Author Topic: A few Specialist Problems (Read 20055 times) Tweet Share

GerrySly · « **on:** January 27, 2009, 10:37:15 pm »

Questions I have had trouble with (or need a little push to get started)

fredrick · « **Reply #1 on:** January 28, 2009, 12:16:06 pm »

all those questions can be/must be done by calc. one way is you can sketch both equations and find(x-value) where they intercept. Dont forget the domain

Mao · « **Reply #2 on:** January 28, 2009, 02:43:55 pm »

question 3: http://vcenotes.com/forum/index.php/topic,2171.0.html

question 5: substitute values for A and r then solve using calculator.

GerrySly · « **Reply #3 on:** February 10, 2009, 08:53:34 pm »

Find correct to two decimal places, the angle between the line $y = \frac{1}{2}x+1$ and the tangent to the graph of $x=y^2-y$ at the point of intersection $(2,2)$

Mao · « **Reply #4 on:** February 10, 2009, 09:26:46 pm »

it is difficult to deal with $x=y^2-y$ , hence, we take the inverse of both functions [which does not change the angle between the two slopes]

$y_1^{-1}: y=2x-2$

$y_2^{-1}: y=x^2-x$

point of intersection is (2,2) [take the inverse of the point]

slope of first graph: $\frac{dy}{dx}=2\implies \tan (\theta_1)=2 \implies \theta_1 = \tan^{-1} (2)$

slope of second graph: $\frac{dy}{dx} \right| _{x=2} = 2\times(2) - 1 = 3\implies \theta_2 = \tan^{-1} (3)$

angle in between: $|\theta_2 - \theta_1| \approx 0.142^c = 8.13^o$

GerrySly · « **Reply #5 on:** February 11, 2009, 09:25:17 pm »

$\frac {d}{dx} -4\cos^{-1} (\frac{5x-3}{2})$

I came up with the following answer $\frac {4}{\sqrt {-(5x-4)(5x-2)}}$ , but the answer is actually $\frac {20}{\sqrt {-5(5x^2 - 6x + 1)}}$

I seem to not be able to do the questions with a constant in front of x, when there is no constant I am able to differentiate. This is the following equation I used to differentiate $cos^{-1}$ with no constant in front of x.

$\frac {d}{dx} \cos^{-1} (\frac{x}{a}) = \frac {-1}{\sqrt {a^2 - x^2}}$

There seems to be something I am missing

Thanks

Mao · « **Reply #6 on:** February 11, 2009, 10:02:44 pm »

instead, try using the chain rule:

$\frac{d}{dx} \left(-4\cos^{-1}\left(\frac{5x-3}{2}\right)\right) = -4\frac{d}{du}\cos^{-1}(u)\cdot \frac{d}{dx}\left(\frac{5x-3}{2}\right) = \frac{4}{\sqrt{1-u^2}}\cdot \frac{5}{2} = \frac{10}{\sqrt{1-\frac{(5x-3)^2}{4}}} = \frac{20}{\sqrt{4-(5x-3)^2}}$

expand that --> simply would give you the correct answer

GerrySly · « **Reply #7 on:** March 03, 2009, 07:46:57 pm »

$\int \cot^2{(x)} dx$

$\int \frac{\cos^2 (x)}{\sin^2 (x)} dx$

$\int \frac{1 - \sin^2(x)}{\sin^2 (x)} dx$

$\int \left [ \frac{1}{\sin^2 (x)} - \frac{\sin^2(x)}{\sin^2(x)} \right ] dx$

$\int \left [ \sin^{-2} {(x)} - 1 \right ] dx$

$-\cot (x) - x + c$

Alright, I managed to $\LaTeX$ hack my way through most of the equation but how do I integrate $\sin^{-2} {(x)}$ ?

Mao · « **Reply #8 on:** March 03, 2009, 08:26:23 pm »

firstly, $\frac{1}{\sin x} = \csc x \neq \sin^{-1} x$ , I was a little confused reading your post

to integrate $\csc^2(x)$ and $\sec^2(x)$ , the only way [that I know of] is by recognition:

$\frac{d}{dx}\tan (x) = \sec^2(x) + C$

$\frac{d}{dx}\cot (X) = -\csc^2 (x) + C$

GerrySly · « **Reply #9 on:** March 06, 2009, 10:34:02 pm »

$\int cosec^2 \left ( x - \frac{\pi}{2} \right )$

and

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

And if any help can be given on $\int \cot^2 (x)$ , our teacher says he's not giving any help on how to integrate it lol

