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Thursday, 26 January 2012
Monday, 23 January 2012
© Nur Al Hayat's question for Yahoo Answers
7(a) (i) Solve the differential equation: \[ \frac{dx}{dt}=\sqrt{x}\sin(t/2) \] to find \(x \) in terms of \(t\).
Well we may observe that this DE is of variables separable type so we may write: \[\int \frac{1}{\sqrt{x}} \; dx = \int \sin(t/2)\; dt \]
We may now integrate to get: \[ \sqrt{x}= -\cos(t/2)+C \] and squaring we get \[x=(C-\cos(t/2))^2\]
Now the remaining parts of this question can be solved by applying the relevant initial conditions etc.
Well we may observe that this DE is of variables separable type so we may write: \[\int \frac{1}{\sqrt{x}} \; dx = \int \sin(t/2)\; dt \]
We may now integrate to get: \[ \sqrt{x}= -\cos(t/2)+C \] and squaring we get \[x=(C-\cos(t/2))^2\]
Now the remaining parts of this question can be solved by applying the relevant initial conditions etc.
Friday, 20 January 2012
Sample Question (pulled off Yahoo Answers): Linear Algebra Question: Finding the Dimension?
Sample Question (pulled off Yahoo Answers): Linear Algebra Question: Finding the Dimension? (Original question posted as comment on the Questions Page)
If S is a subspace in R^4 and S = {(a,b,c,d) in R^4 , a+c = b , a+c+d=0} what is the dimension of S? I have the answers and it says that S is generated by the vectors (1,0,-1,0) and (1,1,0,-1) and that the dimension of S is 2. How do we find those 2 vectors that generate S? Please explain!!
I trust that it is obvious that \( S \) is of dimension 2? This is because we can assign values to \(a\) and \(c\) essentially arbitarily and construct a unique element of \(S\).
In fact we can construct the mapping:
\[ s=(a,b,c,d)=(u,v)A \]
where: \( A= \left[ \begin{array}{cccc}1&1&0&-1\\0&1&1&-1 \end{array} \right] \), which gives us a 1-1 linear mapping from \( \mathbb{R}^2 \) to \(S\).
Hence the image of any basis of \( \mathbb{R}^2 \) is a basis of \(S\) so in particular is:
\( \{ (1,0)A, (0,1)A \}=\{ (1,1,0,-1),(0,1,1,-1) \} \).
One of the elements in the basis found above is in the basis you quote and the other is not. This is because the basis is not unique, any pair of linearly independent vectors in \( \mathbb{R}^2 \) will produce a basis for \(S\). The other element of your quoted basis is \( (1,-1)A=(1,0,-1,0) \).
(If you are more used to working with column vectors just take the transpose of everything above)
If S is a subspace in R^4 and S = {(a,b,c,d) in R^4 , a+c = b , a+c+d=0} what is the dimension of S? I have the answers and it says that S is generated by the vectors (1,0,-1,0) and (1,1,0,-1) and that the dimension of S is 2. How do we find those 2 vectors that generate S? Please explain!!
I trust that it is obvious that \( S \) is of dimension 2? This is because we can assign values to \(a\) and \(c\) essentially arbitarily and construct a unique element of \(S\).
In fact we can construct the mapping:
\[ s=(a,b,c,d)=(u,v)A \]
where: \( A= \left[ \begin{array}{cccc}1&1&0&-1\\0&1&1&-1 \end{array} \right] \), which gives us a 1-1 linear mapping from \( \mathbb{R}^2 \) to \(S\).
Hence the image of any basis of \( \mathbb{R}^2 \) is a basis of \(S\) so in particular is:
\( \{ (1,0)A, (0,1)A \}=\{ (1,1,0,-1),(0,1,1,-1) \} \).
One of the elements in the basis found above is in the basis you quote and the other is not. This is because the basis is not unique, any pair of linearly independent vectors in \( \mathbb{R}^2 \) will produce a basis for \(S\). The other element of your quoted basis is \( (1,-1)A=(1,0,-1,0) \).
(If you are more used to working with column vectors just take the transpose of everything above)
Thursday, 19 January 2012
First Post
This is the opening post for this blog. The purpose of which is to provide free help for maths homework. It is not here to help you cheat on assessed work, if something will constitute part of your course assessment don't post it here.
Post your question in a comment on the top post on the Questions Needing Answers Page
This blog is not here to help with significant work related problems, for that you need to employ a paid for consultant (with appropriate insurance cover?). A question that can be dealt with in a few minutes or if it interests me sufficiently may be answered..
LaTeX test:
[Tex] \int_0^{\xi} \frac{1}{\sin(x)}\; dx[/Tex]
well that's not working! For some reason it needs the page reloading to display properly on this machine/browser! Will need to check this on another machine/browser. Seems to work under Chrome :) but not under Opera :( or IE8 :((
\[ \int_0^{\xi} \frac{1}{\sin(x)}\; dx \]
\( \int_0^{\xi} \frac{1}{\sin(x)}\; dx \)
OK seems to work with the MathJax default delimiters!
Post your question in a comment on the top post on the Questions Needing Answers Page
This blog is not here to help with significant work related problems, for that you need to employ a paid for consultant (with appropriate insurance cover?). A question that can be dealt with in a few minutes or if it interests me sufficiently may be answered..
LaTeX test:
[Tex] \int_0^{\xi} \frac{1}{\sin(x)}\; dx[/Tex]
well that's not working! For some reason it needs the page reloading to display properly on this machine/browser! Will need to check this on another machine/browser. Seems to work under Chrome :) but not under Opera :( or IE8 :((
\[ \int_0^{\xi} \frac{1}{\sin(x)}\; dx \]
\( \int_0^{\xi} \frac{1}{\sin(x)}\; dx \)
OK seems to work with the MathJax default delimiters!