What is the directional derivative of F at point P(1,2,1) with given direction?

[Drain Brain]

For a direction determined by $dx=2dy=-2dz$, find the directional derivative of $F=x^2+y^2+z^2$ at P(1,2,1)

I had no problem getting the gradient of F and evaluating it at P but when I take the directional derivative I'm stuck! I don't know how come up with a unit vector that should be dotted to the gradient of F. please help!

$\nabla F = 2\hat{a_{x}}+4\hat{a_{y}}+2\hat{a_{z}}$ at P(1,2,1)

 

[Fantini]

Hello. :) You can think of the equations $$dx = 2 \, dy = - 2 \, dz$$ as $$\frac{dx}{dt} = 2 \frac{dy}{dt} = - 2 \frac{dz}{dt} = \operatorname{constant} = k.$$ From this you can obtain that $$x = kt, \quad y = \frac{kt}{2}, \quad z = - \frac{kt}{2}.$$ Thus the direction is $$\vec{r} = k t \left( 1, \frac{1}{2}, - \frac{1}{2} \right).$$ Can you take it from here?

 

[Drain Brain]

Hello. :) You can think of the equations $$dx = 2 \, dy = - 2 \, dz$$ as $$\frac{dx}{dt} = 2 \frac{dy}{dt} = - 2 \frac{dz}{dt} = \operatorname{constant} = k.$$ From this you can obtain that $$x = kt, \quad y = \frac{kt}{2}, \quad z = - \frac{kt}{2}.$$ Thus the direction is $$\vec{r} = k t \left( 1, \frac{1}{2}, - \frac{1}{2} \right).$$ Can you take it from here?

Hi Fantini!

Why do you equate it to a constant k? And does that have to be with respect to t?

$\frac{dx}{dt} = 2 \frac{dy}{dt} = - 2 \frac{dz}{dt} = \operatorname{constant} = k.$

And in this

$x = kt, \quad y = \frac{kt}{2}, \quad z = - \frac{kt}{2}$ where did t come from?

 

[Fantini]

He is saying that the derivatives are all proportional to a certain number. I chose to name it $k$. I also chose, somewhat randomly, a name for the variable with respect to which all coordinates are taking their derivatives. It could be $\frac{dx}{d \alpha}$, it doesn't change.

For the rest, notice that if $$\frac{dx}{dt} = k,$$ how do you solve this differential equation? One solution is $x=kt$. Same applies in the other cases. :)

 

[Drain Brain]

Hi Everyone! :)

from the given info provided $\vec{r}$ is

$\vec{r}=kt\hat{a_{x}}+\frac{kt}{2}\hat{a_{y}}-\frac{kt}{2}\hat{a_{z}}$

taking its magnitude $|\vec{r}|=\sqrt{(kt)^2+\frac{(kt)^2}{4}+\frac{(kt)^2}{4}} = \frac{kt\sqrt{6}}{2}$

the unit vector of $\frac{\vec{r}}{|\vec{r}|}=\frac{kt\hat{a_{x}}+\frac{kt}{2}\hat{a_{y}}-\frac{kt}{2}\hat{a_{z}}}{\frac{kt\sqrt{6}}{2}} $

$\frac{\vec{r}}{|\vec{r}|}=\frac{2\hat{a_{x}}}{\sqrt{6}}+\frac{\hat{a_{y}}}{\sqrt{6}}-\frac{\hat{a_{z}}}{\sqrt{6}}$

dotting the unit vector $\vec{r}$ to $\nabla{F}$

$\vec{r}\cdot \nabla{F}= \left(2\hat{a_{x}}+4\hat{a_{y}}+2\hat{a_{z}}\right)\cdot \left(\frac{2\hat{a_{x}}}{\sqrt{6}}+\frac{\hat{a_{y}}}{\sqrt{6}}-\frac{\hat{a_{z}}}{\sqrt{6}}\right)$

$\vec{r}\cdot \nabla{F}=\left(\frac{4}{\sqrt{6}}+\frac{4}{\sqrt{6}}-\frac{2}{\sqrt{6}}\right)=\sqrt{6}$

note that I already evaluated $\nabla{F}$ at point P before I dot it to unit vector of $\vec{r}$

$\frac{dF}{dl}\bigg|_{1,2,1}=\sqrt{6}$---->>FINAL ANSWER.

But the answer in my book is 2. Can you tell where did I go wrong? Thanks!

 

[Fantini]

I believe your answer is correct, not the book's. Let's hope others come in and check. :)

 

[Drain Brain]

There's another thing that I need to understand about post #2 where you equate the derivatives to a constant, what's the reason behind that? I want to know it more deeply. please bear with me.

 

[Fantini]

Consider a vector field ${\mathbf A}({\mathbf r})$ and a trajectory of this field ${\mathbf r}(s)$. The parameter $s$ controls where you are in the trajectory, and is usually the arc length.

You want this vector field to be tangent to the trajectory at every point. Thus, the tangent vector and the vector field must be collinear. If they are collinear then they are proportional (and vice-versa). This means they have to satisfy the condition $$d {\mathbf r} \times {\mathbf A} = 0.$$ But $$d {\mathbf r} = (dx, dy, dz)$$ and $${\mathbf A} = (A_1, A_2, A_3),$$ and so this condition means $$d {\mathbf r} = k {\mathbf A}.$$ Hence $$(dx,dy,dz) = k(A_1,A_2,A_3),$$ and therefore $$\frac{dx}{A_1} =k, \quad \frac{dy}{A_2} = k, \quad \frac{dz}{A_3} =k.$$ We usually shorten this to $$\frac{dx}{A_1} = \frac{dy}{A_2} = \frac{dz}{A_3}.$$

In our case $A_1 = 1, A_2 = \frac{1}{2},$ and $A_3 = - \frac{1}{2}$. The vector field is constant.

 

[Euge]

I disagree with the book's answer -- it should be $\sqrt{6}$. Since $$dF = 2x\, dx + 2y\, dy + 2z\, dz = (2x + y - z)\, dx$$ and $$ds = \sqrt{dx^2 + dy^2 + dz^2} = \sqrt{1 + \frac{1}{4} + \frac{1}{4}}\, dx = \sqrt{\frac{3}{2}}\, dx$$

we have $$\frac{dF}{ds} = (2x + y - z)\sqrt{\frac{2}{3}}$$ as the directional derivative of $F$ at the point $(x,y,z)$. Substituting $x = 1$, $y = 2$, and $z = 1$ yields $$\frac{dF}{ds}\bigg|_{(1,2,1)} = 3\sqrt{\frac{2}{3}} = \sqrt{6}$$