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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: lanvins on August 14, 2008, 07:15:42 pm

Title: integration questions from essential math methods
Post by: lanvins on August 14, 2008, 07:15:42 pm: 1. differentiate f(x)= loge(x) / (x^2+1)

2. Find the equation of the tangent of y= loge(x) at the point (e,1)

3. upper limit 36, lower limit 0 dx/(2x+9)= loge(k), then k is:

Thanks
Title: Re: integration questions from essential math methods
Post by: Collin Li on August 14, 2008, 07:28:50 pm: $f(x)= \frac{\log_ex}{x^2+1}$

Using the quotient rule:

$f'(x) = \frac{ \frac{1}{x}(x^2+1) - 2x\log_ex}{(x^2+1)^2}$

$= \frac{x^2(1-2\log_ex) + 1}{x(x^2+1)^2}$
Title: Re: integration questions from essential math methods
Post by: lanvins on August 14, 2008, 08:39:07 pm: thanks,
Title: Re: integration questions from essential math methods
Post by: xox.happy1.xox on August 14, 2008, 09:19:04 pm: 2. y= loge(x), with point (e,1)

Find derivative,

dy/dx = 1/x

let x=e

dy/dx = 1/e = m

Use the formula y - y1 = m (x - x1), substitute y1 = 1, x1 = e, m = 1/e

y - 1 = 1/e (x - e)

y - 1 = x/e - 1

y = x/e

(Need to learn to use Latex, first try at helping somebody at maths... I wouldn't be surprised if it was wrong :()
Title: Re: integration questions from essential math methods
Post by: lanvins on August 14, 2008, 09:24:00 pm: nah it's right, thanks
Title: Re: integration questions from essential math methods
Post by: dcc on August 14, 2008, 09:51:46 pm: Quote from: lanvins on August 14, 2008, 07:15:42 pm
3. $\int_0^{36} \frac{dx}{2x + 9} = \log_{e} k,\ k = ?$

$\implies \frac{1}{2} \int_0^{36} \frac{2}{2x + 9} \ dx = \log_{e} k$

$\implies \frac{1}{2} \left[ \log_{e} \left(2x + 9\right) \right]_0^{36} = \log_{e} k$

$\implies \frac{1}{2} \left( \log_{e} 81 - \log_{e} 9 \right) = \log_{e} k$

$\implies \log_{e} \sqrt{\frac{81}{9}} = \log_{e} k$

$\implies \boxed{k = 3}$
Title: Re: integration questions from essential math methods
Post by: lanvins on August 14, 2008, 09:59:22 pm: Quote from: dcc on August 14, 2008, 09:51:46 pm
$\implies \frac{1}{2} \int_0^{36} \frac{2}{2x + 9} \ dx = \log_{e} k$

where did the 2 come from?
Title: Re: integration questions from essential math methods
Post by: dcc on August 14, 2008, 10:07:05 pm: Quote from: lanvins on August 14, 2008, 09:59:22 pm
Quote from: dcc on August 14, 2008, 09:51:46 pm
$\implies \frac{1}{2} \int_0^{36} \frac{2}{2x + 9} \ dx = \log_{e} k$

where did the 2 come from?

I put the 2 there so that the integral could be expressed in the form $\int \frac{f'(x)}{f(x)} \ dx = \log_{e} f(x) + c$ . Notice the $\frac{1}{2}$ term out the front of the integral, so the 2 and the 0.5 cancel each other out, leaving the answer the same.
Title: Re: integration questions from essential math methods
Post by: lanvins on August 14, 2008, 10:13:48 pm: oic thanks,