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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: oneoneoneone on November 08, 2012, 12:02:23 pm

Title: Integrating in respect to a function.
Post by: oneoneoneone on November 08, 2012, 12:02:23 pm: $a=\frac{\delta}{\delta x}(\frac{1}{2}v^{2})=4-2(\frac{1}{2}v^{2})$

If we replace 1/2v^2 with a variable (say b) and then solve it for x, then replace the b with 1/2v^2, is this mathematically right?
Title: Re: Integrating in respect to a function.
Post by: BubbleWrapMan on November 08, 2012, 12:16:43 pm: Uh it depends, show me
Title: Re: Integrating in respect to a function.
Post by: oneoneoneone on November 08, 2012, 01:32:22 pm: $a=4-v^{2}$
$a=\frac{\delta}{\delta x}(\frac{1}{2}v^{2})=4-2(\frac{1}{2}v^{2})$
$\therefore 2a=\frac{\delta}{\delta x}(v^{2})=8-2v^{2}$
$\frac{\delta x}{\delta v^2}=\frac{1}{8-2v^{2}}<br />x=\int \frac{\delta x}{\delta v^2}\delta v^2=\frac {1}{2}\int \frac{1}{4-v^{2}}\delta v^2<br />$
$x=-\frac{1}{2}ln\left | 4-v^2 \right |+c$
when $x=0, v=\sqrt{3}$
$\therefore c=0$
$x=-\frac{1}{2}ln(4-v^2)$ as $(v^2<4)$
Title: Re: Integrating in respect to a function.
Post by: paulsterio on November 08, 2012, 01:43:17 pm: btw,

$\frac{\delta}{\delta x}$

AND

$\frac{d}{dx}$

mean two different things.
Title: Re: Integrating in respect to a function.
Post by: BubbleWrapMan on November 08, 2012, 02:03:19 pm: Well, it gets the right answer. If you want another method I'll show you.