Understanding gradient concept

[PainterGuy]

<Moderator's note: Moved from a technical forum and thus no template.>

Hi

I was trying to understand the concept of gradient. I'm using Thomas's Calculus 12th Ed.

Please have a look here. Using the Definition 1, the answer came to be 3.54.

Then, I tried to attempt the same problem using Theorem 1 shown here. My attempt is shown below and the answer is 4.123. For some reason there are few mistakes in my code. For clarity you can refer to this image too.

I'm getting different answers using Definition 1 and Theorem 1. In my opinion, the answers should have been the same. Could you please guide me? Thank you.

$f(x,y)=z=x^{2}+(x\cdot y)$

$\frac{\partial z}{\partial x}=2x+y$

$\frac{\partial z}{\partial y}=x$

$\nabla f=\frac{\partial z}{\partial x}i+\frac{\partial z}{\partial y}j$

$\nabla f=\left( 2x+y\right) i+\left( x\right) j$

At point $P_{0}=(1,2)$

$\nabla f=\left( 2x+y\right) i+\left( x\right) j=\left[ 2(1)+2\right] i+j=4i+j$

$\left( \frac{df}{ds}\right) _{u,P_{0}}=\left( \nabla f\right) _{P_{0}}\cdot u =\left\Vert \nabla f\right\Vert \left\Vert u\right\Vert \cos \theta$

Unit vector in the direction $4i+j$:

$\frac{4i}{4.123}+\frac{j}{4.123}=0.97i+0.243j$

$\left\Vert \nabla f\right\Vert \left\Vert u\right\Vert \cos \theta =0.97(4)+0.243(1)=3.88+0.243=4.123$

$\left( 4i+j\right) \left( \frac{1}{\root{2}\of{2}}i+\frac{1}{\root{2}\of{2}}j\right) =2.83+0.707=3.54$

 

[LCKurtz]

Your Tex is unreadable, but anyway, why are you making a unit vector out of 4i + j? The unit vector u is made from i + j.

 

[Orodruin]

Why are you surprised that you get different directional derivatives when using different directions?

 

[PainterGuy]

Thank you.

There is only error in my code and that is \QTRbfu. I was just trying to boldface unit vector u. I also attached a copy of my code in a picture form so that there is no confusion. Please check the attachment grad33.

why are you making a unit vector out of 4i + j? The unit vector u is made from i + j.

According to Theorem 9 the unit vector should be in the direction of ∇f and ∇f=4i + j.

Why are you surprised that you get different directional derivatives when using different directions?

I think that I understand it now. Definition 1 is for a directional derivative but Theorem 9 is particularly about gradient always points in the direction of maximum increase. The unit vector 1/√2 i + 1/√2 j does not point in the direction of maximum increase or in the direction of 4i + j. Please let me know that if I'm wrong.

Thanks a lot for your help.

 

[fresh_42]

LaTeX code corrected.

 

[Orodruin]

According to Theorem 9 the unit vector should be in the direction of ∇f and ∇f=4i + j.

No, there is nothing in the theorem that states this.

 

[Orodruin]

The unit vector 1/√2 i + 1/√2 j does not point in the direction of maximum increase or in the direction of 4i + j.

There is nothing in the theorem you quote that restricts the derivative to the maximal increase. It is a theorem about the relation between the directional derivative in an arbitrary direction and the gradient.

 

[PainterGuy]

Thank you. I might have misinterpreted it. But to get maximum increase, you need a unit vector in the direction of 4i+j because it gives you cos(theta)=0.

I was thinking that Definition 1 and Theorem 9 stand for the same thing. I think that they do mean the same thing. Theorem 9 will give you a maximum value only if unit vector is the direction of nabla(f).