}\left (x ,y\right ) =\lim _{h \rightarrow 0}\frac{f(x ,y +h) -f\left (x ,y\right )}{h} \end{equation*}

if these limits exist. Now if we use an arbitrary direction, say a unit vector \boldsymbol{u} =\left \langle u_{1} ,u_{2}\right \rangle, then both x and y will have changes. We define the directional derivative of f with respect to an arbitrary unit vector \boldsymbol{u} in xy-plane by

(3)   \begin{equation*}D_{\boldsymbol{u}}f\left (x ,y\right ) =\lim _{h \rightarrow 0}\frac{f(x +hu_{1} ,y +hu_{2}) -f\left (x ,y\right )}{h} \end{equation*}

if this limit exist! In fact, it can be shown that

(4)   \begin{align*}D_{u}f(x ,y) =f_{x}(x ,y)u_{1} +f_{y}\left (x ,y\right )u_{2} \\ =\langle f_{x}(x,y),f_{y}(x,y)\rangle \cdot \boldsymbol{u}\end{align*}

Let us see why. Define

(5)   \begin{equation*}F\left (h\right ) =f\left (x +hu_{1} ,y +hu_{2}\right ) \end{equation*}

and by Chain Rule,

(6)   \begin{align*}\frac{dF}{dh} =\frac{ \partial f}{ \partial (x +hu_{1})}\frac{d(x +hu_{1})}{dh} +\frac{ \partial f}{ \partial (y +hu_{2})}\frac{d(y +hu_{2})}{dh} \\ =f_{\left (x +hu_{1}\right )}u_{1} +f_{\left (y +hu_{2}\right )}u_{2} \\ F^{ \prime }\left (0\right ) =f_{x}u_{1} +f_{y}u_{2}\end{align*}

and of course

(7)   \begin{align*}F^{ \prime }\left (0\right ) =\lim _{h \rightarrow 0}\frac{F\left (h\right ) -F\left (0\right )}{h -0} \\ =\lim _{h \rightarrow 0}\frac{f(x +hu_{1} ,y +hu_{2}) -f\left (x ,y\right )}{h} \\ =D_{\boldsymbol{u}}f\left (x ,y\right )\end{align*}

The vector \left \langle f_{x} ,f_{y}\right \rangle used above is called the gradient of f and is shown by \nabla f .

Example 1.

We want to find the directional derivative of f\left (x ,y\right ) =x^{2} -2y^{2} with respect to the standard position angle 45^{ \circ } at \left (0 ,0\right ). The unit vector used is \left \langle \cos 45^{ \circ } ,\sin 45^{ \circ }\right \rangle =\left \langle \frac{1}{\sqrt{2}} ,\frac{1}{\sqrt{2}}\right \rangle. The gradient of f is

(8)   \begin{equation*} \nabla f =\left \langle 2x , -4y\right \rangle \end{equation*}

Now we compute the directional derivative of f with respect to the given direction

(9)   \begin{equation*}D_{\left \langle \frac{1}{\sqrt{2}} ,\frac{1}{\sqrt{2}}\right \rangle }f =\langle 2x , -4y\rangle \cdot \langle \frac{1}{\sqrt{2}} ,\frac{1}{\sqrt{2}}\rangle =\sqrt{2}(x -2y) \end{equation*}

At the point \left (0 ,0\right ) we get

(10)   \begin{equation*}D_{\left \langle \frac{1}{\sqrt{2}} ,\frac{1}{\sqrt{2}}\right \rangle }f\left (0 ,0\right ) =0 \end{equation*}

Observation 1.

From the construction of directional derivative of f in direction of \boldsymbol{u} we get

(11)   \begin{align*}D_{\boldsymbol{u}}f = \nabla f \cdot \boldsymbol{u} \\ =\left \Vert \nabla f\right \Vert \left \Vert \boldsymbol{u}\right \Vert \cos \theta \\ =\left \Vert \nabla f\right \Vert \cos \theta \end{align*}

Since maximum value of \cos \theta is equal to 1 , we see immediately that the maximum value of directional derivative is \left \Vert \nabla f\right \Vert. On top of this, \cos \theta =1 when \theta =0 and so \boldsymbol{u} must be parallel to \nabla f to get the maximum directional derivative.

Example 2.

