Multi-part question involving a vector field

[CaliforniaRoll88]

Homework Statement

A vector field is specified as $\vec G=24xy\hat a_x+12(x^2+2)\hat a_y+18z^2\hat a_z$. Given two points, $P(1,2-1)$ and $Q(-2,1,3)$, find (a) $\vec G$ at $P$; (b) a unit vector in the direction of $\vec G$ at $Q$; (c) a unit vector directed from $Q$ toward $P$; (d) the equation of the surface on which $|\vec G|=60$
answers:
(a) $48\hat a_x + 36\hat a_y + 18\hat a_z$
(b) $−0.26\hat a_x + 0.39\hat a_y + 0.88\hat a_z$
(c) $0.59\hat a_x + 0.20\hat a_y − 0.78\hat a_z$
(d) $100 = 16x^2 y^2 + 4x^4 + 16x^2 + 16 + 9z^4$
Source: Problem 1.5; Engineering Electromagnetics, 8th Edition, William Hayt, John Buck

Relevant Equations

$\vec G=24xy\hat a_x+12(x^2+2)\hat a_y+18z^2\hat a_z$

(a)
$\vec G=24xy\hat a_x+12(x^2+2)\hat a_y+18z^2\hat a_z$ @ $P(1,2-1)$
$\vec G=24(1)(2)\hat a_x+12(1^2+2)\hat a_y+18(-1)^2\hat a_z$
$\vec G=48\hat a_x+36\hat a_y+18\hat a_z$
(b)
I am not sure how to get this part started. Could someone point me in the right direction?

 

[scottdave]

Start by finding the regular vector at Q, just like you found for P. Then think about what a unit vector is.

How would you take any arbitrary vector, and give a unit vector in the same direction?

 

[CaliforniaRoll88]

How would you take any arbitrary vector, and give a unit vector in the same direction?

Let $\vec A$ be any vector:
It's unit vector would be $\hat a_A=\frac{\vec A}{A}$
$\vec G|_{Q(-2,1,3)}=24(-2)(1)\hat a_x+12[(-2)^2+2)]\hat a_y+18(3)^2\hat a_z$
$\vec G|_{Q(-2,1,3)}=48\hat a_x+48\hat a_y+162\hat a_z$
Is inputing coordinates of a point into the vector field equation the same as constructing a vector from that point to the vector field?

 

[PeroK]

Let $\vec A$ be any vector:
It's unit vector would be $\hat a_A=\frac{\vec A}{A}$
$\vec G|_{Q(-2,1,3)}=24(-2)(1)\hat a_x+12[(-2)^2+2)]\hat a_y+18(3)^2\hat a_z$

Okay.

$\vec G|_{Q(-2,1,3)}=48\hat a_x+48\hat a_y+162\hat a_z$

That's not what i get.

Is inputing coordinates of a point into the vector field equation the same as constructing a vector from that point to the vector field?

A vector field is a vector at each point in space, where the vector depends on the point. E.g. wind velocity.

Saying "that point to the vector field" suggests you misunderstand what a vector field is.

You could simplify your notation and, for example, write:
$$\vec G(P)$$These questions you are doing seem heavy on elementary arithmetic, IMO. Perhaps that's distracting you from the conceptual side of things?

 

[PeroK]

PS a vector field is the next level of complexity up from a scalar field, where you have a scalar at each point in space. E.g air temperature or air density.

 

[CaliforniaRoll88]

These questions you are doing seem heavy on elementary arithmetic, IMO. Perhaps that's distracting you from the conceptual side of things?

I would agree. Thank you for the replies.

 

[PeroK]

Or, if we consider electrodynamics, charge density and electric potential are scalar fields; and, the electric and magnetic fields are vector fields.

 

[scottdave]

The wind velocity example that @PeroK gave is a good one. Plugging in values will give a vector representing direction and strength of the wind at that point in space, after correcting the arithmetic mistake in your calculation.

Then, as you said, divide by the length of that vector to get the unit vector.

