Author Topic: TrueTears question thread (Read 66165 times) Tweet Share

TrueTears · « **Reply #90 on:** January 02, 2009, 10:58:51 pm »

thx /0 XD

and just this Q im half way through but im stuck...

$\int \frac{x}{x^2+2x+3} dx$

/0 · « **Reply #91 on:** January 02, 2009, 11:36:00 pm »

Quote from: TrueTears on January 02, 2009, 10:58:51 pm

thx /0 XD

and just this Q im half way through but im stuck...

$\int \frac{x}{x^2+2x+3} dx$

$\int \frac{x}{x^2+2x+3}dx = \int \frac{x}{(x+1)^2+2}dx$

Let $u=x+1 \Rightarrow \frac{du}{dx}=1$

$\Rightarrow \int \frac{u-1}{u^2+2}du$

$=\int \frac{u}{u^2+2}du - \int \frac{1}{u^2+2}du$

$=\frac{1}{2} \int \frac{2u}{u^2+2}du - \int \frac{1}{u^2+2}$

Since $\int \frac{f'(x)}{f(x)}dx = \log_e|f(x)|+C$

$\Rightarrow \frac{1}{2}\log_e|u^2+2|- \frac{1}{\sqrt{2}}\int \frac{\sqrt{2}}{u^2+(\sqrt{2})^2}$

$=\frac{1}{2}\log_e|u^2+2|-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right)+C$

$=\frac{1}{2}\log_e|x^2+2x+3|-\frac{\sqrt{2}}{2}\tan^{-1} \left(\frac{\sqrt{2}(x+1)}{2}\right)+C$

TrueTears · « **Reply #92 on:** January 02, 2009, 11:37:57 pm »

yeah i got to this step

= $\frac{1}{2} \int \frac{2u}{u^2+2}du - \int \frac{1}{u^2+2}$

$\Rightarrow \frac{1}{2}\log_e|u^2+2|- \frac{1}{\sqrt{2}}\int \frac{\sqrt{2}}{u^2+(\sqrt{2})^2}$

how u know to use log here, is it just trial and error?

ell · « **Reply #93 on:** January 02, 2009, 11:40:07 pm »

Quote from: TrueTears on January 02, 2009, 11:37:57 pm

yeah i got to this step

= $\frac{1}{2} \int \frac{2u}{u^2+2}du - \int \frac{1}{u^2+2}$

$\Rightarrow \frac{1}{2}\log_e|u^2+2|- \frac{1}{\sqrt{2}}\int \frac{\sqrt{2}}{u^2+(\sqrt{2})^2}$

how u know to use log here, is it just trial and error?

When it's in the form $\frac{f'(x)}{f(x)}$

edit: oops just realised /0 already pointed this out. you can get that answer by substituting $v=u^2+2$

vce08 · « **Reply #94 on:** January 02, 2009, 11:42:37 pm »

Quote from: TrueTears on January 02, 2009, 11:37:57 pm

yeah i got to this step

= $\frac{1}{2} \int \frac{2u}{u^2+2}du - \int \frac{1}{u^2+2}$

$\Rightarrow \frac{1}{2}\log_e|u^2+2|- \frac{1}{\sqrt{2}}\int \frac{\sqrt{2}}{u^2+(\sqrt{2})^2}$

how u know to use log here, is it just trial and error?

the other integral thingy is in the form of tan^-1 i think

TrueTears · « **Reply #95 on:** January 02, 2009, 11:43:47 pm »

ahh i see, u can also use another substitution for u there and get same thing.

TrueTears · « **Reply #96 on:** January 03, 2009, 01:22:42 am »

are we required to know how to sketch something like $y = \frac{1}{\sqrt{9-4x^2}}$ in the spesh course without a calculator?

Matt The Rat · « **Reply #97 on:** January 03, 2009, 01:49:30 am »

You could be asked to. It's a reciprocal graph for a square root function graph and reciprocal graphs are part of the spec course. Unlikely to be asked, but could be.

