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| bit4bit |
 Posted: Mon Apr 07, 2008 6:28 am Post subject: Gradient vector question |
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Forum Bachelors Degree
Joined: 14 Jul 2007
Posts: 492
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The question is: "For the function f(x,y) = 2x+3y find a unit vector that points in the direction of the maximum rate of change at the point (1,1)"
first of all, I found the gradient vector, ∇f = <2,3>, and this should be the same for any co-ordinates. Then I used this formula:
V.∇f = |V||∇f|cosθ
if |V| = 1, and cosθ = 1 (same direction as gradient vector), then
V.∇f=|∇f|
<a,b>.<2,3>=√22+32
2a+3b = √13
Now I'm lost on how to solve for a and b. If I square the expression on the left, and re-arrange, I just get a polynomial with two variables, which i can't solve.
The answer the book gives is:
<(2√13)/13, (3√13)/13>
Thanks
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| serpicojr |
| Posted: Mon Apr 07, 2008 2:26 pm Post subject: |
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Forum Ph.D.
Joined: 17 Jul 2007
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| So what does it mean for two vectors to be in the same direction? |
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| bit4bit |
| Posted: Mon Apr 07, 2008 5:12 pm Post subject: |
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Forum Bachelors Degree
Joined: 14 Jul 2007
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I think I got it, <a,b>=<2,3> / √13 = <2/√13, 3/√13>...I can't believe i didn't realise that, i was trying all sorts. Looks like the answer is right, since I checked its magnitude, and it's 1. Looks like another wrong answer in the book. 
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| serpicojr |
| Posted: Mon Apr 07, 2008 5:46 pm Post subject: |
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Forum Ph.D.
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| Hey hey hey check to see if they're equal! |
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| bit4bit |
| Posted: Tue Apr 08, 2008 10:03 am Post subject: |
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Forum Bachelors Degree
Joined: 14 Jul 2007
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Oh yeh, 1/√13 = √13/13, so dividing both by √13 gives:
1/√13√13 = 1/13
1/13=1/13
I don't know why they used √13/13 though, when surely 1/√13 is much more straigtforward, since a unit vector is just the vector divided by its magnitude, √13.
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| serpicojr |
| Posted: Tue Apr 08, 2008 10:19 am Post subject: |
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Forum Ph.D.
Joined: 17 Jul 2007
Posts: 871
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Somebody came up with the convention once that the denominator should always be rationalized. I bet it was developed by or at the behest of a teacher who was sick of having to check whether various expressions were equal. And, now that I think about this, I'm once again frightened that a lot of math teachers aren't that good at math.
You're right, 1/sqrt(13) is a lot more brief than sqrt(13)/13, and it displays the algebraic property of the number--namely that its square is 1/13--more clearly.
In general, you should make an effort to become fluent in manipulating expressions. It'll be useful for checking your answers, and it's a good strategy for solving a lot of problems. |
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| bit4bit |
| Posted: Tue Apr 08, 2008 10:32 am Post subject: |
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Forum Bachelors Degree
Joined: 14 Jul 2007
Posts: 492
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Thanks for the help. I just need to keep doing problems over and over until I get used to it. I always find with maths, that if I'm not actively solving problems, then within the space of a year or so, I'll be extremely rusty. That has morealess been the case with me before I started reading this book....I finished calculus, did almost no challenging maths for a year, and then started this book. I also hear that universities/companies don't like maths students/employees to be, to take gap years because of this reason....I heard that was the case over here anyway.
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| serpicojr |
| Posted: Tue Apr 08, 2008 10:35 am Post subject: |
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Forum Ph.D.
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| Well, then, make sure you're always doing some math! We're always here to support you in such endeavors. |
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| bit4bit |
| Posted: Tue Apr 08, 2008 10:43 am Post subject: |
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Forum Bachelors Degree
Joined: 14 Jul 2007
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Thanks. Actually I feel quite honoured to get help from a Phd maths student! 
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