 

[CaliforniaRoll88]

Correction:
$\vec G(P)=-48\hat a_x+48\hat a_y+162\hat a_z$
Part b) Continued:
$\hat a_{\vec G(P)}=\frac{-48\hat a_x+48\hat a_y+162\hat a_z}{\sqrt{48^2+48^2+162^2}}$
$\hat a_{\vec G(P)}=\frac{-48\hat a_x+48\hat a_y+162\hat a_z}{\sqrt{30852}}$
$\hat a_{\vec G(P)}=-0.2745\hat a_x+0.2745\hat a_y+0.9263\hat a_z$

Then, as you said, divide by the length of that vector to get the unit vector.

I didn't get the answer. Do you know where I went wrong?

 

[PeroK]

Correction:
$\vec G(P)=-48\hat a_x+48\hat a_y+162\hat a_z$
Part b) Continued:
$\hat a_{\vec G(P)}=\frac{-48\hat a_x+48\hat a_y+162\hat a_z}{\sqrt{48^2+48^2+162^2}}$
$\hat a_{\vec G(P)}=\frac{-48\hat a_x+48\hat a_y+162\hat a_z}{\sqrt{30852}}$
$\hat a_{\vec G(P)}=-0.2745\hat a_x+0.2745\hat a_y+0.9263\hat a_z$

I didn't get the answer. Do you know where I went wrong?

The y calculation is wrong. Also, this is point $Q$.

 

[PeroK]

Note that it only took me a minute or so to put these calculations on a spreadsheet. E.g: this is the calculation of the field and unit vector at the point $P$. Then, I just copied that line and changed the coordinates to Q.

	x	y	x	G_x	G_y	G_z	|G|	a_x	a_y	a_z
P	1​	2​	-1​	48​	36​	18​	62.64​	0.77​	0.57​	0.29​

 

[CaliforniaRoll88]

Correction:
$\vec G(Q)=-48\hat a_x+72\hat a_y+162\hat a_z$
$\hat a_{\vec G(Q)}=\frac{-48\hat a_x+72\hat a_y+162\hat a_z}{\sqrt{48^2+72^2+162^2}}$
$\hat a_{\vec G(Q)}=\frac{-48\hat a_x+72\hat a_y+162\hat a_z}{\sqrt{33732}}$
$\hat a_{\vec G(Q)}=-0.261\hat a_x+0.392\hat a_y+0.882\hat a_z$

 

[CaliforniaRoll88]

Note that it only took me a minute or so to put these calculations on a spreadsheet. E.g: this is the calculation of the field and unit vector at the point $P$. Then, I just copied that line and changed the coordinates to Q.

	x	y	x	G_x	G_y	G_z	|G|	a_x	a_y	a_z
P	1​	2​	-1​	48​	36​	18​	62.64​	0.77​	0.57​	0.29​

Wow that's fast. How did you input the formulas so fast?

 

[PeroK]

Wow that's fast.

Computers tend to do calculations quite quickly.

 

[CaliforniaRoll88]

Part c)
$\vec G(P)=48\hat a_x+36\hat a_y+18\hat a_z$
$\vec G(Q)=-48\hat a_x+72\hat a_y+162\hat a_z$
$\vec G(P)-\vec G(Q)=[48-(-48)]\hat a_x+(36-72)\hat a_y+(18-162)\hat a_z=96\hat a_x-36\hat a_y+144\hat a_z$
$\hat a_{\vec G(P)-\vec G(Q)}=\frac{96\hat a_x-36\hat a_y+144\hat a_z}{\sqrt {96^2+36^2+144^2}}$
$\hat a_{\vec G(P)-\vec G(Q)}=0.543\hat a_x-0.203\hat a_y+0.815\hat a_z$
Can someone tell me where I went wrong?

 

[PeroK]

Part c)
$\vec G(P)=48\hat a_x+36\hat a_y+18\hat a_z$
$\vec G(Q)=-48\hat a_x+72\hat a_y+162\hat a_z$
$\vec G(P)-\vec G(Q)=[48-(-48)]\hat a_x+(36-72)\hat a_y+(18+162)\hat a_z=96\hat a_x-36\hat a_y+180\hat a_z$
$\hat a_{\vec G(P)-\vec G(Q)}=0.463\hat a_x-0.147\hat a_y+0.869\hat a_z$
Can someone tell me where I went wrong?