Mao · « **Reply #10 on:** March 07, 2009, 05:40:53 pm »

$\int \csc^2 \left( x-\frac{\pi}{2}\right) \; dx = -\cot \left( x-\frac{\pi}{2}\right) + C$

$\begin{align*} \int \sin^2(x)\cos^4(x)\; dx &= \int \cos^4(x) - \cos^6(x)\; dx \\ &= \int \left(\frac{\cos(2x)+1}{2}\right)^2 - \left(\frac{\cos(2x)+1}{2}\right)^3\; dx\\ &= \int \frac{\cos^2(2x) + 2\cos(2x) + 1}{4} - \frac{\cos^3(2x) +3\cos^2(2x) + 3\cos(2x) +1}{8}\; dx\\ &= \frac{1}{4}\int \frac{\cos(4x)+1}{2}+ 2\cos(2x) + 1\; dx - \frac{1}{8}\int (1-\sin^2(2x))\cos(2x) + 3\frac{\cos(4x)+1}{2} + 3\cos(2x) + 1\; dx\\ &= \frac{\sin(4x)}{32}+ \frac{x}{8} + \frac{\sin(2x)}{4} + \frac{x}{4} - \frac{1}{16}\left(\sin(2x) - \frac{\sin^3(2x)}{3}\right) - \frac{3}{64}\sin(4x) -\frac{3}{16}x - \frac{3}{16}\sin(2x) - \frac{x}{16} + C \\ &= -\frac{\sin(4x)}{64} + \frac{\sin^3(2x)}{48} + \frac{x}{8} + C\\ \end{align*}$
(checked)

$\int \cot^2(x)\; dx = \int \csc^2 (x) - 1\; dx = -\cot(x) - x + C$

GerrySly · « **Reply #11 on:** March 23, 2009, 04:02:40 pm »

A few more that I came across...

Find the asymptotes of the graph with equation $\frac{x^2+x+2}{x}$

$\int_{0}^{\sqrt{3}}\frac{2x + 3}{x^2 + 9}\, dx$

GerrySly · « **Reply #12 on:** March 23, 2009, 06:30:42 pm »

Oh also I wanted to know how do I write up a multi choice SAC? Our teacher told us to get the correct answer and then explain why the others are incorrect. How do I write that though?

GerrySly · « **Reply #13 on:** March 23, 2009, 07:19:18 pm »

Quote from: GerrySly on March 23, 2009, 04:02:40 pm

A few more that I came across...

Find the asymptotes of the graph with equation $\frac{x^2+x+2}{x}$

$\int_{0}^{\sqrt{3}}\frac{2x + 3}{x^2 + 9}\, dx$

Just made attempt number 3 on the second problem and got this

$\begin{align*} \int_{0}^{\sqrt{3}}\frac{2x + 3}{x^2 + 9}\, dx &= \int_{0}^{\sqrt{3}}\frac{2x}{x^2 + 9} + \frac{3}{x^2 + 9}\, dx \\ &= \begin{bmatrix} \log_e{(x^2 + 9)} + \tan^{-1} {(x/3)} \end{bmatrix}_{0}^{\sqrt{3}} \\ &= \left [ \log_e {(12)} + \frac{\pi}{6} \right ] - \left [ \log_e {(9)} \right ]\\ &= \log_e { \left ( \frac{4}{3} \right )} - \frac{\pi}{6}\\ &\approx 0.811 \end{align*}$

Gosh I love how LaTeX makes me see things more clearer, now just to find out how I find asymptotes, I am sure I have learnt it before but I just can't figure out what it is heh

Mao · « **Reply #14 on:** March 23, 2009, 07:42:31 pm »

Quote from: GerrySly on March 23, 2009, 04:02:40 pm

Find the asymptotes of the graph with equation $\frac{x^2+x+2}{x}$

dividing through by x to get
$\implies (x + 1) + \frac{2}{x}$

as $x\to -\infty$ , $\frac{2}{x}\to 0^-$ , so on the left, the graph becomes $(x+1) + 0^- = x+1$ . Do the same thing on the right to obtain $(x+1)+0^+ = x+1$ . Hence $y=x+1$ is a slant asymptote.

as $x\to 0^-$ (approach from the left), $\frac{2}{x}\to -\infty$ , $(x+1)+\frac{2}{x}\to -\infty$ . as $x\to 0^+$ , $\frac{2}{x}\to \infty$ , $(x+1)+\frac{2}{x}\to \infty$ , so there is a verticle asymptote $x=0$ .

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