Gradient, unit normal in vector calculus

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Homework Help Overview

The discussion revolves around finding the unit normal to the surface defined by the equation xy3z2 = 4 at the point (-1, -1, 2). Participants explore the concept of the gradient as it relates to the normal vector of a scalar field.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the gradient and its implications for determining the unit normal vector. Questions arise about the correctness of the gradient calculation and the definition of a unit normal vector.

Discussion Status

There is an ongoing examination of the gradient values and the process of normalizing the vector. Some participants express confusion about the definition of a unit vector, while others provide clarifications and corrections. The conversation reflects a mix of attempts to resolve misunderstandings and validate calculations.

Contextual Notes

Participants note discrepancies between their calculations and textbook answers, leading to discussions about potential errors in computation and the understanding of vector magnitudes. There is acknowledgment of varying interpretations of the term "unit normal."

msslowlearner
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Homework Statement



there is a surface xy3z2=4. What is the unit normal to this surface at a pt in the surface (-1,-1,2)??

Homework Equations


what is a unit normal to a scalar region? how can it be calculated?


The Attempt at a Solution


i calculated the gradient (del operator) of this surface at the given point to be 3i-12j+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 don't know how to go further.
 
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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.
 
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.
\nablaS = y3z2i+3xy2z2j+2xy3zk.
now,\nablaS at the point (-1,-1,2) = 3i-12j+4k

the unit normal as i got is 3i-12j+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
 
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 x-component, for example, should be (-1)322=-4, not 3. (I didn't check the rest.)
 
Hi msslowlearner! :smile:

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.
 
I like Serena said:
Hi msslowlearner! :smile:

Furthermore, your problem statement asks for a unit normal.
This means your vector should have length 1.

i don't think it should have length 1. its the vector/mod vector, which is not 1. its the smallest multipliable unit of the vector, when multipled by the length of the vector gives the actual vector .
 
yes, the x-component 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 !
 
hang on hang on ! i got it .. simple math.. sqrt(176) = 4*sqrt(11). sorry guys.. my mistake.. thanks for the help :)
 
msslowlearner said:
i don't think it should have length 1. its the vector/mod vector, which is not 1. its the smallest multipliable unit of the vector, when multipled by the length of the vector gives the actual vector .

Huh? :confused:
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:
(\frac {-1} {\sqrt {11}}, \frac {-3} {\sqrt {11}}, \frac {1} {\sqrt {11}})
 
  • #10
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
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
somehow, i don't 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
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
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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