How Do You Calculate the Directional Derivative of a Function at a Point?

[Destroxia]

Homework Statement

Find the directional derivative of $f$ at $P$ in the direction of $a$.

$f(x,y) = 2x^3y^3 ; P(3,4) ; a = 3i - 4j$

Homework Equations

$D_u f(x_0, y_0, z_0) = f_x(x_0, y_0, z_0)u_1 + f_y(x_0, y_0, z_0)u_2$

The Attempt at a Solution

$f_x (x,y) = 6x^2y^3$
$f_y (x,y) = 6x^3y^2$

$f_x (3,4) = 3456$
$f_y (3,4) = 2592$

$D_u f(x_0, y_0) = 3456u_1 +2592 u_2$

$u = \frac {a} {||a||} = \frac {\langle 3,4 \rangle} {5} = \langle \frac {3} {5}, \frac {4} {5} \rangle$

$D_u f(x_0, y_0) = 3456(\frac {3} {5}) + 2592(\frac {4} {5})$

$D_u f(x_0, y_0) = \frac {20736} {5}$

Now, my program wants this an exact number, no tolerance. It won't accept division either, so I don't know how to put in 20736/5. Just wondering if I made a mishap somewhere within the solution.

 

[SteamKing]

Homework Statement

Find the directional derivative of $f$ at $P$ in the direction of $a$.

$f(x,y) = 2x^3y^3 ; P(3,4) ; a = 3i - 4j$

Homework Equations

$D_u f(x_0, y_0, z_0) = f_x(x_0, y_0, z_0)u_1 + f_y(x_0, y_0, z_0)u_2$

The Attempt at a Solution

$f_x (x,y) = 6x^2y^3$
$f_y (x,y) = 6x^3y^2$

$f_x (3,4) = 3456$
$f_y (3,4) = 2592$

$D_u f(x_0, y_0) = 3456u_1 +2592 u_2$

$u = \frac {a} {||a||} = \frac {\langle 3,4 \rangle} {5} = \langle \frac {3} {5}, \frac {4} {5} \rangle$

$D_u f(x_0, y_0) = 3456(\frac {3} {5}) + 2592(\frac {4} {5})$

$D_u f(x_0, y_0) = \frac {20736} {5}$

Now, my program wants this an exact number, no tolerance. It won't accept division either, so I don't know how to put in 20736/5. Just wondering if I made a mishap somewhere within the solution.

You can't enter in a decimal number?

 

[Destroxia]

You can't enter in a decimal number?

Wouldn't that not be exact, but approximate form though? Or if I decimal isn't repeating is it considered exact?

 

[SteamKing]

Wouldn't that not be exact, but approximate form though? Or if I decimal isn't repeating is it considered exact?

So, you're saying that (1/2) = 0.5 is only an approximation and not an exact representation? Interesting.

 

[Destroxia]

So, you're saying that (1/2) = 0.5 is only an approximation and not an exact representation? Interesting.

So, I'm guessing it's not an approximation? Makes sense, good to learn something new. I will attempt to insert my answer.

 

[Destroxia]

So, you're saying that (1/2) = 0.5 is only an approximation and not an exact representation? Interesting.

I attempted the answer of 4147.2, and it was incorrect. Therefore, my work must be incorrect somewhere.

 

[SteamKing]

I attempted the answer of 4147.2, and it was incorrect. Therefore, my work must be incorrect somewhere.

What if the wrong answer has been programmed into the software you're using?

 

[Destroxia]

What if the wrong answer has been programmed into the software you're using?

I've spoken to other students who have the same problem, but with different numbers, so I'm pretty positive that the programs solution is correct, but now that I am at home I don't have access to see any of their solutions.

 

[Ray Vickson]

Homework Statement

Find the directional derivative of $f$ at $P$ in the direction of $a$.

$f(x,y) = 2x^3y^3 ; P(3,4) ; a = 3i - 4j$

Homework Equations

$D_u f(x_0, y_0, z_0) = f_x(x_0, y_0, z_0)u_1 + f_y(x_0, y_0, z_0)u_2$

The Attempt at a Solution

$f_x (x,y) = 6x^2y^3$
$f_y (x,y) = 6x^3y^2$

$f_x (3,4) = 3456$
$f_y (3,4) = 2592$

$D_u f(x_0, y_0) = 3456u_1 +2592 u_2$

$u = \frac {a} {||a||} = \frac {\langle 3,4 \rangle} {5} = \langle \frac {3} {5}, \frac {4} {5} \rangle$

$D_u f(x_0, y_0) = 3456(\frac {3} {5}) + 2592(\frac {4} {5})$

$D_u f(x_0, y_0) = \frac {20736} {5}$

Now, my program wants this an exact number, no tolerance. It won't accept division either, so I don't know how to put in 20736/5. Just wondering if I made a mishap somewhere within the solution.

$u_y \neq 4/5$; go back and check your work.

 

[Destroxia]

$u_y \neq 4/5$; go back and check your work.

Ah, thank you I missed that. I have remodeled my work, and the solution turns out to be 0, and is correct. Thank you both for your time.