# directional derivative

• Jul 6th 2008, 08:51 PM
chris25
directional derivative
Find the directional derivative of f x, y = 3 x^2 y at the point (1,2) in the direction of v= 3 i+4 j .

after i find the unit vector i am lost...can anyone help???
• Jul 6th 2008, 09:16 PM
Chris L T521
Quote:

Originally Posted by chris25
Find the directional derivative of f x, y = 3 x^2 y at the point (1,2) in the direction of v= 3 i+4 j .

after i find the unit vector i am lost...can anyone help???

The directional derivative can be defined in terms of the gradient:

$\bold{D}_{\bold u} f(x_0,y_0)=\nabla f(x_0,y_0)\cdot \bold{u}$

Well, once we get the unit vector $\bold{u}=\frac{3}{5}\bold{i} +\frac{4}{5}\bold{j}$, we need to find $\nabla f(1,2)$

Note that the $\nabla$ operator [in this case] refers to $\frac{\partial}{\partial x}\bold i +\frac{\partial}{\partial y}\bold j$

Thus $\nabla f(x,y)=\frac{\partial}{\partial x}(3x^2y)\bold i+\frac{\partial}{\partial y}(3x^2y)\bold j=6xy\bold i +3x^2 \bold j$

Now we can see that:

$\nabla f(1,2)=6(1)(2)\bold i +3(1)^2 \bold j=12\bold i +6 \bold j$

Thus,

$\bold{D}_{\bold u} f(1,2)=\nabla f(1,2)\cdot \bold{u}=\underbrace{\bigg(12\bold i +6\bold j\bigg)}_{\nabla f(1,2)}\cdot \underbrace{\bigg(\frac{3}{5}\bold i +\frac{4}{5}\bold j\bigg)}_{\bold u}=\frac{36}{5}+\frac{24}{5}=\frac{60}{5}=\color{b lue}\boxed{12}$

Does this make sense?

--Chris
• Jul 13th 2008, 03:53 PM
Mathstud28
Quote:

Originally Posted by Chris L T521
The directional derivative can be defined in terms of the gradient:

$\bold{D}_{\bold u} f(x_0,y_0)=\nabla f(x_0,y_0)\cdot \bold{u}$

Well, once we get the unit vector $\bold{u}=\frac{3}{5}\bold{i} +\frac{4}{5}\bold{j}$, we need to find $\nabla f(1,2)$

Note that the $\nabla$ operator [in this case] refers to $\frac{\partial}{\partial x}\bold i +\frac{\partial}{\partial y}\bold j$

Thus $\nabla f(x,y)=\frac{\partial}{\partial x}(3x^2y)\bold i+\frac{\partial}{\partial y}(3x^2y)\bold j=6xy\bold i +3x^2 \bold j$

Now we can see that:

$\nabla f(1,2)=6(1)(2)\bold i +3(1)^2 \bold j=12\bold i +6 \bold j$

Thus,

$\bold{D}_{\bold u} f(1,2)=\nabla f(1,2)\cdot \bold{u}=\underbrace{\bigg(12\bold i +6\bold j\bigg)}_{\nabla f(1,2)}\cdot \underbrace{\bigg(\frac{3}{5}\bold i +\frac{4}{5}\bold j\bigg)}_{\bold u}=\frac{36}{5}+\frac{24}{5}=\frac{60}{5}=\color{b lue}\boxed{12}$

Does this make sense?

--Chris

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

$\nabla{f}=\left\langle{6xy,3x^2}\right\rangle$

$\nabla{f}(1,2)=\left\langle{6(1)(2),3(1)^2}\right\ rangle=\left\langle{12,{\color{red}3}}\right\rangl e$

$\nabla{f}(1,2)\cdot\bold{\hat{u}}=12\cdot\frac{3}{ 5}+3\cdot\frac{4}{5}=\frac{48}{5}$
$\therefore{\bold{D}_{\bold{u}}}=\frac{48}{5}\quad\ because\nabla{f}(1,2)\cdot\bold{\hat{u}}=\frac{48} {5}$