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Sarahisme

aghh...its like beating my head into a brick wall!!

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hey peoples , if anyone is willing could the please give me few hints for this question....

 

i think i have done part a)

 

i get for part a) ~5.271 units

 

but i have no idea how to "use" the extra information given in part b) ...

 

anyways aghh been tring for hours and hours, lol , you know the feeling ;):P

 

Sarah :)

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Given a point (u,v) on the curve what is the equation of the tangent line at that point? In particular what is its gradient. Now consider the line from (8,1) to (u,v) what is its gradient? When are those two lines perpendicular, ie when is that line feom (8,1) to (u,v) normal to the curve? You have done Normal lines, right?

 

The tangent is [math]3u^{1/2}/2[/math] and the line from 8,1 to u,v has slope [math]\frac{v-1}{u-8}[/math] but of course v=1+u^{3/2}, or v-1=u^{3/2}

 

Recall that if the tangent's gradient is D that the normal has gradient -1/D, so we must need the u.v such that

 

 

[math]\frac{u^{3/2}}{u-8}= \frac{-2}{3u^{1/2}}[/math]

 

rearrange and solve 3u^2=-2(u-8)

 

Edit: what's wrong with my tex?

 

here it is in ascii:

 

 

The tangent is 3u^{1/2}/2 and the line from 8,1 to u,v has slope (v-1)/(u-8) but of course v=1+u^{3/2}, or v-1=u^{3/2}

 

Recall that if the tangent's gradient is D that the normal has gradient -1/D, so we must need the u.v such that

 

 

(u^{3/2})/(u-8)= (-2)/)3u^{1/2}

 

rearrange and solve 3u^2=-2(u-8)

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I've no idea, I didn't do a) - who cares what the actual number is, all we know si that we need to minimize the distance between (u,v) and 8,1 where u,v is in the curve. We know the distance, and we know minimizing it is the same as minimizing the square fothe distance, namely

 

minimize (u-8)^2+(v-1)^2

 

and we know that v-1=u^{3/2} thus w minimize

 

(u-8)^2 + u^3

 

differentiate it to find the critical points and we get the same thing as in b.

 

I don't even own a calculator so i can't find the critical points, as they won't be at integers by the look of it.

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hmm ok, i did part a) a different way, but i suppose it was the wrong way then :(

 

oh well, thanks for you help matty :) (do you mind being called matty?)

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3u^2=-2(u-8) doesn't solving this just give you the (x,y) corodinates on the oringial curve, and so you would not actually be finding the distance by just doing that?

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Obviously - i only ever said I wqs finding the critical points on the curve. As I really don't own a calculator I cannot find them and really can't then work out the distance.

 

I'm ignoring the matty thing.

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oh can i ask one other quick question/example.... about finding the null space of a matrix?

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given a matrix A you need to solve the equations Ax=0 and see what x can be - often it will give you a "system of paramters" (not standard notation) and anything satisfying that "system" is avector in the null space.

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ok, can i ask you about a particular example about the null space of a matrix? cause there is something strange with it i think...?

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this is the question, i get a funny answer for the null space bit, the column space part seems ok though... sorry i don't have enough time at the mo to post answers, will do in a few hours when i get some time...

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for the basis for the null space i get:

 

x3[0,0,1,0] (pretend that the stuff between the "[ ]" is actually vertical)

 

so is that right or should there be no x3 there or what?

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What is the x3? a basis is just some (finite) set of vectors, eg (0,0,1,0), perhaps.

 

In order to state what the null space is i need to know whether you're acting on the left or right, ie is this a map from R^3 to R^4, or R^4 to R^3?

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"i need to know whether you're acting on the left or right, ie is this a map from R^3 to R^4, or R^4 to R^3?"

 

i am not sure about that stuff, the question does not explicitely say anything other than what i have posted. so can i just say for the answer to this question that the basis for the null space of matrix A is

 

[0,0,1,0] (meant to be vertical of course :P)

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also i have a linear algebra question, if you are willinging to help me a bit with it , that'd be great..... the question is...

 

Suppose A is a 5 × 4 matrix and there exists a 4 × 5 matrix C such that C A = I4 (the 4 × 4 identity matrix). Suppose further that for some given b in R^5, the equation Ax = b has at least one solution. Show that this solution is unique.

 

 

i started using the inverse matrix theorem, but then realised that was only for square matrices.... any suggestions?

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No, it works fine:

 

if Ax=Ay, the applying C we get x=CAx=CAy=y.

 

And this answers my own question from before, you are writing Ax so you're acting on the left, so the null space is the set of all vectors such that Ax=0. And no, I'm not going to check your answer to that question, sorry.

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