Question:

A transformation T is represented by

$\displaystyle \begin{pmatrix}x'\\ y'\end{pmatrix} = \begin{pmatrix}b &0 \\ 0 & 1/b \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} $ where $\displaystyle b \epsilon \mathbb{R}^{+} $

a) Draw a diagram showing the unit square and its image under T

[b) Show that area is invariant under T

c) Show that T maps the curve $\displaystyle y =1/x $ onto itself

d) Show that T maps the region bounded by the curve $\displaystyle y = 1/x $, the lines x=1 and x =a and the x-axis onto the region bounded by the same curve the lines x =b and x =ab and the x-axis.

e) hence show that $\displaystyle \int_{1}^{a} \frac{1}{x} dx = \int_{a}^{ab} \frac{1}{x} dx $

f)Given that $\displaystyle F(t) = \int_{1}^{t} \frac{1}{x} dx $ show that $\displaystyle F(ab)=F(a)+F(b) $

my attempt:a)

the unit square $\displaystyle = \begin{pmatrix} 0 &1 & 1 &0 \\ 0& 0 & 1 &1 \end{pmatrix} $

transformed by T ;

$\displaystyle \begin{pmatrix} x'\\ y' \end{pmatrix}= \begin{pmatrix} b &0 \\ 0 & 1/b \end{pmatrix} \begin{pmatrix} 0 &1 & 1 &0 \\ 0& 0 & 1 &1 \end{pmatrix} = \begin{pmatrix} x'\\ y' \end{pmatrix}= \begin{pmatrix} 0 &b &b & 0\\ 0 &0 & 1/b &1/b \end{pmatrix} $

did i draw the right diagram?

b) the area is invariant because the area of the transformed image is $\displaystyle \frac{1}{b} \times b = 1 $

the determinant of the unit square = 1

Since the area of the transformed = 1 and the determinant of the unit square = 1 therefore, the area is invariant.

c)$\displaystyle \begin{pmatrix}x'\\ y'\end{pmatrix} = \begin{pmatrix}b &0 \\ 0 & 1/b \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} $

since it maps $\displaystyle y = \frac{1}{x}$ onto itself thus

$\displaystyle \begin{pmatrix}x' \\ \frac{1}{x}' \end{pmatrix} = \begin{pmatrix}b &0 \\ 0 & 1/b \end{pmatrix}\begin{pmatrix}x \\ \frac{1}{x}\end{pmatrix} $

$\displaystyle \begin{pmatrix}x' \\ \frac{1}{x}' \end{pmatrix} = \begin{pmatrix}bx\\ \frac{b}{x}\end{pmatrix} $

T maps points on y =1/x to new points on the curve y =1/x

d) I need help this part i dont know what to do

e)

$\displaystyle \int_{1}^{a} \frac{1}{x} dx = \int_{a}^{ab} \frac{1}{x} dx $

we know that $\displaystyle \int \frac{1}{x}dx = ln \left | x \right |+c $

thus, $\displaystyle \int_{1}^{a} \frac{1}{x} dx = ln \left | a \right | - ln \left | 1 \right | =ln \left | a \right | - 0 = ln \left | a \right |$

$\displaystyle \int_{a}^{ab} \frac{1}{x} dx = ln \left | ab \right | - ln \left | a \right | = \frac{ln \left | ab \right |}{ln \left | a \right |}= \frac{ln \left | a \right |ln \left | b \right |}{ln \left | a \right |}= ln \left | b \right | $

a =b

i am not confident if i did this part right. I would like to know if i did this part correctly

f) can i please have some help doing part F?

Please I would like help with this question