It asks for a unit vector from $Q$ to $P$. Nothing to do with the vector field.

 

[CaliforniaRoll88]

It asks for a unit vector from $Q$ to $P$. Nothing to do with the vector field.

Aren't I creating a displacement vector between the two points and finding it's unit vector?

 

[PeroK]

Aren't I creating a displacement vector between the two points and finding it's unit vector?

No. You're calculating a difference in the vector field values.

 

[CaliforniaRoll88]

No. You're calculating a difference in the vector field values.

I am stumped. Can you please give me a hint?

 

[erobz]

I am stumped. Can you please give me a hint?

How do you write a vector between two points?

 

[kuruman]

Forget the field. For part (c) you are asked

Given two points, $P(1,2-1)$ and $Q(-2,1,3)$, find $\dots~$ (c) a unit vector directed from $Q$ toward $P$

 

[CaliforniaRoll88]

$\vec {PQ}=\left< 1-(-2),2-1,-1-3 \right>$
$\vec {PQ}=\left< 3,1,-4 \right>$
$a_{\vec {PQ}}=\frac {\vec {PQ}}{PQ}=\left< \frac {3}{\sqrt 26},\frac {1}{\sqrt 26},\frac {-4}{\sqrt 26} \right>=\left<.588,.196,-.784\right>$

 

[kuruman]

Looks good.

 

[CaliforniaRoll88]

Looks good

Could you please advise me on how to start part d)?

 

[erobz]

Could you please advise me on how to start part d)?

How do you find a the magnitude of a vector?

 

[CaliforniaRoll88]

How do you find a the magnitude of a vector?

Sum the squares of the components of the vector and takes it's square root.

 

[erobz]

Sum the squares of the components of the vector and takes it's square root.

Ok, based on that what is the magnitude $| \vec{G}|$?

 

[kuruman]

I am teaching myself how to represent such surfaces using Apple's app Grapher. This is what it looks like (two views) if I taught myself correctly. There are hardly any instructions, only examples to modify by trial and error. I changed the $y$-component from $12(x^2+2)$ to $12(x^2+2r^2)$ in order to make it dimensionally correct.

 

[CaliforniaRoll88]

Part d)
$\vec G=24xy\hat a_x+12(x^2+2)\hat a_y+18z^2\hat a_z$
$60^2=(24xy)^2+[12(x^2+2)]^2+(18z^2)^2$
$60^2=24^2x^2y^2+[12x^2+24]^2+18^2z^4$
$60^2=24^2x^2y^2+12^2x^4+2(12x^2)(24)+24^2+18^2z^4$
$60^2=24^2x^2y^2+12^2x^4+24^2x^2+24^2+18^2z^4$
$60^2=6^2[4^2x^2y^2+2^2x^4+4^2x^2+4^2+3^2z^4]$
$100=4x^4+16x^2+16x^2y^2+9z^4+16$

 

[CaliforniaRoll88]

I am teaching myself how to represent such surfaces using Apple's app Grapher.

Is that compatible with windows?

 

[kuruman]

Is that compatible with windows?

I don’t think so.

 

[kuruman]

Part d)
$\vec G=24xy\hat a_x+12(x^2+2)\hat a_y+18z^2\hat a_z$
$60^2=(24xy)^2+[12(x^2+2)]^2+(18z^2)^2$
$60^2=24^2x^2y^2+[12x^2+24]^2+18^2z^4$
$60^2=24^2x^2y^2+12^2x^4+2(12x^2)(24)+24^2+18^2z^4$
$60^2=24^2x^2y^2+12^2x^4+24^2x^2+24^2+18^2z^4$
$60^2=6^2[4^2x^2y^2+2^2x^4+4^2x^2+4^2+3^2z^4]$
$100=4x^4+16x^2+16x^2y^2+9z^4+16$

Here is the surface above. It's a bit different from mine but, like I said, I'm learning how to do this.

 

[CaliforniaRoll88]

That's awesome. I have my own surface XD. Thank you.