ATAR Notes: Forum

HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Advanced => Topic started by: Keerit on February 13, 2017, 09:15:03 pm

Title: PLEASE HELP!! MATHS
Post by: Keerit on February 13, 2017, 09:15:03 pm: (http://uploads.tapatalk-cdn.com/20170213/92a6fc249f215a36c38509084fffc81d.jpg)

Both b & c please!

Mod Edit: Merged your triple post.
Title: Re: PLEASE HELP!! MATHS
Post by: RuiAce on February 13, 2017, 09:30:12 pm: $\text{From a) the point of intersection is at }(1,0)$
$\text{It is easy to find the tangent of }y=\ln x\text{ at }(1,0)\\ \frac{dy}{dx}=\frac{1}{x}=1\text{ when }x=1\\ \text{So the equation of the tangent is just }y-0=1(x-1)\\ \text{i.e. }y=x-1$
$\text{As for }y=\log_{10}x\text{, we will have to use the change of base rule:}\\ \begin{align*}y&=\frac{\ln x}{\ln 10}\\ \implies \frac{dy}{dx}&=\frac{1}{x\ln 10}\end{align*}$
$\text{When }x=1\text{, we have }\frac{dy}{dx}=\frac{1}{\ln 10}\\ \text{So the equation of the tangent to this curve will be}\\ y=\frac{1}{\ln 10}(x-1)$
______________________________
$\text{The }y\text{-intercept occurs when }x=0\\ \text{So for }X:\quad y=-1\\ \text{And for }Y:\quad y=-\frac{1}{\ln 10}$
$\text{So clearly the distance }XY\text{ will be }1-\frac{1}{\ln 10}$

ATAR Notes | SMF © 2014, Simple Machines