Directional Derivative Calculator + Online Solver With Free Steps

The directional derivative calculator is used to calculate the directional derivative of a function in terms of two variables $x$ and $y$ at a given point.

The derivative of a function is the rate of change of the function. Directional derivative is commonly defined as the rate of change of the function in any given direction.

Directional derivatives have a wide range of applications in real life as the inputs are continuously changing. The calculator also computes the gradient vector of the given function. The gradient defines the slope of the function.

What Is a Directional Derivative Calculator?

The Directional Derivative Calculator is an online calculator that solves for the directional derivative of a two-variable function f( $x$ , $y$ ) at a point ( $x$, $y$ ) along the unit vector U and also outputs the gradient $grad$ $f$($x$,$y$) of the input function.

The direction is determined by the unit vector:

$\overrightarrow{U} = (U_{1})\hat{e_{x}} + (U_{2})\hat{e_{y}}$

$U_{1}$ specifies the direction along the $x$-axis and $U_{2}$ specifies the direction along the $y$-axis.

The calculator computes the directional derivative of a function at a given point. The $x$-coordinate specifies the point on the $x$-axis and the $y$-coordinate specifies the point on the $y$-axis for which the directional derivative needs to be calculated.

It also calculates the gradient of the function. The gradient of a function is the rate of change or slope of the function.

For the two-variable function, we need to determine the rate of change of function $f$ along the $x$-axis and $y$-axis. This gives the concept of partial derivative.

The partial derivative along the $x$-axis is the rate of change of the function $f$($x$,$y$) in the $x$ direction and the partial derivative along the $y$-axis is the rate of change of the function $f$($x$,$y$) in the $y$ direction.

The partial derivative of the function $f$($x$,$y$) with respect to $x$ is represented as:

$f^{(1,0)}$

And the partial derivative of $f$($x$,$y$) with respect to $y$ is represented as:

$f^{(0,1)}$

The partial derivative is different from the directional derivative.

The partial derivative gives the instantaneous rate of change of a function only along the three perpendicular axes, which are the $x$-axis, $y$-axis, and the $z$-axis at a given point.

On the other hand, the directional derivative gives the instantaneous rate of change in any direction at a certain point.

How To Use the Directional Derivative Calculator?

You can use the Directional Derivative calculator by selecting the desired function and specifying the values of $U1$ and $U2$ along with the $x$ and $y$ coordinates.

The following steps are required to use the directional derivative calculator.

Step 1

Enter the function in terms of two variables $x$ and $y$ in the block labeled $f$( $x$ , $y$ ). The calculator shows the following function:

$f ( x , y ) = 3x^2.y$

by default.

Step 2

Enter the part of the unit vector which shows the direction along the $x$-axis. This is $U_{1}$ in the calculator’s input window. The calculator shows $U_{1}$ as $(\dfrac{3}{5})$ by default.

Step 3

Enter the value of $U_{2}$ which is the part of the unit vector showing the direction along the $y$-axis. The calculator displays $U_{2}$ as $(\dfrac{4}{5})$ by default.

Step 4

The calculator also requires the point ($x$,$y$) for which the directional derivative and gradient are to be determined.

Enter the x-coordinate in the calculator’s input window, which shows the position of the point along the $x$-axis. The $x$-coordinate by default is $1$.

Step 5

Enter the y-coordinate, which is the location of the point along the $y$-axis for which the user requires the directional derivative. The $y$-coordinate by default is $2$.

Step 6

The user should press Submit after entering all the required input data for the results.

The output window opens in front of the user, which shows the following windows. If the user’s input is incorrect or incomplete, the calculator prompts “Not a valid input, please try again.”

Input Interpretation

The calculator interprets the input and displays it in this window. First, it shows the function $f$( $x$,$y$ ) for which the directional derivative is required.

Then, it shows the direction ( $U_{1}$ , $U_{2}$ ) and the point ( $x$-coordinate , $y$-coordinate ) which the user entered.

Result

This window shows the resultant directional derivative after placing the point ( $x$-coordinate, $y$-coordinate ) in the directional derivative function.

It shows the equation of directional derivative in an open form which shows the values of the partial derivatives concerning $x$ and $y$.

Gradient

This window shows the gradient $grad$ $f$ ($x$,$y$) of the input function $f$. It also displays $x$, which is the first Cartesian coordinate, and $y$, which is the second Cartesian coordinate.

Also,

$\frac{\partial f(x,y)}{\partial x}$

in the gradient equation represents the partial derivative of $f$($x$,$y$) with respect to $x$ and

$\frac{\partial f(x,y)}{\partial y}$

represents the partial derivative of $f$($x$,$y$) with respect to $y$.

Solved Examples

The following examples are solved through the directional derivative calculator.

Example 1

Calculate the directional derivative of the given function:

$f ( x , y ) = 4x^3 – 3xy^2$

At the point ($1$ , $2$)

Where,

$U_{1} = \frac{1}{2}$

and

$U_{2} = \frac{\sqrt{3}}{2}$

Also, evaluate the gradient vector of the given function.

