Author Topic: TrueTears question thread (Read 66994 times) Tweet Share

TrueTears · « **Reply #390 on:** June 14, 2009, 01:16:49 am »

$\int \sqrt{9-x^2}dx$

Thanks.

ryley · « **Reply #391 on:** June 14, 2009, 01:34:18 am »

I think if you make the substitution x = 3cos(theta) it should work.

/0 · « **Reply #392 on:** June 14, 2009, 02:03:30 am »

Yeah or $x=3\sin{\theta}$ .
In general when you get stuff like $\sqrt{a^2 - x^2}$ you can substitute $x = a\sin{\theta}$ or $a\cos{\theta}$

$x = 3\sin{\theta} \Rightarrow dx = 3\cos{\theta}d\theta$

$\Rightarrow \int \sqrt{9-x^2}dx = \int \sqrt{9-(3\sin{\theta})^2}dx = \int 3\sqrt{1-\sin^2{\theta}}dx = \int 3\cos{\theta}dx$

$= \int (3\cos\theta)(3\cos\theta d\theta) =9 \int \cos^2\theta d\theta = 9\int \frac{\cos{2\theta}+1}{2}d\theta = \frac{9}{2}\int(\cos{2\theta}+1)d\theta = \frac{9}{2}(\frac{1}{2}\sin{2\theta}+\theta) = \frac{9}{4}\sin2\theta + \frac{9}{2}\theta +C$

$\theta = \sin^{-1}\left(\frac{x}{3}\right)$
$\sin{2\theta} = 2\cos\theta\sin\theta = 2\sqrt{1-\left(\frac{x}{3}\right)^2}(\frac{x}{3})=\frac{2}{9}x\sqrt{9-x^2}$

$\therefore \int \sqrt{9-x^2}dx = \frac{9}{4}\left(\frac{2}{9}x\sqrt{9-x^2}\right)+\frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right)=\frac{1}{2}x\sqrt{9-x^2}+\frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right)+C$

I don't think this is in the course though

TrueTears · « **Reply #393 on:** June 14, 2009, 02:07:24 am »

Alright cool thanks ryley and /0.

TrueTears · « **Reply #394 on:** June 14, 2009, 02:19:00 am »

Quote from: /0 on June 14, 2009, 02:03:30 am

Yeah or $x=3\sin{\theta}$ .
In general when you get stuff like $\sqrt{a^2 - x^2}$ you can substitute $x = a\sin{\theta}$ or $a\cos{\theta}$

$x = 3\sin{\theta} \Rightarrow dx = 3\cos{\theta}d\theta$

$\Rightarrow \int \sqrt{9-x^2}dx = \int \sqrt{9-(3\sin{\theta})^2}dx = \int 3\sqrt{1-\sin^2{\theta}}dx = \int 3\cos{\theta}dx$

$= \int (3\cos\theta)(3\cos\theta d\theta) =9 \int \cos^2\theta d\theta = 9\int \frac{\cos{2\theta}+1}{2}d\theta = \frac{9}{2}\int(\cos{2\theta}+1)d\theta = \frac{9}{2}(\frac{1}{2}\sin{2\theta}+\theta) = \frac{9}{4}\sin2\theta + \frac{9}{2}\theta +C$

$\theta = \sin^{-1}\left(\frac{x}{3}\right)$
$\sin{2\theta} = 2\cos\theta\sin\theta = 2\sqrt{1-\left(\frac{x}{3}\right)^2}(\frac{x}{3})=\frac{2}{9}x\sqrt{9-x^2}$

$\therefore \int \sqrt{9-x^2}dx = \frac{9}{4}\left(\frac{2}{9}x\sqrt{9-x^2}\right)+\frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right)=\frac{1}{2}x\sqrt{9-x^2}+\frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right)+C$

I don't think this is in the course though

Wait isn't $\sqrt{1-sin^2\theta} = \sqrt{cos^2\theta} = |cos\theta|$ ?

What about the $\sqrt{9} = \pm 3$ ? Discard the negative?

/0 · « **Reply #395 on:** June 14, 2009, 02:51:03 am »

The function we are analysing is defined between $x=-3$ and $x=3$ . Therefore, $\theta$ is defined between $3\sin{\theta} = -3\Rightarrow \theta = -\frac{\pi}{2}$ and $3\sin{\theta}=3 \Rightarrow \theta = \frac{\pi}{2}$ .
For this domain, $\cos{\theta}$ is positive, so there is no need for absolute value signs or the negative part of the square root.

TrueTears · « **Reply #396 on:** June 14, 2009, 11:10:35 pm »

thx

The shaded region shown in the diagram above is rotated around the x axis to form a solid of revolution. $g'(x) > 9$ and $g"(x) > 0$ for all $x \in [a,b]$ and the volume of the solid of revolution is V cubic units.

Which of the following is FALSE:

a) $V < \pi (g(b))^2(b-a)$
b) $V > \pi (g(a))^2(b-a)$
c) $V = \pi \int_a^b (g(x))^2 dx$
d) $V = \pi((G(b))^2 - (G(a))^2)$ , where $G'(x) = g(x)$
e) $V < \pi ((g(b))^2b - (g(a))^2a)$

/0 · « **Reply #397 on:** June 14, 2009, 11:55:35 pm »

I say d) because while $\int_a^b g(x) dx = [G(x)]_a^b$ , this does not imply that $\int_a^b [g(x)]^2dx = [\ [G(x)]^2\ ]_a^b$

TrueTears · « **Reply #398 on:** June 14, 2009, 11:55:57 pm »

Thanks I just got that as well.