TrueTears · « **Reply #98 on:** January 03, 2009, 07:23:24 pm »

ah kk

Also just this stupid Q >.<

How do u solve the following simultaneous equation for x and y

$y = 9-x^2$ ......(1)

$y = \frac{1}{\sqrt{9-x^2}}$ ......(2)

ell · « **Reply #99 on:** January 03, 2009, 07:44:36 pm »

Quote from: TrueTears on January 03, 2009, 07:23:24 pm

ah kk

Also just this stupid Q >.<

How do u solve the following simultaneous equation for x and y

$y = 9-x^2$ ......(1)

$y = \frac{1}{\sqrt{9-x^2}}$ ......(2)

From (1): $x^2=9-y$ (3)

From (2): $x^2=9-\frac{1}{y^2}$ (4)

(3) = (4):

$9-y=9-\frac{1}{y^2}$

$\implies \frac{1}{y^2}-y=0$

Multiply both sides by $y^2$ :

$1-y^3=0$

$\implies y=1$

Substitute $y=1$ into (1):

$\implies x=\pm2\sqrt{2}$ (Substituting into (2) also yields the same answer.)

TrueTears · « **Reply #100 on:** January 03, 2009, 07:49:00 pm »

ah thank you so much

TrueTears · « **Reply #101 on:** January 04, 2009, 07:05:39 pm »

just some questions:

1. Find the volume of the solid generated when the region enclosed by $y = \sqrt{3x+1}$ , $y = \sqrt{3x}$ , $y = 0$ and $x = 1$ is rotated about the x axis.

2. The region defined by the inequalities $y \ge x^2-2x+4$ and $y \le 4$ is rotated about the line $y = 4$ . Find the volume generated.

3. The region bounded by the parabola $y^2 = 4(1-x)$ and the y axis is rotated about the a) x axis, b) y axis
prove that the volumes of the solids formed are in the ratio 15:16

For this question, i can do part b), which is just u solve for x, and then u have the intergral: $\pi \int^{2}_{0} (1 - \frac{1}{4}y^2)^2 dy$ = $\frac{16\pi}{15}$

However, how do u do part a) ?

4. The region bounded by the graph $y = \frac{1}{\sqrt{x^2+9}}$ , the x axis, the y axis and the line $y = 4$ is rotated about the y axis. Find the volume of the solid formed.

Many thanks again XD

ell · « **Reply #102 on:** January 04, 2009, 07:39:12 pm »

1. Graph the functions first to visualise what volume you need and to determine what terminals your integrals will use. The top function is $\sqrt{3x+1}$ and the bottom function is $\sqrt{3x}$ so find the difference in volumes between the two functions.

$\begin{alignat}{2} \mbox{Volume} & = \pi\int_{-\frac{1}{3}}^{1}3x+1\ dx-\pi\int_{0}^{1}3x\ dx \\ & = \pi\left[\frac{3x^2}{2}+x\right]_{-\frac{1}{3}}^{1}-\pi\left[\frac{3x^2}{2}\right]_{0}^{1} \\ & = \pi\left[\frac{3}{2}+1-\frac{1}{6}+\frac{1}{3}\right]-\pi\left[\frac{3}{2}\right] \\ & = \frac{7\pi}{6}\ \mbox{cubic units} \\ \end{alignat} $

TrueTears · « **Reply #103 on:** January 05, 2009, 12:22:06 am »

ah yes thanks ell, just new to this was kinda confused.

I get the all the other questions now cept for Q2.
2. The region defined by the inequalities $y \ge x^2-2x+4$ and $y \le 4$ is rotated about the line $y = 4$ . Find the volume generated. What do u do when it says rotated around y = 4? I've only met questions which says rotate around x/y axis.

Also just this new question:

if $\frac{dx}{dy}$ is inversely proportional to y, and y = 2 when x = 0 and y = 4 when x = 2. Find y when x = 3.

/0 · « **Reply #104 on:** January 05, 2009, 12:32:31 am »

It is the same as rotating the region defined by $y \geq x^2-2x$ and $y \leq 0$ about the line $y=0$ . You just translated it down by 4 but the volume is the same.

$\frac{dx}{dy} = \frac{k}{y}\Rightarrow x = klog_e|y|+C$ , differentiating with respect to y. Then solve for k and C.

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