Solution

The calculator displays $f$($x$,$y$), which is the given function.

It also displays the direction and the point ($1$,$2$) at which the directional derivative is required. This is shown in the input interpretation window of the calculator’s output.

The calculator computes the directional derivative and shows the result as follows:

$\frac{1}{2}(\sqrt{3}(f^{(0,1)}(1,2)) = -12) + (f^{(1,0)}(1,2) = 0 )$

Here:

$f^{(0,1)} = \frac{\partial f(x,y)}{\partial y}$

$f^{(1,0)} = \frac{\partial f(x,y)}{\partial x}$

The calculator also calculates the gradient $grad$ $f$($x$,$y$) of the entered function $f$.

For the gradient, the calculator first computes the partial derivatives of the function $f$.

For the partial derivative of $f$($x$,$y$) with respect to $x$:

$\frac{\partial f(x,y)}{\partial x} = 12x^2 – 3y^2$

$\frac{\partial f(x,y)}{\partial x} + 3y^2 = 12x^2$

The calculator shows the above equation in the gradient result.

For the partial derivative of $f$($x$,$y$) with respect to $y$:

$\frac{\partial f(x,y)}{\partial y} = – 6xy$

The gradient of the function is:

$grad f(x,y) = \Big\{ \frac{\partial f(x,y)}{\partial x} + 3y^2 = 12x^2 \Big\} .e_{x} + \Big\{ \frac{\partial f(x,y)}{\partial y} = – 6xy \Big\} .e_{y}$

Where $e_{x}$ and $e_{y}$ represents the unit vectors along the direction of $x$ and $y$ axis, respectively.

Example 2

Evaluate the directional derivative of the function:

$f ( x , y ) = x.y^2 – 2.x^3$

At the point ($3$ , $2$)

Where,

$U_{1} = \frac{1}{2}$

and

$U_{2} = \frac{1}{4}$

Also, find the gradient vector of the function.

Solution

The calculator displays the given function, the direction ( $\dfrac{1}{2}$, $\dfrac{1}{4}$ ) and the point ($3$,$2$) for which the directional derivative is required. The input interpretation window shows this result.

The calculator computes the directional derivative and shows the result as follows:

$\frac{1}{\sqrt{5}} ((f^{(0,1)}(3,2) = 12) + 2(f^{(1,0)}(3,2) = -50 )$

Here,

$f^{(0,1)} = \frac{\partial f(x,y)}{\partial y}$

$f^{(1,0)} = \frac{\partial f(x,y)}{\partial x}$

The calculator also computes the gradient vector grad $f$($x$,$y$) of the input function $f$.

It computes the partial derivatives of the function $f$ with respect to $x$ and $y$, which are used in the gradient vector.

For the partial derivative of $f$($x$,$y$) with respect to $x$:

$\frac{\partial f(x,y)}{\partial x} = – 6x^2 + y^2$

$\frac{\partial f(x,y)}{\partial x} + 6x^2 = y^2$

The calculator shows the above equation in the gradient vector.

For the partial derivative of $f$($x$,$y$) with respect to $y$:

$\frac{\partial f(x,y)}{\partial y} = 2xy$

The gradient of the function is:

$grad f ( x, y ) = \Big\{ 6x^2 + \frac{\partial f(x,y)}{\partial x} = y^2 \Big\} .e_{x} + \Big\{ 2xy = \frac{\partial f(x,y)}{\partial y} \Big\} .e_{y}$

Where $e_{x}$ and $e_{y}$ are the unit vectors along the $x$-axis and $y$-axis, respectively.

Example 3

Evaluate the directional derivative of the function:

$f ( x , y ) = x^2 – y^2$

At the point ($1$ , $3$)

Where,

$U_{1} = \frac{1}{3}$

and

$U_{2} = \frac{1}{2}$

Also, find the gradient vector of the function.

Solution

The calculator displays the input function, the direction ( $U_{1}$ , $U_{2}$ ), and the point ($3$,$2$).

The input interpretation window of the calculator shows these specifications.

The result for the directional derivative is:

$\frac{1}{\sqrt{13}} (3(f^{(0,1)}(1,3) = – 6 ) + 2(f^{(1,0)}(1,3) = 2 )$

The calculator then computes the gradient vector of the input function $f$.

But first, the partial derivatives of the function $f$ concerning $x$ and $y$ are computed for the gradient.

For the partial derivative of $f$($x$,$y$) with respect to $x$:

$\frac{\partial f(x,y)}{\partial x} = 2x$

For the partial derivative of $f$($x$,$y$) with respect to $y$:

$\frac{\partial f(x,y)}{\partial y} = – 2y$

The gradient of the function is:

$grad f ( x, y ) = \Big\{ \frac{\partial f(x,y)}{\partial x} = 2x \Big\} .e_{x} + \Big\{ \frac{\partial f(x,y)}{\partial y} = – 2y \Big\} .e_{y}$

Where $e_{x}$ and $e_{y}$ are the unit vectors with magnitude $1$ pointing in the direction of $x$-axis and $y$-axis, respectively.

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