TrueTears · « **Reply #399 on:** June 28, 2009, 01:26:21 pm »

Quote from: kevinn on June 28, 2009, 01:08:46 pm

Hi guys, I have a mind blank
Can someone please do this for me
It's question 4g from Essential, exercise 7E

$\int sin(x)^2cos(x)^4$

Thanks

As a general rule next time please create another thread to post your questions to avoid ambiguity

Mao · « **Reply #400 on:** June 28, 2009, 11:42:52 pm »

Quote from: Flaming_Arrow on June 28, 2009, 01:17:33 pm

Quote from: kevinn on June 28, 2009, 01:08:46 pm
Hi guys, I have a mind blank
Can someone please do this for me
It's question 4g from Essential, exercise 7E

$\int sin(x)^2cos(x)^4$

Thanks

$\int (1-\cos^2 x) cos ^4 (x)$

let u = cos x

i think u can do it from there

No, that won't do it, you will have a sin(x) term from substitution which will be difficult (but not impossible) to get rid of.

$\begin{align*} \int \sin^2(x)\cos^4(x)\; dx &= \int (\sin (x)\cos(x))^2 \cdot \cos^2(x)\; dx \\ &= \int \frac{\sin^2(2x)}{4}\cdot \frac{1+\cos(2x)}{2}\; dx \\ &= \frac{1}{8}\int \sin^2 (2x) + \sin^2(2x)\cos(2x)\; dx\\ &= \frac{1}{8}\left(\int \frac{1-\cos(4x)}{2}\; dx + \int \sin^2(2x)\cos(2x)\; dx\right)\\ &= \frac{1}{16}\left(x - \frac{\sin(4x)}{4}\right) + \frac{\sin^3(2x)}{48} + C \\ I & = -\frac{\sin(4x)}{64} + \frac{\sin^3(2x)}{48} + \frac{x}{16} + C\\ \end{align*}$

Combinations of $\sin(2\theta) = 2\sin(\theta)\cos(\theta),\; \cos^2(\theta) = \frac{1}{2}(1+\cos(2\theta)),\; \sin^2 (\theta)=\frac{1}{2}(1-\cos(2\theta))$ and the substitution method were used.

Flaming_Arrow · « **Reply #401 on:** June 28, 2009, 11:45:27 pm »

Quote from: Mao on June 28, 2009, 11:42:52 pm

Quote from: Flaming_Arrow on June 28, 2009, 01:17:33 pm
Quote from: kevinn on June 28, 2009, 01:08:46 pm
Hi guys, I have a mind blank
Can someone please do this for me
It's question 4g from Essential, exercise 7E

$\int sin(x)^2cos(x)^4$

Thanks

$\int (1-\cos^2 x) cos ^4 (x)$

let u = cos x

i think u can do it from there

No, that won't do it, you will have a left over sin(x) term which will be difficult (but not impossible) to get rid of.

$\begin{align*} \int \sin^2(x)\cos^4(x)\; dx &= \int (\sin (x)\cos(x))^2 \cdot \cos^2(x)\; dx \\ &= \int \frac{\sin^2(2x)}{4}\cdot \frac{1+\cos(2x)}{2}\; dx \\ &= \frac{1}{8}\int \sin^2 (2x) + \sin^2(2x)\cos(2x)\; dx\\ &= \frac{1}{8}\left(\int \frac{1-\cos(4x)}{2}\; dx + \int \sin^2(2x)\cos(2x)\; dx\\ &= \frac{1}{16}\left(x - \frac{\sin(4x)}{4}\right) + \frac{\sin^3(2x)}{48} + C \\ I & = -\frac{\sin(4x)}{64} + \frac{\sin^3(2x)}{48} + \frac{x}{16} + C\\ \end{align*}$

oops didnt read the question prlly

TrueTears · « **Reply #402 on:** July 10, 2009, 02:56:04 am »

Consider the 2nd order DE.

$a(\frac{d^2y}{dx^2}) + b(\frac{dy}{dx}) + cy = 0$ ...[A] where a,b,c $\in R$ and $a \neq 0$

$am^2 + bm + c = 0$ ...(B).

$y = e^{mx}$ is a solution to the above DE.

(B) could have 2 distinct real roots, say $m_1$ and $m_2$ . Without differentiating show that $y = C_1e^{m_1x} + C_2e^{m_2x}$ , where $C_1$ and $C_2$ are real numbers, is a general solution of [A] in this case.

TrueTears · « **Reply #403 on:** July 10, 2009, 04:26:41 am »

Hmmm I did this:

Let $y_{1} = e^{m_{1}x} \mbox{and} y_{2}= e^{m_{2}x}$ both be soutions to the DE.

Then $a y''_{1} + b y'_{1} + cy_{1} = 0 \mbox{and} a y''_{2} + b y'_{2} + cy_{2} = 0$

Let $y = C_{1}y_{1} + C_{2}y_{2}$

$a (C_{1}y_{1} + C_{2}y_{2})'' + b ( C_{1}y_{1} + C_{2}y_{2})' + c (C_{1}y_{1} + C_{2}y_{2} )$

$= a (C_{1}y''_{1} + C_{2}y''_{2}) + b (C_{1}y'_{1} + C_{2}y'_{2}) + c(C_{1}y_{1} + C_{2}y_{2})$

$= C_{1}(ay''_{1} + by'_{1}+ cy_{1}) + C_{2}((ay''_{2} + by'_{2}+ cy_{2})$

$= C_{1}(0) + C_{2}(0) = 0$

Any other way to do it?

TrueTears · « **Reply #404 on:** July 10, 2009, 04:53:49 am »

Next part of the question says:

(B) could also have 1 repeated root, say m.

Show that 2am+b=0

Do I just simply implicit differentiate both sides of equation (B) ?

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