1 vectors and 1 limits problem

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  • #1
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Hi guys,

Starting revision for my exams and I can't seem to solve these problems. ):

1. Given that three non zero vectors A, B and C are perpendicular to each other and that the vector R = pA + qB + rC is a zero vector, show that p = q = r = 0


2. In another question, given that f(x) is differenciatable at x = 0 and [itex]\frac{Lim}{x->0}[/itex] [itex]\frac{f(x) - f(kx)}{x}[/itex] = a, wjere a and k are constants.

b. Show that f'(0) = a/1-k

What I did was take the defination of differenciation:
[itex]\frac{Lim}{h->0}[/itex] [itex]\frac{f(x+h) - f(x)}{h}[/itex]

and I let x = kx, and h = (1-k)x,
[itex]\frac{Lim}{(1-k)x->0}[/itex] [itex]\frac{f(x) - f(kx)}{(1-k)x}[/itex]

and I did this (assuming that I can write [itex]\frac{Lim}{(1-k)x->0}[/itex] as [itex]\frac{Lim}{x->0}[/itex]

1/(1-k)[itex]\frac{Lim}{x->0}[/itex] [itex]\frac{f(x) - f(kx)}{x}[/itex] = a/1-k

Is this correct?

Thanks in advance for your help guys!

*Edit* Just realized that I misread the forum title and posted in the wrong one. ):
 
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Answers and Replies

  • #2
22,089
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Hi guys,

Starting revision for my exams and I can't seem to solve these problems. ):

1. Given that three non zero vectors A, B and C are perpendicular to each other and that the vector R = pA + qB + rC is a zero vector, show that p = q = r = 0
What can you tell us about the inner products R.A, R.B and R.C??

2. In another question, given that f(x) is differenciatable at x = 0 and [itex]\frac{Lim}{x->0}[/itex] [itex]\frac{f(x) - f(kx)}{x}[/itex] = a, wjere a and k are constants.

b. Show that f'(0) = a/1-k

What I did was take the defination of differenciation:
[itex]\frac{Lim}{h->0}[/itex] [itex]\frac{f(x+h) - f(x)}{h}[/itex]

and I let x = kx, and h = (1-k)x,
[itex]\frac{Lim}{(1-k)x->0}[/itex] [itex]\frac{f(x) - f(kx)}{(1-k)x}[/itex]

and I did this (assuming that I can write [itex]\frac{Lim}{(1-k)x->0}[/itex] as [itex]\frac{Lim}{x->0}[/itex]

1/(1-k)[itex]\frac{Lim}{x->0}[/itex] [itex]\frac{f(x) - f(kx)}{x}[/itex] = a/1-k

Is this correct?

Thanks in advance for your help guys!

*Edit* Just realized that I misread the forum title and posted in the wrong one. ):
[/QUOTE]

No, you can't do that. You took

[itex]\frac{Lim}{h->0}[/itex] [itex]\frac{f(x+h) - f(x)}{h}[/itex]

but the x in there is a constant!! But later

[itex]\frac{Lim}{(1-k)x->0}[/itex] [itex]\frac{f(x) - f(kx)}{(1-k)x}[/itex]

your x is suddenly a limit variable.

I'll illustrate it by taking x=2, then you write

[itex]\frac{Lim}{h->0}[/itex] [itex]\frac{f(2+h) - f(2)}{h}[/itex]

this is perfectly valid and it equals f'(2). But then you write

[itex]\frac{Lim}{(1-k)2->0}[/itex] [itex]\frac{f(2) - f(k2)}{(1-k)2}[/itex]

which obviously makes no sense. (1-k)2 -> 0 makes no sense!!

I'll give you a hint on how to solve it:

\begin{align}
f^\prime(0)
&= \lim_{h\rightarrow 0}{\frac{f(h)-f(0)}{h}}\\
&= \lim_{h\rightarrow 0}{ \frac{\frac{f(h)-f(kh)}{h}}{ \frac{f(h)-f(kh)}{f(h)-f(0)}}}\\
&= \lim_{h\rightarrow 0}{ \frac{\frac{f(h)-f(kh)}{h}}{ \frac{(f(h)-f(0)) - (f(kh)-f(0))}{f(h)-f(0)}}}\\
\end{align}
 
  • #3
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What can you tell us about the inner products R.A, R.B and R.C??
Oh!

R.R = R.A + R.B + R.C
Since R.R = 0,
R= p = q = r?

I'll give you a hint on how to solve it:

\begin{align}
f^\prime(0)
&= \lim_{h\rightarrow 0}{\frac{f(h)-f(0)}{h}}\\
&= \lim_{h\rightarrow 0}{ \frac{\frac{f(h)-f(kh)}{h}}{ \frac{f(h)-f(kh)}{f(h)-f(0)}}}\\
&= \lim_{h\rightarrow 0}{ \frac{\frac{f(h)-f(kh)}{h}}{ \frac{(f(h)-f(0)) - (f(kh)-f(0))}{f(h)-f(0)}}}\\
\end{align}
How did you get from the first to the second step? )(Actually I'm lost from that step onwards also)

Anyway thanks for your help so far micromass!
 
  • #4
mathman
Science Advisor
7,898
461
R.A = pA.A Since R=0 then R.A=0, however A.A > 0, therefore p=0. Similarly for q and r.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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Oh!

R.R = R.A + R.B + R.C
Since R.R = 0,
R= p = q = r?
Why do you say "R.R= 0"?? And "R= p= q= r" makes no sense. R is a vector while p, q, and r are numbers.
 
  • #6
mathman
Science Advisor
7,898
461
Why do you say "R.R= 0"?? And "R= p= q= r" makes no sense. R is a vector while p, q, and r are numbers.
R=0, therefore R.R=0

The second line is a typo - I meant p=q=r=0.
 
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