We want to find the direction of maximum rate of change and its value for the function f\left (x ,y ,z\right ) =x\arctan yz at the point \left (1 ,0 ,1\right ). We start by writing

(12)   \begin{align*} \nabla f =\left \langle f_{x} ,f_{y} ,f_{z}\right \rangle \\ =\left \langle \arctan yz ,\frac{xz}{1 +\left (yz\right )^{2}} ,\frac{xy}{1 +\left (yz\right )^{2}}\right \rangle \end{align*}

At \left (1 ,0 ,1\right ) the direction which gives the maximum rate of change is \nabla f(1 ,0 ,1) =\langle 0 ,1 ,0\rangle and the maximum rate of change is \left \Vert \left \langle 0 ,1 ,0\right \rangle \right \Vert =1.

Observation 2.

Suppose a function of two variables z =f\left (x\left (t\right ) ,y\left (t\right )\right ) is given where z is differentiable with respect to x and y and also each x and y are differentiable with respect to t. An arbitrary level curve of this function associated to c is identified by f\left (x\left (t\right ) ,y\left (t\right )\right ) =c which is a curve in xy-plane. We can differentiate both side of this equation to get

(13)   \begin{equation*}\frac{ \partial f}{ \partial x}\frac{dx}{dt} +\frac{ \partial f}{ \partial y}\frac{dy}{dt} =0 \end{equation*}

This means

(14)   \begin{equation*}\left \langle f_{x} ,f_{y}\right \rangle \cdot \left \langle x^{ \prime } ,y^{ \prime }\right \rangle =0 \end{equation*}

We know \left \langle x^{ \prime } ,y^{ \prime }\right \rangle is a tangent vector at any given point. This dot product being equal to zero implies that a tangent vector to the level curve and the gradient vector are perpendicular to each other.

Observation 3.

Suppose a function of three variables w =f\left (x\left (t\right ) ,y(t) ,z(t)\right ) is given where w is differentiable with respect to x ,y ,z and each of x ,y ,z is differentiable with respect to t . Then an arbitrary level surface of w associated to a constant c is

(15)   \begin{equation*}f\left (x\left (t\right ) ,y\left (t\right ) ,z\left (t\right )\right ) =c \end{equation*}

Similar to previous case, observation 2, we can write

(16)   \begin{equation*}\frac{ \partial f}{ \partial x}\frac{dx}{dt} +\frac{ \partial f}{ \partial y}\frac{dy}{dt} +\frac{ \partial f}{ \partial z}\frac{dz}{dt} =0 \end{equation*}

Of course like before this means that the gradient vector and the tangent vector at any point are perpendicular to each other; we also see the equation of the tangent plane can be written using the gradient vector as the normal that is for the point \left (x_{0} ,y_{0} ,z_{0}\right )

(17)   \begin{equation*}f_{x}\left (x_{0} ,y_{0} ,z_{0}\right )\left (x -x_{0}\right ) +f_{y}\left (x_{0} ,y_{0} ,z_{0}\right )\left (y -y_{0}\right ) +f_{z}(x_{0},y_{0},z_{0})(z -z_{0}) =0 \end{equation*}

More, we can write equation of the normal line to the level surface at any point \left (x_{0} ,y_{0} ,z_{0}\right )

(18)   \begin{equation*}\frac{x -x_{0}}{f_{x}\left (x_{0} ,y_{0} ,z_{0}\right )} =\frac{y -y_{0}}{f_{y}(x_{0} ,y_{0} ,z_{0})} =\frac{z -z_{0}}{f_{z}\left (x_{0} ,y_{0} ,z_{0}\right )} \end{equation*}

Example 3.

To find the equation of the tangent plane to the paraboloid z =x^{2} +y^{2} we define w =f\left (x ,y ,z\right ) =x^{2} +y^{2} -z . The level surface of w associated to c =0 is precisely the given paraboloid. And so

(19)   \begin{equation*}2x\left (x -x_{0}\right ) +2y\left (y -y_{0}\right ) -\left (z -z_{0}\right ) =0 \end{equation*}

is the desired tangent plane. Next we find the equation of the normal line:

(20)   \begin{equation*}\frac{x -x_{0}}{2x} =\frac{y -y_{0}}{2y} =\frac{z -z_{0}}{ -1} \end{equation*}

To see more sample problems, visit directional derivative sample problems.

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