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Gradient, unit normal in vector calculus 
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#1
Aug2611, 11:13 AM

P: 33

1. The problem statement, all variables and given/known data
there is a surface xy^{3}z^{2}=4. What is the unit normal to this surface at a pt in the surface (1,1,2)?? 2. Relevant equations what is a unit normal to a scalar region? how can it be calculated? 3. The attempt at a solution i calculated the gradient (del operator) of this surface at the given point to be 3i12j+4k . this is in the direction of the unit normal along which max. rate of increase occurs. So far so good. But how dowe get the unit normal from this ? i dont know how to go further. 


#2
Aug2611, 11:16 AM

P: 33

here, all points in the surface has a constant value =4, irrespective of the location of the points in the surface. does this have something to do with the calculation of the unit normal? it does sound absurd..but i'm equally blank..so just throwing all that's on my mind.



#3
Aug2611, 12:32 PM

P: 33

by the way, i'm writing down the gradient of the surface as i calculated it, so that any mistakes i've made can be corrected too.
[itex]\nablaS[/itex] = y^{3}z^{2}i+3xy^{2}z^{2}j+2xy^{3}zk. now,[itex]\nablaS[/itex] at the point (1,1,2) = 3i12j+4k the unit normal as i got is 3i12j+4k  13 but my textbook says it is (1/rt 11, 3/rt 11, 1/rt 11) i've no idea how to arrive at this one .. any help is appreciated 


#4
Aug2611, 12:50 PM

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Gradient, unit normal in vector calculus
There's a theorem that says whenever you have a surface defined by f(x,y,z) = c, where c is a constant, the normal to the surface is given by grad f. This is exactly the case you have here, so taking the gradient will indeed give you the normal.
The reason you're not getting the right answer is that you simply calculated the numbers wrong. The xcomponent, for example, should be (1)^{3}2^{2}=4, not 3. (I didn't check the rest.) 


#5
Aug2611, 12:50 PM

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Hi msslowlearner!
When I calculate your gradient I get a different result. Your gradient formula is correct, but filling in (1, 1, 2) should give a different result. Furthermore, your problem statement asks for a unit normal. This means your vector should have length 1. 


#6
Aug2611, 01:12 PM

P: 33




#7
Aug2611, 01:15 PM

P: 33

yes, the xcomponent was wrong. now, i get the gradient as 4i 12j +4k
whose magnitude is sqrt(176). if the unit normal vector is supposed to be (gradient/magnitude of gradient), i still am not arrivng at the answer !! 


#8
Aug2611, 01:16 PM

P: 33

hang on hang on !! i got it .. simple math.. sqrt(176) = 4*sqrt(11). sorry guys.. my mistake.. thanks for the help :)



#9
Aug2611, 01:17 PM

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I don't understand you. Could you elaborate? Note that your textbook says: (1/rt 11, 3/rt 11, 1/rt 11) This should be read as: [tex](\frac {1} {\sqrt {11}}, \frac {3} {\sqrt {11}}, \frac {1} {\sqrt {11}})[/tex] 


#10
Aug2611, 01:26 PM

P: 33

the magnitude of my vector was 4*rt(11). when i multiply my unit vector with the magnitude, i get my vector again = 4i 12j +4k :)
the terms i used to explain may have been obscure,but what i intended to say was tat the unit vector here has a length which is not equal to 1, but it is still a unit vector for a given vector. 


#11
Aug2611, 01:29 PM

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I'm sorry to say, that what you say is still obscure.
A unit vector is defined to be a vector of length 1. To obtain a unit vector from a vector it needs to be divided by its length. 


#12
Aug2611, 01:44 PM

P: 33

somehow, i dont get it .. i still think it need not be necessarily 1. but since ur a PF contributor, what you say should be right ..i'm just a beginner .. i'll look my concepts again.. thanks for the help :)



#13
Aug2611, 01:53 PM

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You mean the normal vector you first calculated, (4, 12, 4), isn't length 1. By dividing it by its magnitude, you get the unit normal, (1/√11, 3/√11, 1/√11), which is length 1. As ILS noted earlier, that's why it's called the unit normal.



#14
Aug2711, 02:04 AM

P: 33

i'm sorry people .. i see it now .. unit normal vector has a length 1 ... personal apologies to "i like serena" .. i kinda questioned ur intelligence.. sorry ... sometimes i miss out the so obvious tiny little things tat matter the most .. sorry guys



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